Chemistry Steps
General Chemistry
Stoichiometry.
This is a comprehensive, end-of-chapter set of practice problems on stoichiometry that covers balancing chemical equations, mole-ratio calculations, limiting reactants, and percent yield concepts.
The links to the corresponding topics are given below.
- The Mole and Molar Mass
- Molar Calculations
- Percent Composition and Empirical Formula
- Stoichiometry of Chemical Reactions
Limiting Reactant
- Reaction/Percent Yield
- Stoichiometry Practice Problems
Balance the following chemical equations:
a) HCl + O 2 → H 2 O + Cl 2
b) Al(NO 3 ) 3 + NaOH → Al(OH) 3 + NaNO 3
c) H 2 + N 2 → NH 3
d) PCl 5 + H 2 O → H 3 PO 4 + HCl
e) Fe + H 2 SO 4 → Fe 2 (SO 4 ) 3 + H 2
f) CaCl 2 + HNO 3 → Ca(NO 3 ) 2 + HCl
g) KO 2 + H 2 O → KOH + O 2 + H 2 O 2
h) Al + H 2 O → Al 2 O 3 + H 2
i) Fe + Br 2 → FeBr 3
j) Cu + HNO 3 → Cu(NO 3 ) 2 + NO 2 + H 2 O
k) Al(OH) 3 → Al 2 O 3 + H 2 O
l) NH 3 + O 2 → NO + H 2 O
m) Ca(AlO 2 ) 2 + HCl → AlCl 3 + CaCl 2 + H 2 O
n) C 5 H 12 + O 2 → CO 2 + H 2 O
o) P 4 O 10 + H 2 O → H 3 PO 4
p) Na 2 CrO 4 + Pb(NO 3 ) 2 → PbCrO 4 + NaNO 3
q) MgCl 2 + AgNO 3 → AgCl + Mg(NO 3 ) 2
r) KClO 3 → KClO 4 + KCl
s) Ca(OH) 2 + H 3 PO 4 → Ca 3 (PO 4 ) 2 + H 2 O
4HCl + O 2 → 2H 2 O + 2Cl 2
Al(NO 3 ) 3 + 3NaOH → Al(OH) 3 + 3NaNO 3
3H 2 + N 2 → 2NH 3
PCl 5 + 4H 2 O → H 3 PO 4 + 5HCl
2Fe + 3H 2 SO 4 → Fe 2 (SO 4 ) 3 + 3H 2
CaCl 2 + 2HNO 3 → Ca(NO 3 ) 2 + 2HCl
2KO 2 + 2H 2 O → 2KOH + O 2 + H 2 O 2
2Al + 3H 2 O → Al 2 O 3 + 3H 2
2Fe + 3Br 2 → 2FeBr 3
Cu + 4HNO 3 → Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O
2Al(OH) 3 → Al 2 O 3 + 3H 2 O
4NH 3 + 5O 2 → 4NO + 6H 2 O
Ca(AlO 2 ) 2 + 8HCl → 2AlCl 3 + CaCl 2 + 4H 2 O
C 5 H 12 + 8 O 2 → 5CO 2 + 6H 2 O
P 4 O 10 + 6H 2 O → 4H 3 PO 4
Na 2 CrO 4 + Pb(NO 3 ) 2 → PbCrO 4 + 2NaNO 3
MgCl 2 + 2AgNO 3 → 2AgCl + Mg(NO 3 ) 2
4KClO 3 → 3KClO 4 + KCl
3Ca(OH) 2 + 2H 3 PO 4 → Ca 3 (PO 4 ) 2 + 6H 2 O
Consider the balanced equation:
Complete the table showing the appropriate number of moles of reactants and products.
The moles of other molecules are calculated based on the stoichiometric ratio. For the first row the moles are calculated as follows:
\[{\rm{n}}\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{8}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}}}\;{\rm{ = }}\;{\rm{16}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;C{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}}}\;{\rm{ = }}\;{\rm{10}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{6}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}}}\;{\rm{ = }}\;{\rm{12}}\;{\rm{mol}}\]
For the second row :
\[{\rm{n}}\left( {{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.5 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}{{8\;\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.3125}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.5 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}{{{\rm{8}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.5625}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.5 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{6}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{8}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.875}}\;{\rm{mol}}\]
For the third row :
\[{\rm{n}}\left( {{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.6}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{8}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.8}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{6}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.6}}\;{\rm{mol}}\]
For the fourth row:
\[{\rm{n}}\left( {{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.9}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\;{\rm{ \times }}\;\frac{{{\rm{8}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.2}}\;{\rm{mol}}\]
\[{\rm{n}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.5}}\;{\rm{mol}}\]
How many grams of CO 2 and H 2 O are produced from the combustion of 220. g of propane (C 3 H 8 )?
C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g)
660. g CO 2
360. g H 2 O
The amount of product, for a given amount of reactant, is calculated based on their molar/stoichiometric ratio. So, the first step is to determine the moles of propane. The overall plan is:
The moles of C 3 H 8 are calculated using its molar mass:
\[{\rm{n(}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{)}}\;{\rm{ = }}\;{\rm{220}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{44}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.00}}\;{\rm{mol}}\]
Next, we find the moles of CO 2 based on its molar ratio with C 3 H 8 :
\[{\rm{n}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}}}\;{\rm{ = }}\;{\rm{15}}{\rm{.0}}\;{\rm{mol}}\]
And finally, the mass of the CO 2 is determined using its moles and molar mass:
\[{\rm{m}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{15}}{\rm{.0 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{44}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{660}}{\rm{.}}\;{\rm{g}}\]
Follow the same procedure to find the mass of H 2 O:
\[{\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}}}\;{\rm{ = }}\;{\rm{20}}{\rm{.0}}\;{\rm{mol}}\]
\[{\rm{m}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{20}}{\rm{.0 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{18}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{360}}{\rm{.}}\;{\rm{g}}\]
How many grams of CaCl 2 can be produced from 65.0 g of Ca(OH) 2 according to the following reaction,
Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O
Here is the conceptual plan for solving this problem:
The moles of Ca(OH) 2 are calculated using its molar mass:
\[{\rm{n}}\;{\rm{(Ca}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{65}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{74}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.877}}\;{\rm{mol}}\]
Next, we find the moles of CaCl 2 based on its molar ratio with Ca(OH) 2 :
\[{\rm{n}}\;\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.877 }}\cancel{{{\rm{mol}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{CaC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.877}}\;{\rm{mol}}\]
Notice that moles of CaCl 2 are equal to the moles of Ca(OH) 2 . This is because of their 1:1 molar ratio in the equation.
The last step is converting the moles to mass of the CaCl 2 :
\[{\rm{m}}\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.877 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{111}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{97}}{\rm{.3}}\;{\rm{g}}\]
This process can, and often is shown as a one-step conversion as we do in dimensional analysis. It would be as follows:
\[{\rm{m}}\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{65}}{\rm{.0}}\;{\cancel{{{\rm{g}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}_{\rm{\;}}}{\rm{ \times }}\;\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}{{{\rm{74}}{\rm{.1}}\;\cancel{{{\rm{g}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{CaC}}{{\rm{l}}_{\rm{2}}}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\; \times \;\frac{{{\rm{111}}\;{\rm{g}}\;{\rm{CaC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\;}}}}\;\;{\rm{ = }}\;{\rm{97}}{\rm{.3}}\;{\rm{g}}\]
This method is completely fine, however, I prefer doing these conversions stepwise to better see what happens in each step, and therefore, most exercises will be solved by doing one conversion at a time.
How many moles of oxygen are formed when 75.0 g of Cu(NO 3 ) 2 decomposes according to the following reaction?
2Cu(NO 3 ) 2 → 2CuO + 4NO 2 + O 2
The amount of product, for a given amount of reactant, is calculated based on their molar/stoichiometric ratio.
So, the conceptual plan is:
\[{\rm{n}}\;{\rm{(Cu}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{\;)}}\;{\rm{ = }}\;{\rm{75}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{187}}{\rm{.5}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.400}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{{\rm{O}}_{{\rm{2\;}}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.400}}\;\cancel{{{\rm{mol}}\;{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}{\rm{\;}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{\cancel{{{\rm{2}}\;{\rm{mol}}\;{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}{\rm{\;}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.200}}\;{\rm{mol}}\]
How many grams of MnCl 2 can be prepared from 52.1 grams of MnO 2 ?
MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O
The moles of MnO 2 are calculated using its molar mass:
\[{\rm{n}}\;{\rm{(Mn}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{52}}{\rm{.1}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{86}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\]
Next, we find the moles of MnCl 2 based on its molar ratio with MnO 2 :
\[{\rm{n}}\;\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.600 }}\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MnC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\]
And finally, the mass of the MnCl 2 is determined using its moles and molar mass:
\[{\rm{m}}\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.600 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{125}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{75}}{\rm{.5g}}\]
Determine the mass of oxygen that is formed when an 18.3-g sample of potassium chlorate is decomposed according to the following equation:
2KClO 3 (s) → 2KCl(s) + 3O 2 (g).
The amount of product, for a given amount of reactant, is calculated based on their molar/stoichiometric ratio. So, the first step is to determine the moles of propane. The conceptual plan is:
\[{\rm{n}}\;{\rm{(KCl}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{18}}{\rm{.3}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{122}}{\rm{.6}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.149}}\;{\rm{mol}}\]
Next, we find the moles of O 2 based on its molar ratio with KClO 3 :
\[{\rm{n}}\;\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.149 }}\cancel{{{\rm{mol}}\;{\rm{KCl}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{KCl}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.224}}\;{\rm{mol}}\]
\[{\rm{m}}\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.224 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.16}}\;{\rm{g}}\]
How many grams of H 2 O will be formed when 48.0 grams H 2 are mixed with excess hydrogen gas?
2H 2 + O 2 → 2H 2 O
The moles of H 2 O are calculated using its molar mass:
\[{\rm{n}}\;{\rm{(}}{{\rm{H}}_2}{\rm{)}}\;{\rm{ = }}\;{\rm{48}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{2}}{\rm{.00}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{24}}{\rm{.0}}\;{\rm{mol}}\]
Next, we find the moles of H 2 O based on its molar ratio with H 2 :
\[{\rm{n}}\;\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{24}}{\rm{.0 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{24}}{\rm{.0}}\;{\rm{mol}}\]
And finally, the mass of the H 2 O is determined using its moles and molar mass:
\[{\rm{m}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{24}}{\rm{.0 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{18}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{432}}\;{\rm{g}}\]
Consider the chlorination reaction of methane (CH4):
CH 4 (g) + 4Cl 2 (g) → CCl 4 (g) + 4HCl(g)
How many moles of CH 4 were used in the reaction if 51.9 g of CCl4 were obtained?
The amount of a reactant, for a given amount of product, is calculated based on their molar/stoichiometric ratio. So, the first step is to determine the moles of carbon tetrachloride, CCl 4 . The conceptual plan is:
The moles of CCl 4 are calculated using its molar mass:
\[{\rm{n}}\;{\rm{(CC}}{{\rm{l}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{51}}{\rm{.9}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{153}}{\rm{.8}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.337}}\;{\rm{mol}}\]
Next, we find the moles of CH 4 based on its molar ratio with CCl 4. It is a 1:1 ratio, so there is going to be 0.337 mol CH 4 :
\[{\rm{n}}\;\left( {{\rm{C}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.337 }}\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{C}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.337}}\;{\rm{mol}}\]
How many grams of Ba(NO 3 ) 2 can be produced by reacting 16.5 g of HNO 3 with an excess of Ba(OH) 2 ?
We need to first write the balanced chemical equation to do the calculations:
Ba(OH) 2 + 2HNO 3 → Ba(NO 3 ) 2 + 2H 2 O
The conceptual plan would be:
So, let’s start with the moles of HNO 3 :
\[{\rm{n}}\;{\rm{(HN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{16}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{63}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.262}}\;{\rm{mol}}\]
Once we have the moles, we can determine the moles of Ba(NO 3 ) 2 based on the molar ratio, and then convert the moles to the mass:
\[{\rm{n}}\;\left( {{\rm{Ba}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.262 }}\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.131}}\;{\rm{mol}}\]
\[{\rm{m}}\left( {{\rm{Ba}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.131 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{261}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{34}}{\rm{.2}}\;{\rm{g}}\]
Ethanol can be obtained by fermentation – a complex chemical process breaking down glucose to ethanol and carbon dioxide.
C 6 H 12 O 6 → 2C 2 H 5 OH + 2CO 2 glucose ethanol
How many mL of ethanol (d =0.789 g/mL) can be obtained by this process starting with 286 g of glucose?
This problem has an additional step of converting the mass of the product to volume. Other than that, we follow the same strategy and steps as in the previous problems:
\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{)}}\;{\rm{ = }}\;{\rm{286}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{180}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.59}}\;{\rm{mol}}\]
Now, we can find the moles of C 2 H 5 OH:
\[{\rm{n}}\;\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.59 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.18}}\;{\rm{mol}}\]
\[{\rm{m}}\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}} \right)\;{\rm{ = }}\;{\rm{3}}{\rm{.18 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{46}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{147}}\;{\rm{g}}\]
The last step is converting the mass to volume:
\[{\rm{v}}\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}} \right)\;{\rm{ = }}\;{\rm{147}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mL}}}}{{{\rm{0}}{\rm{.789}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\;{\rm{186}}\;{\rm{mL}}\]
36.0 g of butane (C 4 H 10 ) was burned in an excess of oxygen and the resulting carbon dioxide (CO 2 ) was collected in a sealed vessel.
2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O
How many grams of LiOH will be necessary to consume all the CO 2 from the first reaction?
2LiOH + CO 2 → Li 2 CO 3 + H 2 O
There are two reactions here, and the link between them is the CO 2 that is formed in the first reaction and used in the second.
So, the plan would be to determine the moles of CO 2 in the first reaction, and use it to calculate the moles, and consequently the mass of LiOH.
\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{)}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{58}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.620}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.620 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{8}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.48}}\;{\rm{mol}}\]
This is the amount of CO 2 that will be reacted with LiOH in a 1:2 ratio. So, we can determine the moles of LiOH and convert them to the mass:
\[{\rm{n}}\;\left( {{\rm{LiOH}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.48 }}\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{LiOH}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.96}}\;{\rm{mol}}\]
\[{\rm{m}}\;\left( {{\rm{LiOH}}} \right)\;{\rm{ = }}\;{\rm{4}}{\rm{.96}}\;\cancel{{{\rm{mol}}\;{\rm{LiOH}}}}\;{\rm{ \times }}\;\frac{{{\rm{24}}{\rm{.0}}\;{\rm{g}}\;{\rm{LiOH}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{LiOH}}}}}}\;{\rm{ = }}\;{\rm{119}}\;{\rm{g}}\;\]
13. Which statement about limiting reactant is correct?
a) The limiting reactant is the one in a smaller quantity.
b) The limiting reactant is the one in greater quantity.
c) The limiting reactant is the one producing less product.
d) The limiting reactant is the one producing more product.
c) The limiting reactant is the one producing less product than any of the other reactants.
Find the limiting reactant for each initial amount of reactants.
a) 2 mol of NH 3 and 2 mol of O 2
b) 2 mol of NH 3 and 3 mol of O 2
c) 3 mol of NH 3 and 3 mol of O 2
d) 3 mol of NH 3 and 2 mol of O 2
Note: This is not a multiple-choice question. Each row represents a separate question where you need to determine the limiting reactant.
The limiting reactant is the one producing less product. So, to determine the limiting reactant for each pair, we need to pick a product (NO or H 2 O) and see whether NH 3 or O 2 will produce less product for the given mole ratio.
Let’s do the calculations based on the amount of NO that can be formed.
\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{2}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.6}}\;{\rm{mol}}\]
So, for (a), O 2 is the limiting reactant.
\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4}}\;{\rm{mol}}\]
So, for (b), NO is the limiting reactant.
\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{3}}\;{\rm{mol}}\]
So, for (c), O 2 is the limiting reactant.
So, for (d), O 2 is the limiting reactant.
The shorter way of determining the limiting reactant is to divide the moles of a given reactant by its coefficient in the chemical reaction. This is the effective quantity of the reactant as it takes into consideration not only the amount of the reactant but also its molar ratio in the reaction.
Let’s put the equation and the moles here and determine the LR by the ratio of the moles and coefficient:
4 NH 3 + 5 O 2 → 4NO + 6H 2 O
So, for (a), we are comparing 2 / 4 mol of NH 3 with 2 / 5 mol of O 2 , and therefore, O 2 is the LR.
For (b), we are comparing 2 / 4 mol of NH 3 with 3 / 5 mol of O 2 , and therefore, NH 3 is the LR.
For (c), we are comparing 3 / 4 mol of NH 3 with 3 / 5 mol of O 2 , and therefore, O 2 is the LR.
For (d), 3 mol of NH 3 and 3 / 4 mol of NH 3 with 2 / 5 mol of O 2 , and therefore, O 2 is the LR.
How many g of hydrogen are left over in producing ammonia when 14.0 g of nitrogen is reacted with 8.0 g of hydrogen?
N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)
When the amounts of both reactants are given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed. This is the standard procedure when working on examples covering limiting reactant.
Now, in this problem, we are not asked to determine the amount of any product, but rather we need to determine how much hydrogen is left out once the reaction is complete. This already tells us that the nitrogen is the LR because some of the hydrogen remains unreacted, so it must’ve been in excess.
The first step is, as always, to determine the moles of the reactants, then using the mole ratio, calculate how much hydrogen has reacted. The remaining mass of the hydrogen is calculated by subtracting this amount from the initial mass.
\[{\rm{n}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{14}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{28}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.500}}\;{\rm{mol}}\]
\[{\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{8}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{2}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.0}}\;{\rm{mol}}\]
The moles of hydrogen reacted are calculated based on the moles of the nitrogen which is the limiting reactant:
\[{\rm{n}}\;\left( {{{\rm{H}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.500 }}\cancel{{{\rm{mol}}\;{{\rm{N}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_2}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{mol}}\]
So, there are:
4.0 – 1.50 = 2.5 mol H 2 left out
\[{\rm{m}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.5}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{2}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.0}}\;{\rm{g}}\]
How many grams of PCl 3 will be produced if 130.5 g Cl 2 is reacted with 56.4 g P 4 according to the following equation?
6Cl 2 (g) + P 4 (s) → 4PCl 3 (l)
When the amounts of both reactants are given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed.
The moles of Cl 2 and P 4 are:
\[{\rm{n}}\;{\rm{(C}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;130.5\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{70}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.84}}\;{\rm{mol}}\]
\[{\rm{n}}\;{\rm{(}}{{\rm{P}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{56}}{\rm{.4}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{123}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.455}}\;{\rm{mol}}\]
Next, we calculate how much PCl 3 can be formed from each reactant:
\[{\rm{n}}\;\left( {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.84 }}\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.23}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.455 }}\cancel{{{\rm{mol}}\;{{\rm{P}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{P}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.82}}\;{\rm{mol}}\]
Cl 2 gives less PCl 3 and therefore, it is the LR and 1.23 mol of PCl 3 can be produced in this reaction.
The mass of PCl 3 is:
\[{\rm{m}}\;{\rm{(PC}}{{\rm{l}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.23}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{137}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{169}}\;{\rm{g}}\]
How many grams of sulfur can be obtained if 12.6 g H 2 S is reacted with 14.6 g SO 2 according to the following equation?
2H 2 S(g) + SO 2 (g) → 3S(s) + 2H 2 O(g)
When the amount of both reactants is given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed.
The moles of H 2 S and SO 2 are:
\[{\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S)}}\;{\rm{ = }}\;{\rm{12}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{34}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.370}}\;{\rm{mol}}\]
\[{\rm{n}}\;{\rm{(S}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{14}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{64}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.228}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.370 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.228 }}\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.684}}\;{\rm{mol}}\]
H 2 S gives less S and therefore, it is the LR and 0.555 mol of S can be produced in this reaction.
The mass of S is:
\[{\rm{m}}\;{\rm{(S)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{17}}{\rm{.8}}\;{\rm{g}}\]
The following equation represents the combustion of octane, C 8 H 18 , a component of gasoline:
2C 8 H 18 (g) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O(g)
Will 356 g of oxygen be enough for the complete combustion of 954 g of octane?
356 g of oxygen is not enough for the complete combustion of 954 g of octane.
First, calculate the moles of the reactants:
\[{\rm{n}}\;{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{356}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{11}}{\rm{.4}}\;{\rm{mol}}\]
\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{)}}\;{\rm{ = }}\;{\rm{954}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{114}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{8}}{\rm{.35}}\;{\rm{mol}}\]
And now let’s see how many moles of O 2 are needed to react with 8.35 mol of C 8 H 18 :
\[{\rm{n}}\;\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{8}}{\rm{.35 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{25}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}}}}}\;{\rm{ = }}\;{\rm{104}}\;{\rm{mol}}\]
104 moles of O 2 are needed to react with only 8.35 mol of C 8 H 18 and this is because of the 25:2 mole ratio. So, 356 g of oxygen is not enough for the complete combustion of 954 g of octane.
When 140.0 g of AgNO 3 was added to an aqueous solution of NaCl, 86.0 g of AgCl was collected as a white precipitate. Which salt was the limiting reactant in this reaction? How many grams of NaCl were present in the solution when AgNO 3 was added?
AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq)
NaCl was the limiting reactant and 35.0 g of it were present in the solution when AgNO 3 was added.
We need to find the moles of AgCl which will allow us to calculate how much AgNO 3 and NaCl had reacted in the reaction.
\[{\rm{n}}\;{\rm{(AgCl)}}\;{\rm{ = }}\;{\rm{86}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{143}}{\rm{.3}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\]
The stoichiometric ratio of all the reactants and products is1:1, and therefore, there were 0.600 mol of AgNO 3 and 0.600 mol of NaCl participating in the reaction.
Now, let’s calculate the moles of 140.0 g of AgNO 3 and see whether it corresponds to 0.600 mol. If it is more than that, then AgNO 3 was added in excess and NaCl must be the limiting reactant:
\[{\rm{n}}\;{\rm{(AgN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{140}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{169}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.824}}\;{\rm{mol}}\]
So, 0.824 mol of AgNO 3 were added to the solution, but only 0.600 mol had reacted which means that is how much NaCl was present in the solution, and it was not enough to react with all the AgNO 3.
And this confirms that AgNO 3 was added in excess and NaCl is the limiting reactant .
In the last step, we determine the mass of 0.600 mol NaCl that was in the solution when AgNO 3 was added:
\[{\rm{m}}\;{\rm{(NaCl)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{58}}{\rm{.4}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{35}}{\rm{.0}}\;{\rm{g}}\]
Consider the reaction between MnO 2 and HCl:
MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O
What is the theoretical yield of MnCl 2 in grams when 165 g of MnO 2 is added to a solution containing 94.2 g of HCl?
0.645 mol or 81.1 g MnCl 2
The theoretical yield is the amount of product that can be formed, so this problem is similar to the ones we have been working on.
The amounts of both reactants are given, so we determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed.
The moles of MnO 2 and HCl are:
\[{\rm{n}}\;{\rm{(Mn}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{165}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{86}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.90}}\;{\rm{mol}}\]
\[{\rm{n}}\;{\rm{(HCl)}}\;{\rm{ = }}\;{\rm{94}}{\rm{.2}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{36}}{\rm{.5}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.58}}\;{\rm{mol}}\]
Next, we calculate how much MnCl 2 can be formed from each reactant:
\[{\rm{n}}\;\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.90 }}\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MnC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.90}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.58 }}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MnC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.645}}\;{\rm{mol}}\]
HCl gives less product so, it is the LR, and therefore, 0.645 mol MnCl 2 is the theoretical yield of the reaction.
If you are asked to find the theoretical yield in grams, then convert the moles to grams using the molar mass of MnCl 2:
\[{\rm{m}}\;{\rm{(MnC}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.645}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{125}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{81}}{\rm{.1}}\;{\rm{g}}\]
Percent Yield
21. In a chemistry experiment, a student obtained 5.68 g of a product. What is the percent yield of the product if the theoretical yield was 7.12 g?
The percent yield of the reaction is the ratio of the actual over the theoretical yield. What we obtain is the actual yield and the maximum amount of product that can be obtained from the given amount of the limiting reactant is the theoretical yield .
So, in this example, the actual yield is 5.68 g, and the theoretical yield is 7.12 g.
\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{5}}{\rm{.68}}\;{\rm{g}}}}{{{\rm{7}}{\rm{.12}}\;{\rm{g}}}}\; \times \;100\% \; = \;79.8\;\% \]
When 38.45 g CCl 4 is reacted with an excess of HF, 21.3 g CCl 2 F 2 is obtained. Calculate the theoretical and percent yields of this reaction.
CCl 4 + 2HF → CCl 2 F 2 + 2HCl
To find the theoretical yield, we need to calculate the moles of CCl 4 and, using the mole ratio, determine how much CCl 2 F 2 can be formed.
\[{\rm{n}}\;{\rm{(CC}}{{\rm{l}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{38}}{\rm{.45}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{153}}{\rm{.8}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;{\rm{mol}}\]
The amount of CCl 2 F 2 that can be formed from this is also 0.250 mol because of the 1:1 molar ratio:
\[{\rm{n}}\;\left( {{\rm{CC}}{{\rm{l}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.25 }}\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.25}}\;{\rm{mol}}\]
The mass, which is also the theoretical yield of CCl 2 F 2 is:
\[{\rm{m}}\;{\rm{(CC}}{{\rm{l}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{120}}{\rm{.9}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{30}}{\rm{.2}}\;{\rm{g}}\]
The percent yield of the reaction is the ratio of the actual over the theoretical yield:
\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{21}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{30}}{\rm{.2}}\;{\rm{g}}}}\; \times \;100\% \; = \;70.5\;\% \]
Iron(III) oxide reacts with carbon monoxide according to the equation:
Fe 2 O 3 ( s ) + 3CO( g ) → 2Fe( s ) + 3CO 2 ( g )
What is the percent yield of this reaction if 623 g of iron oxide produces 341 g of iron?
The actual yield of the reaction is given as 341 g Fe so, to determine the percent yield, we need to first calculate the theoretical yield. This is the amount of Fe that can be formed if all the starting material (in this case Fe 2 O 3 ) converts into a product according to the chemical equation.
So, we are going to calculate the moles of 623 g Fe 2 O 3 and determine the moles and the mass of the iron based on the stoichiometric ratio:
\[{\rm{n}}\;{\rm{(F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{623}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{159}}{\rm{.7}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.90}}\;{\rm{mol}}\]
The amount of Fe that can be formed from this is calculated using the mole ratio of the oxide and iron:
\[{\rm{n}}\;\left( {{\rm{Fe}}} \right)\;{\rm{ = }}\;{\rm{3}}{\rm{.90 }}\cancel{{{\rm{mol}}\;{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{Fe}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.80}}\;{\rm{mol}}\]
The mass, which is also the theoretical yield of Fe is:
\[{\rm{m}}\;{\rm{(Fe)}}\;{\rm{ = }}\;{\rm{7}}{\rm{.80}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{435}}\;{\rm{g}}\]
\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{341}}\;{\rm{g}}}}{{{\rm{435}}\;{\rm{g}}}}\; \times \;100\% \; = \;78.4\;\% \]
Determine the percent yield of the reaction if 77.0 g of CO 2 are formed from burning 2.00 moles of C 5 H 12 in 4.00 moles of O 2 .
C 5 H 12 + 8 O 2 → 5CO 2 + 6H 2 O
When the amount of both reactants is given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed. This would be the theoretical yield of CO 2.
Now, we are given the moles of the reactants, so we calculate how much CO 2 can each produce:
\[{\rm{n}}\;\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}}}\;{\rm{ = }}\;{\rm{10}}{\rm{.0}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{4}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{8}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.50}}\;{\rm{mol}}\]
Because oxygen gives less product, it is the LR, and therefore, 2.50 mol CO 2 could be formed in this reaction. This is the theoretical yield which is grams is:
\[{\rm{m}}\;{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.50}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{44}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{110}}{\rm{.}}\;{\rm{g}}\]
\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{77}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{110}}{\rm{.}}\;{\rm{g}}}}\; \times \;100\% \; = \;70.0\;\% \]
The percent yield for the following reaction was determined to be 84%:
N 2 ( g ) + 2H 2 ( g ) → N 2 H 4 ( l )
How many grams of hydrazine (N 2 H 4 ) can be produced when 38.36 g of nitrogen reacts with 6.68 g of hydrogen?
The problem asks us to find the actual yield of the reaction given that the percent yield is 80%.
When the amounts of both reactants are given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed. This would be the theoretical yield of hydrazine, and we can use it with the percent yield to find the actual yield of the reaction.
\[{\rm{n}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{38}}{\rm{.36}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{28}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.37}}\;{\rm{mol}}\]
\[{\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{6}}{\rm{.68}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{2}}{\rm{.00}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.34}}\;{\rm{mol}}\]
Now, we are given the moles of the reactants, so we calculate how much N 2 H 4 can each produce:
\[{\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.37 }}\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.37}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{3}}{\rm{.34 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.67}}\;{\rm{mol}}\]
N 2 gives less hydrazine so, it is the LR, and therefore the theoretical yield of N 2 H 4 is 1.37 mol. The mass of N 2 H 4 is:
\[{\rm{m}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}_{\rm{2}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.37}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{43}}{\rm{.8}}\;{\rm{g}}\]
The percent yield of the reaction is the ratio of the actual over the theoretical yield, so rearranging this equation we can calculate the actual yield:
\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;\]
A = T x % Yield
A = 43.8 x 0.84 = 36.8 g
Another approach for determining the actual yield is to convert the moles of limiting reactant to the percentage of the percent yield. So, if the yield is 84%, we could have multiplied the moles of N 2 by 0.84 and determine the mass of N 2 H 4 based on that.
We would have 1.37 mole x 0.84 = 1.1508 mol N 2 which is the moles of nitrogen that actually convert into hydrazine. By the mole ratio, the moles of N 2 H 4 would be:
\[{\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.1508 }}\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.1508}}\;{\rm{mol}}\]
And the mass would be:
\[{\rm{m}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}_{\rm{2}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.1508}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.8}}\;{\rm{g}}\]
Silver metal can be prepared by reducing its nitrate, AgNO 3 with copper according to the following equation:
Cu( s ) + 2AgNO 3 ( aq ) → Cu(NO 3 ) 2 ( aq ) + 2Ag( s )
What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 132.5 grams of AgNO 3 ?
The actual yield of the reaction is given as 71.5 g Ag so, to determine the percent yield, we need to first calculate the theoretical yield. This is the amount of Ag that can be formed if all the starting material (in this case AgNO 3 ) converts into product according to the chemical equation.
So, we are going to calculate the moles of 132.5 g AgNO 3 and determine the moles and the mass of the silver based on their stoichiometric ratio:
\[{\rm{n}}\;{\rm{(AgN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{132}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{169}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.780}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{\rm{Ag}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.780 }}\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{Ag}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.780}}\;{\rm{mol}}\]
The mass, which is also the theoretical yield of Ag is:
\[{\rm{m}}\;{\rm{(Ag)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.780}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{107}}{\rm{.9}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{84}}{\rm{.2}}\;{\rm{g}}\]
\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{71}}{\rm{.5}}\;{\rm{g}}}}{{{\rm{84}}{\rm{.2}}\;{\rm{g}}}}\; \times \;100\% \; = \;85\;\% \]
Industrially, nitric acid is produced from ammonia by the Ostwald process in a series of reactions:
4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2 O( l )
2NO( g ) + O 2 ( g ) → 2NO 2 ( g )
2NO 2 ( g ) + H 2 O( l ) → HNO 3 ( aq ) + HNO 2 ( aq )
Considering that each reaction has an 85% percent yield, how many grams of NH 3 must be used to produce 25.0 kg of HNO 3 by the above procedure?
2.20 x 10 4 g
Because we are given the mass of the product and need to determine the mass of the starting material, the calculations are going to be in reverse order. So, first, we can calculate the moles of HNO 3 and using the mole ratio, find the moles of NO 2 which is the link between the 3 rd and 2 nd reactions.
Before using the mass of HNO 3 , convert it to grams because the molar mass is given in grams per mole:
\[{\rm{m}}\left( {{\rm{HN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{25}}{\rm{.0}}\;\cancel{{{\rm{kg}}}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.5}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\]
\[{\rm{n}}\;{\rm{(HN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.50}}\;{\rm{ \times }}\;{10^4}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{63}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{397}}\;{\rm{mol}}\]
\[{\rm{n}}\;\left( {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{397 }}\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{794}}\;{\rm{mol}}\]
Now, this would have been the conversion if it was a process with a 100% yield. However, because each step of this process has an 85% percent yield, we need to multiply each mole conversion by a factor of
\[\frac{{{\rm{100\% }}}}{{{\rm{85\% }}}}\;{\rm{or}}\;{\rm{simply}}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\]
So, the moles of NO 2 that produced 397 mol of HNO 3 would be:
\[{\rm{n}}\;\left( {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{794}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\;{\rm{ = }}\;{\rm{934}}\;{\rm{mol}}\]
Now, we go to the second reaction and see how much NO was needed to produce 934 mol NO 2 considering that the yield of the reaction is again 85%.
\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{934 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}}}\; \times \;\frac{{100}}{{85}}\;{\rm{ = }}\;{\rm{1098}}\;{\rm{mol}}\]
Next, we do the same for the first reaction and determine how much NH 3 was needed to produce 1098 mol NO considering that the yield of the reaction is again 85%.
\[{\rm{n}}\;\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{1098 }}\cancel{{{\rm{mol}}\;{\rm{NO}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{NO}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\;{\rm{ = }}\;{\rm{1293}}\;{\rm{mol}}\]
In the last step, we convert the moles to the mass of ammonia:
\[{\rm{m}}\;{\rm{(N}}{{\rm{H}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{1293}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{17}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{21,981}}\;{\rm{g}}\]
Rounding off to three significant figures, we get:
Aspirin (acetylsalicylic acid) is widely used to treat pain, fever, and inflammation. It is produced from the reaction of salicylic acid with acetic anhydride. The chemical equation for aspirin synthesis is shown below:
In one container, 10.00 kg of salicylic acid is mixed with 10.00 kg of acetic anhydride.
a) Which reactant is limiting? Which is in excess? b) What mass of excess reactant is left over? c) What mass of aspirin is formed assuming 100% yield (Theoretical yield)? d) What mass of aspirin is formed if the reaction yield is 70.0% ? e) If the actual yield of aspirin is 11.2 kg, what is the percent yield? f) How many kg of salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?
Salicylic acid is the limiting reagent
2.608 x 10 3 g Acetic anhydride
1.305 x 10 4 g Aspirin
9.14 x 10 3 g Aspirin
a) Which reactant is limiting? Which is in excess?
The limiting reactant is the one producing less product, so we need to calculate the moles of 00 kg salicylic acid and 10.00 kg acetic anhydride and see which one produces less aspirin. Because the mole ratio of all the reactants and products is 1:1, whichever has a smaller number of moles, it is the limiting reactant.
Before using the masses, convert them to grams because the molar mass is given in grams per mole:
\[{\rm{m}}\left( {{\rm{salicylic acid}}} \right)\;{\rm{ = }}\;{\rm{10}}{\rm{.00}}\;\cancel{{{\rm{kg}}}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\]
\[{\rm{m}}\left( {{\rm{acetic anhydride}}} \right)\;{\rm{ = }}\;{\rm{10}}{\rm{.00}}\;\cancel{{{\rm{kg}}}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\]
\[{\rm{n}}\;{\rm{(salicylic acid)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{138}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}\]
\[{\rm{n}}\;{\rm{(acetic anhydride)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.000}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{102}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{97}}{\rm{.9}}\;{\rm{mol}}\]
Salicylic acid is present in a smaller quantity, and because of the 1:1 molar ratio of all the reactants and products, it is going to produce less aspirin, and therefore, it is the limiting reactant.
b) What mass of excess reactant is left over?
We know that salicylic acid is the limiting reactant, so the calculations are going to be based on its moles (72.4 mol). Using the mole ratio, we can calculate how much acetic anhydride has reacted. The remaining mass of the acetic anhydride is calculated by subtracting this amount from the initial mass.
Again, because of the 1:1 molar ratio, there is going to be 72.4 mol acetic anhydride reacted:
\[{\rm{n}}\;\left( {{\rm{acetic anhydride}}} \right)\;{\rm{ = }}\;{\rm{72}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{acetic anhydride}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}\]
This corresponds to:
\[{\rm{m}}\;{\rm{(acetic anhydride)}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{102}}{\rm{.1}}\;{\rm{g}}\;}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{7392}}\;{\rm{g}}\]
The remaining mass of acetic anhydride is:
10,000 – 7392 = 2608 g = 2.608 kg
c) What mass of aspirin is formed assuming 100% yield (Theoretical yield)?
72.4 mol aspirin will be formed because of the 1:1 molar ratio:
\[{\rm{n}}\;\left( {{\rm{aspirin}}} \right)\;{\rm{ = }}\;{\rm{72}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{aspirin}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{salicylic acid}}}}}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}\]
Using the molar mass of aspirin, we can calculate its mass for 100%-yield reaction:
\[{\rm{m}}\;{\rm{(aspirin)}}\;{\rm{ = }}\;{\rm{72}}{\rm{.4}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{180}}{\rm{.2}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{13,046}}\;{\rm{g}}\]
This corresponds to 13.05 kg when divided by 1000.
d) What mass of aspirin is formed if the reaction yield is 70.0%?
To find the mass of aspirin when the percent yield is 70.0%, we need to multiply the theoretical yield (13.05 kg) by 0.700:
m (aspirin) = 13.05 kg x 0.700 = 9.14 kg
e) If the actual yield of aspirin is 11.2 kg, what is the percent yield?
\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{11}}{\rm{.2}}\;{\rm{kg}}}}{{{\rm{13}}{\rm{.05}}\;{\rm{kg}}}}\;{\rm{ \times }}\;{\rm{100\% }}\;{\rm{ = }}\;{\rm{85}}{\rm{.8}}\;{\rm{\% }}\]
f) How many kg of salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?
First, calculate the moles of 20.0 kg aspirin which is 2.00 x 10 4 g:
\[{\rm{n}}\;{\rm{(aspirin)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.0}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{180}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{111}}\;{\rm{mol}}\]
According to the mole ratio of the chemical equation, this is how much salicylic acid would be needed because of the 1:1 mole ratio.
However, because the reaction has an 85.0% percent yield, we need to multiply the moles by a factor of
\[{\rm{n}}\;{\rm{(salicylic acid)}}\;{\rm{ = }}\;{\rm{111}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\;{\rm{ = }}\;{\rm{131}}\;{\rm{mol}}\]
The mass of salicylic acid would be:
\[{\rm{m}}\;{\rm{(salicylic acid)}}\;{\rm{ = }}\;{\rm{131}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{138}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{18,078}}\;{\rm{g}}\]
This corresponds to 18.1 kg
3 thoughts on “Stoichiometry Practice Problems”
You forgot the subscript 3 for O in the molecular formula for acetic anhydride and the reaction is not balanced as written. For part F) it’s 18.1 kg and not1.81 kg as written in the final line of the solution.
Thanks for letting me know! Fixed.
You’re welcome!
Leave a Comment Cancel reply
WassUp 1.9.4.5 timestamp: 2023-06-08 12:22:11AM UTC (07:22PM) If above timestamp is not current time, this page is cached.
Solving Stoichiometry Problems
Core Concepts
In this tutorial, you will learn what is stoichiometry and the different types of problems involving it. You will go through several examples to practice and master the content!
Topics Covered in Other Articles
Balancing chemical equations.
- What is a Chemical Reaction?
- Calculating Molar Mass
- Percent Yield Calculation
- How to Calculate Molarity
- What Are Significant Figures
Stoichiometry Definition
What is stoichiometry.
Stoichiometry is math having to do with chemical reactions. There are different types of calculations you can perform; stoichiometry with moles is the most common, but you can also do math with masses and even percentages. Read about the origins of stoichiometry here ! Learn what is mole in chemistry .
Stoichiometric Ratio
A stoichiometric ratio comes into play when talking about the relationships of elements or molecules in specific problems. This is the exact ratio between the coefficients of the reactants and products needed for a reaction to proceed normally. Let’s work through some problems you may see when learning about stoichiometry.
Stoichiometry Problems
A very common type of stoichiometric problem is balancing equations. This is an important chemistry skill to have because you have to have the correct ratio of reactants and products in order for a reaction to proceed; this is also an important foundation for organic chemistry. Although we have a tutorial on balancing equations , let’s look at one example.
Balance the following reaction:
The main idea when balancing equations is that there should be the same number of each element on both sides of the reaction. You can balance the carbons and the hydrogens first, then move onto the oxygen. The balanced equation should look like this:
Example – Using Stoichiometric Ratio (Moles)
Use the equation below to solve the problem.
Example – Using Stoichiometric Ratio (Mass)
Let’s use the same equation as the problem above.
Similar to the previous problem, by using the stoichiometric ratio of reactant to product, you can find the answer. Dimensional analysis is used to go from grams of C 2 H 5 OH to molar mass to mole (stoichiometric) ratio and back to grams.
Stoichiometry Practice Problems
CaCl 2 and AgNO 3 react according to the following equation:
Hexane combusts according to the following reaction:
Stoichiometry Practice Problem Solutions
Further Reading
- Percent by Weight Calculation
- Balancing Redox Reactions
- How to Write Net Ionic Equations
- What is Specific Heat?
- Common Ion Effect
- TPC and eLearning
- What's NEW at TPC?
- Read Watch Interact
- Practice Review Test
- Teacher-Tools
- Subscription Selection
- Seat Calculator
- Ad Free Account
- Student Progress Edit
- Task Properties
- Export Student Progress
- Task, Activities, and Scores
- Edit Profile Settings
- Tasks and Courses
- Subscription
- Subscription Locator
- Metric Conversions Questions
- Metric System Questions
- Metric Estimation Questions
- Significant Digits Questions
- Proportional Reasoning
- Acceleration
- Distance-Displacement
- Dots and Graphs
- Graph That Motion
- Match That Graph
- Name That Motion
- Motion Diagrams
- Pos'n Time Graphs Numerical
- Pos'n Time Graphs Conceptual
- Up And Down - Questions
- Balanced vs. Unbalanced Forces
- Change of State
- Force and Motion
- Mass and Weight
- Match That Free-Body Diagram
- Net Force (and Acceleration) Ranking Tasks
- Newton's Second Law
- Normal Force Card Sort
- Recognizing Forces
- Air Resistance and Skydiving
- Solve It! with Newton's Second Law
- Which One Doesn't Belong?
- Component Addition Questions
- Head-to-Tail Vector Addition
- Projectile Mathematics
- Trajectory - Angle Launched Projectiles
- Trajectory - Horizontally Launched Projectiles
- Vector Addition
- Vector Direction
- Which One Doesn't Belong? Projectile Motion
- Forces in 2-Dimensions
- Being Impulsive About Momentum
- Explosions - Law Breakers
- Hit and Stick Collisions - Law Breakers
- Case Studies: Impulse and Force
- Impulse-Momentum Change Table
- Keeping Track of Momentum - Hit and Stick
- Keeping Track of Momentum - Hit and Bounce
- What's Up (and Down) with KE and PE?
- Energy Conservation Questions
- Energy Dissipation Questions
- Energy Ranking Tasks
- LOL Charts (a.k.a., Energy Bar Charts)
- Match That Bar Chart
- Words and Charts Questions
- Name That Energy
- Stepping Up with PE and KE Questions
- Case Studies - Circular Motion
- Circular Logic
- Forces and Free-Body Diagrams in Circular Motion
- Gravitational Field Strength
- Universal Gravitation
- Angular Position and Displacement
- Linear and Angular Velocity
- Angular Acceleration
- Rotational Inertia
- Balanced vs. Unbalanced Torques
- Getting a Handle on Torque
- Torque-ing About Rotation
- Balloon Interactions
- Charge and Charging
- Charge Interactions
- Charging by Induction
- Conductors and Insulators
- Coulombs Law
- Electric Field
- Electric Field Intensity
- Polarization
- Case Studies: Electric Power
- Know Your Potential
- Light Bulb Anatomy
- I = ∆V/R Equations as a Guide to Thinking
- Parallel Circuits - ∆V = I•R Calculations
- Resistance Ranking Tasks
- Series Circuits - ∆V = I•R Calculations
- Series vs. Parallel Circuits
- Equivalent Resistance
- Period and Frequency of a Pendulum
- Pendulum Motion: Velocity and Force
- Energy of a Pendulum
- Period and Frequency of a Mass on a Spring
- Horizontal Springs: Velocity and Force
- Vertical Springs: Velocity and Force
- Energy of a Mass on a Spring
- Decibel Scale
- Frequency and Period
- Closed-End Air Columns
- Name That Harmonic: Strings
- Rocking the Boat
- Wave Basics
- Matching Pairs: Wave Characteristics
- Wave Interference
- Waves - Case Studies
- Color Addition and Subtraction
- Color Filters
- If This, Then That: Color Subtraction
- Light Intensity
- Color Pigments
- Converging Lenses
- Curved Mirror Images
- Law of Reflection
- Refraction and Lenses
- Total Internal Reflection
- Who Can See Who?
- Formulas and Atom Counting
- Atomic Models
- Bond Polarity
- Entropy Questions
- Cell Voltage Questions
- Heat of Formation Questions
- Reduction Potential Questions
- Oxidation States Questions
- Measuring the Quantity of Heat
- Hess's Law
- Oxidation-Reduction Questions
- Galvanic Cells Questions
- Thermal Stoichiometry
- Molecular Polarity
- Quantum Mechanics
- Balancing Chemical Equations
- Bronsted-Lowry Model of Acids and Bases
- Classification of Matter
- Collision Model of Reaction Rates
- Density Ranking Tasks
- Dissociation Reactions
- Complete Electron Configurations
- Enthalpy Change Questions
- Equilibrium Concept
- Equilibrium Constant Expression
- Equilibrium Calculations - Questions
- Equilibrium ICE Table
- Ionic Bonding
- Lewis Electron Dot Structures
- Line Spectra Questions
- Measurement and Numbers
- Metals, Nonmetals, and Metalloids
- Metric Estimations
- Metric System
- Molarity Ranking Tasks
- Mole Conversions
- Name That Element
- Names to Formulas
- Names to Formulas 2
- Nuclear Decay
- Particles, Words, and Formulas
- Periodic Trends
- Precipitation Reactions and Net Ionic Equations
- Pressure Concepts
- Pressure-Temperature Gas Law
- Pressure-Volume Gas Law
- Chemical Reaction Types
- Significant Digits and Measurement
- States Of Matter Exercise
- Stoichiometry - Math Relationships
- Subatomic Particles
- Spontaneity and Driving Forces
- Gibbs Free Energy
- Volume-Temperature Gas Law
- Acid-Base Properties
- Energy and Chemical Reactions
- Chemical and Physical Properties
- Valence Shell Electron Pair Repulsion Theory
- Writing Balanced Chemical Equations
- Mission CG1
- Mission CG10
- Mission CG2
- Mission CG3
- Mission CG4
- Mission CG5
- Mission CG6
- Mission CG7
- Mission CG8
- Mission CG9
- Mission EC1
- Mission EC10
- Mission EC11
- Mission EC12
- Mission EC2
- Mission EC3
- Mission EC4
- Mission EC5
- Mission EC6
- Mission EC7
- Mission EC8
- Mission EC9
- Mission RL1
- Mission RL2
- Mission RL3
- Mission RL4
- Mission RL5
- Mission RL6
- Mission KG7
- Mission RL8
- Mission KG9
- Mission RL10
- Mission RL11
- Mission RM1
- Mission RM2
- Mission RM3
- Mission RM4
- Mission RM5
- Mission RM6
- Mission RM8
- Mission RM10
- Mission LC1
- Mission RM11
- Mission LC2
- Mission LC3
- Mission LC4
- Mission LC5
- Mission LC6
- Mission LC8
- Mission SM1
- Mission SM2
- Mission SM3
- Mission SM4
- Mission SM5
- Mission SM6
- Mission SM8
- Mission SM10
- Mission KG10
- Mission SM11
- Mission KG2
- Mission KG3
- Mission KG4
- Mission KG5
- Mission KG6
- Mission KG8
- Mission KG11
- Mission F2D1
- Mission F2D2
- Mission F2D3
- Mission F2D4
- Mission F2D5
- Mission F2D6
- Mission KC1
- Mission KC2
- Mission KC3
- Mission KC4
- Mission KC5
- Mission KC6
- Mission KC7
- Mission KC8
- Mission AAA
- Mission SM9
- Mission LC7
- Mission LC9
- Mission NL1
- Mission NL2
- Mission NL3
- Mission NL4
- Mission NL5
- Mission NL6
- Mission NL7
- Mission NL8
- Mission NL9
- Mission NL10
- Mission NL11
- Mission NL12
- Mission MC1
- Mission MC10
- Mission MC2
- Mission MC3
- Mission MC4
- Mission MC5
- Mission MC6
- Mission MC7
- Mission MC8
- Mission MC9
- Mission RM7
- Mission RM9
- Mission RL7
- Mission RL9
- Mission SM7
- Mission SE1
- Mission SE10
- Mission SE11
- Mission SE12
- Mission AAA2
- Mission SE4
- Mission SE5
- Mission SE6
- Mission SE7
- Mission SE8
- Mission SE9
- Mission VP1
- Mission VP10
- Mission VP2
- Mission VP3
- Mission VP4
- Mission VP5
- Mission VP6
- Mission VP7
- Mission VP8
- Mission VP9
- Mission WM1
- Mission WM2
- Mission WM3
- Mission WM4
- Mission WM5
- Mission WM6
- Mission WM7
- Mission WM8
- Mission WE1
- Mission WE10
- Mission WE2
- Mission WE3
- Mission WE4
- Mission WE5
- Mission WE6
- Mission WE7
- Mission WE8
- Mission WE9
- Vector Walk Interactive
- Name That Motion Interactive
- Kinematic Graphing 1 Concept Checker
- Kinematic Graphing 2 Concept Checker
- Graph That Motion Interactive
- Rocket Sled Concept Checker
- Force Concept Checker
- Free-Body Diagrams Concept Checker
- Free-Body Diagrams The Sequel Concept Checker
- Skydiving Concept Checker
- Elevator Ride Concept Checker
- Vector Addition Concept Checker
- Vector Walk in Two Dimensions Interactive
- Name That Vector Interactive
- River Boat Simulator Concept Checker
- Projectile Simulator 2 Concept Checker
- Projectile Simulator 3 Concept Checker
- Turd the Target 1 Interactive
- Turd the Target 2 Interactive
- Balance It Interactive
- Go For The Gold Interactive
- Egg Drop Concept Checker
- Fish Catch Concept Checker
- Exploding Carts Concept Checker
- Collision Carts - Inelastic Collisions Concept Checker
- Its All Uphill Concept Checker
- Stopping Distance Concept Checker
- Chart That Motion Interactive
- Roller Coaster Model Concept Checker
- Uniform Circular Motion Concept Checker
- Horizontal Circle Simulation Concept Checker
- Vertical Circle Simulation Concept Checker
- Race Track Concept Checker
- Gravitational Fields Concept Checker
- Orbital Motion Concept Checker
- Balance Beam Concept Checker
- Torque Balancer Concept Checker
- Aluminum Can Polarization Concept Checker
- Charging Concept Checker
- Name That Charge Simulation
- Coulomb's Law Concept Checker
- Electric Field Lines Concept Checker
- Put the Charge in the Goal Concept Checker
- Circuit Builder Concept Checker (Series Circuits)
- Circuit Builder Concept Checker (Parallel Circuits)
- Circuit Builder Concept Checker (∆V-I-R)
- Circuit Builder Concept Checker (Voltage Drop)
- Equivalent Resistance Interactive
- Pendulum Motion Simulation Concept Checker
- Mass on a Spring Simulation Concept Checker
- Particle Wave Simulation Concept Checker
- Boundary Behavior Simulation Concept Checker
- Slinky Wave Simulator Concept Checker
- Simple Wave Simulator Concept Checker
- Wave Addition Simulation Concept Checker
- Standing Wave Maker Simulation Concept Checker
- Color Addition Concept Checker
- Painting With CMY Concept Checker
- Stage Lighting Concept Checker
- Filtering Away Concept Checker
- Young's Experiment Interactive
- Plane Mirror Images Interactive
- Who Can See Who Concept Checker
- Optics Bench (Mirrors) Concept Checker
- Name That Image (Mirrors) Interactive
- Refraction Concept Checker
- Total Internal Reflection Concept Checker
- Optics Bench (Lenses) Concept Checker
- Stopping Distance Preview
- Velocity Time Graphs Preview
- Kinematics Preview
- Friction Preview
- Coffee Filter Lab Preview
- Juggling Preview
- Ball Bat Collision Preview
- Keplers Laws Preview
- Slinky-Experiments Preview
- Gaining Teacher Access
- 1-D Kinematics
- Newton's Laws
- Vectors - Motion and Forces in Two Dimensions
- Momentum and Its Conservation
- Work and Energy
- Circular Motion and Satellite Motion
- Thermal Physics
- Static Electricity
- Electric Circuits
- Vibrations and Waves
- Sound Waves and Music
- Light and Color
- Reflection and Mirrors
- About the Physics Interactives
- Task Tracker
- Usage Policy
- Newtons Laws
- Vectors and Projectiles
- Forces in 2D
- Momentum and Collisions
- Circular and Satellite Motion
- Balance and Rotation
- Waves and Sound
- Forces in Two Dimensions
- Work, Energy, and Power
- Circular Motion and Gravitation
- Sound Waves
- 1-Dimensional Kinematics
- Circular, Satellite, and Rotational Motion
- Einstein's Theory of Special Relativity
- Waves, Sound and Light
- QuickTime Movies
- About the Concept Builders
- Pricing For Schools
- Directions for Version 2
- Measurement and Units
- Relationships and Graphs
- Rotation and Balance
- Vibrational Motion
- Reflection and Refraction
- Teacher Accounts
- Task Tracker Directions
- Kinematic Concepts
- Kinematic Graphing
- Wave Motion
- Sound and Music
- About CalcPad
- 1D Kinematics
- Vectors and Forces in 2D
- Simple Harmonic Motion
- Rotational Kinematics
- Rotation and Torque
- Rotational Dynamics
- Electric Fields, Potential, and Capacitance
- Light Waves
- Units and Measurement
- Stoichiometry
- Molarity and Solutions
- Thermal Chemistry
- Acids and Bases
- Kinetics and Equilibrium
- Solution Equilibria
- Oxidation-Reduction
- Nuclear Chemistry
- About the Science Reasoning Center
- NGSS Activities
- 1D-Kinematics
- Projectiles
- Circular Motion
- Newton's Laws of Motion
- Work and Energy Packet
- Static Electricity Review
- Graphing Practice
- About the ACT
- ACT Preparation
- For Teachers
- Other Resources
- Solutions Guide
- Solutions Guide Digital Download
- Motion in One Dimension
- Work, Energy and Power
- Purchasing the CD
- Purchasing the Digital Download
- About the NGSS Corner
- NGSS Search
- Force and Motion DCIs - High School
- Energy DCIs - High School
- Wave Applications DCIs - High School
- Force and Motion PEs - High School
- Energy PEs - High School
- Wave Applications PEs - High School
- Crosscutting Concepts
- The Practices
- Physics Topics
- NGSS Corner: Activity List
- NGSS Corner: Infographics
- About the Toolkits
- Position-Velocity-Acceleration
- Position-Time Graphs
- Velocity-Time Graphs
- Newton's First Law
- Newton's Second Law
- Newton's Third Law
- Terminal Velocity
- Projectile Motion
- Forces in 2 Dimensions
- Impulse and Momentum Change
- Momentum Conservation
- Work-Energy Fundamentals
- Work-Energy Relationship
- Roller Coaster Physics
- Satellite Motion
- Electric Fields
- Circuit Concepts
- Series Circuits
- Parallel Circuits
- Describing-Waves
- Wave Behavior Toolkit
- Standing Wave Patterns
- Resonating Air Columns
- Wave Model of Light
- Plane Mirrors
- Curved Mirrors
- Teacher Guide
- Using Lab Notebooks
- Current Electricity
- Light Waves and Color
- Reflection and Ray Model of Light
- Refraction and Ray Model of Light
- Subscriptions
- Teacher Resources
- Newton's Laws
- Einstein's Theory of Special Relativity
- About Concept Checkers
- School Pricing
- Legacy Version
- Newton's Laws of Motion
- Newton's First Law
- Newton's Third Law
Chemistry: Molarity and Solutions
- EXPLORE Coupons Tech Help Pro Random Article About Us Quizzes Contribute Train Your Brain Game Improve Your English Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
- HELP US Support wikiHow Community Dashboard Write an Article Request a New Article More Ideas...
- EDIT Edit this Article
- PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Coupons Quizzes Upgrade Sign In
- Browse Articles
- Learn Something New
- Train Your Brain
- Improve Your English
- Explore More
- Support wikiHow
- About wikiHow
- H&M Coupons
- Hotwire Promo Codes
- StubHub Discount Codes
- Ashley Furniture Coupons
- Blue Nile Promo Codes
- NordVPN Coupons
- Samsung Promo Codes
- Chewy Promo Codes
- Ulta Coupons
- Vistaprint Promo Codes
- Shutterfly Promo Codes
- DoorDash Promo Codes
- Office Depot Coupons
- adidas Promo Codes
- Home Depot Coupons
- DSW Coupons
- Bed Bath and Beyond Coupons
- Lowe's Coupons
- Surfshark Coupons
- Nordstrom Coupons
- Walmart Promo Codes
- Dick's Sporting Goods Coupons
- Fanatics Coupons
- Edible Arrangements Coupons
- eBay Coupons
- Log in / Sign up
- Education and Communications
How to Do Stoichiometry
Last Updated: December 2, 2022 References
This article was co-authored by Bess Ruff, MA . Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group. There are 13 references cited in this article, which can be found at the bottom of the page. This article has been viewed 257,258 times.
In a chemical reaction, matter can neither be created nor destroyed according to the law of conservation of mass, so the products that come out of a reaction must equal the reactants that go into a reaction. This means the same amount of each atom that you put in must come back out. Stoichiometry is the measure of the elements within a reaction. [1] X Research source It involves calculations that take into account the masses of reactants and products in a given chemical reaction. Stoichiometry is one half math, one half chemistry, and revolves around the one simple principle above - the principle that matter is never lost or gained during a reaction. The first step in solving any chemistry problem is to balance the equation .
Balancing the Chemical Equation
- Don’t forget to multiply through by a coefficient or subscript if one is present.
- For example, H 2 SO 4 + Fe ---> Fe 2 (SO 4 ) 3 + H 2
- On the reactant (left) side of the equation there are 2 H atoms (H 2 ), 1 S atom, 4 O atoms (O 4 ), and 1 Fe atom.
- On the product (right) side of the equation there are 2H atoms (H 2 ), 3 S atoms (S 3 ), 12 O atoms (O 12 ), and 2 Fe atoms (Fe 2 ).
- For example, the lowest common factor between 2 and 1 is 2 for Fe. Add a 2 in front of the Fe on the left side to balance it.
- The lowest common factor between 3 and 1 is 3 for S. Add a 3 in front of H 2 SO 4 to balance the left and right sides.
- At this stage, our equation looks like this: 3 H 2 SO 4 + 2 Fe ---> Fe 2 (SO 4 ) 3 + H 2
- In our example, we added a 3 in front of H 2 SO 4 and now have 6 hydrogens on the left and only 2 on the right side of the equation. We also have 12 oxygen on the left and 12 oxygen on the right, so it is balanced.
- We can balance hydrogens by adding a 3 in front of H 2 .
- Our final balanced equation is 3 H 2 SO 4 + 2 Fe ---> Fe 2 (SO 4 ) 3 + 3 H 2 .
- Let's check our equation, 3 H 2 SO 4 + 2 Fe ---> Fe 2 (SO 4 ) 3 + 3 H 2 , for balance.
- On the left side of the arrow, there are 6 H, 3 S, 12 O, and 2 Fe.
- On the right side of the arrow, there are 2 Fe, 3 S, 12 O, and 6 H.
- The left and the right sides of the equation match, therefore, it is now balanced.
Converting Between Grams and Moles
- Define the number of atoms of each element in a compound. For example, glucose is C 6 H 12 O 6 , there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
- Identify the atomic mass in grams per mol (g/mol) of each atom. The atomic masses of each element are usually found underneath the element's symbol on a periodic table, usually as a decimal. The atomic masses of the elements in glucose are: carbon, 12.0107 g/mol; hydrogen, 1.007 g/mol; and oxygen, 15.9994 g/mol.
- Multiply each element's atomic mass by the number of atoms present in the compound. Carbon: 12.0107 x 6 = 72.0642 g/mol; Hydrogen: 1.007 x 12 = 12.084 g/mol; Oxygen: 15.9994 x 6 = 95.9964 g/mol.
- Adding these products together yields the molar mass of the compound. 72.0642 + 12.084 + 95.9964 = 180.1446 g/mol. 180.14 grams is the mass of one mole of glucose.
- For example: How many moles are in 8.2 grams of hydrogen chloride (HCl)?
- The atomic mass of H is 1.007 and Cl is 35.453 making the molar mass of the compound 1.007 + 35.453 = 36.46 g/mol.
- Dividing the number of grams of the substance by the molar mass yields: 8.2 g / (36.46 g/mol) = 0.225 moles of HCl.
- For example, what is the molar ratio of KClO 3 to O 2 in the reaction 2 KClO 3 ---> 2 KCl + 3 O 2 .
- First, check to see the equation is balanced. Never forget this step or your ratios will be wrong. In this case there are equal amounts of each element on both sides of the reaction so it is balanced.
- The ratio of KClO 3 to O 2 is 2/3. It doesn’t matter which number goes on top or on bottom as long as you keep the same compounds on the top and bottom throughout the rest of the problem. [11] X Research source
- For example, given the reaction N 2 + 3 H 2 ---> 2 NH 3 how many moles of NH 3 will be produced given 3.00 grams of N 2 reacting with sufficient H 2 ?
- In this example, sufficient H 2 means that there is enough available and you don’t have to take it into account to solve the problem.
- First, convert grams of N 2 to moles. The atomic mass of nitrogen is 14.0067 g/mol so the molar mass of N 2 is 28.0134 g/mol. Dividing mass by molar mass gives you 3.00 g/28.0134 g/mol = 0.107 mol.
- Set up the ratios given by the question: NH 3 : N 2 = x/0.107 mol.
- Cross multiply this ratio by the molar ratio of NH 3 to N 2 : 2:1. x/0.107 mol = 2/1 = (2 x 0.107) = 1x = 0.214 mol.
- The molar mass of NH 3 is 17.028 g/mol. Therefore 0.214 mol x (17.028 grams/mol) = 3.647 grams of NH 3 .
Converting Between Liters of Gas and Moles
- Generally, a reaction will say that it is given at 1 atm and 273 K or will simply say STP.
- For example, convert 3.2 liters of N 2 gas to moles: 3.2 L/22.414 L/mol = 0.143 moles.
- The equation can be rearranged to solve for moles: n = RT/PV.
- The units of the gas constant are designed to cancel out the units of the other variables.
- For example, determine the number of moles in 2.4 liters of O 2 at 300 K and 1.5 atm. Plugging in the variables yields: n = (0.0821 x 300)/(1.5 x 2) = 24.63/3.6 = 6.842 moles of O 2
Converting Between Liters of Liquid and Moles
- If the density is not given within the problem, you may have to look it up in a reference text or online.
- Identify the volume given. For example, let’s say the problem states that you have 1 liter of H 2 O. To convert to mL simply multiply by 1000. There are 1000 milliliters in a liter of water.
- The density of H 2 O, for instance, is approximately 1.0 g/mL.
- Identify the atomic mass in grams per mol (g/mol) of each atom. The atomic masses of the elements in glucose are: carbon, 12.0107 g/mol; hydrogen, 1.007 g/mol; and oxygen, 15.9994 g/mol.
- Multiply each elements atomic mass by the number of atoms present in the compound. Carbon: 12.0107 x 6 = 72.0642 g/mol; Hydrogen: 1.007 x 12 = 12.084 g/mol; Oxygen: 15.9994 x 6 = 95.9964 g/mol.
Community Q&A
Video . By using this service, some information may be shared with YouTube.
You Might Also Like
- ↑ http://www.chemteam.info/Stoichiometry/What-is-Stoichiometry.html
- ↑ https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:chemical-reactions/x2eef969c74e0d802:stoichiometry/a/stoichiometry
- ↑ https://web.ung.edu/media/chemistry/Chapter4/Chapter4-StoichiometryOfChemicalReactions.pdf
- ↑ http://preparatorychemistry.com/Bishop_molar_mass_conversion_factors_help.htm
- ↑ http://www.chemteam.info/Stoichiometry/Mole-Mass.html
- ↑ http://www.chemteam.info/Stoichiometry/Mass-Mass.html
- ↑ https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/stoichiometry-ideal/a/stoichiometry
- ↑ http://www.chemteam.info/Stoichiometry/Molar-Ratio.html
- ↑ http://www.chemteam.info/Stoichiometry/Mole-Mole.html
- ↑ http://www.engineeringtoolbox.com/stp-standard-ntp-normal-air-d_772.html
- ↑ http://www.thegeoexchange.org/chemistry/stoichiometry/liters-to-moles.html
- ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law
- ↑ https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry_and_Balancing_Reactions
About This Article
To do stoichiometry, start by balancing the chemical equation so that the number of atoms on each side of the equal sign are exactly the same. Next, convert the units of measurement into moles and use the mole ratio to calculate the moles of substance yielded by the chemical reaction. Then, convert moles of wanted substance to the desired units of measurement! For tips on converting different units of measurement into moles, read on! Did this summary help you? Yes No
- Send fan mail to authors
Reader Success Stories
Aug 21, 2017
Did this article help you?
Mar 22, 2018
Randy Orton
May 11, 2016
Arnipalli Sateesh
Jan 4, 2017
John Malone
Apr 13, 2016
Featured Articles
Trending Articles
Watch Articles
- Terms of Use
- Privacy Policy
- Do Not Sell or Share My Info
- Not Selling Info
Don’t miss out! Sign up for
wikiHow’s newsletter
IMAGES
VIDEO
COMMENTS
First things first: we need to balance the equation! In this case, we have 1 1 \ce {Na} Na atom and 3 3 \ce {H} H atoms on the reactant side and 2 2 \ce {Na} Na atoms and 2 2 \ce {H} H atoms on the product side.
Given the following reaction: \ce {Zn + CuCl2 -> ZnCl2 + Cu} Zn+CuClX 2 ZnClX 2+Cu How many moles of \ce {ZnCl2} ZnClX 2 will be produced from 23.0 \text { g} 23.0 g of \ce {Zn} Zn, assuming \ce {CuCl2} CuClX 2 is available in excess? moles (round to three significant figures) Show Calculator Show Periodic Table Stuck?
Answer f Click here to see a video of the solution PROBLEM I 2 is produced by the reaction of 0.4235 mol of CuCl 2 according to the following equation: . How many molecules of I 2 are produced? What mass of I 2 is produced? Answer a Answer b PROBLEM Silver is often extracted from ores as K [Ag (CN) 2] and then recovered by the reaction
answer 4HCl + O 2 → 2H 2 O + 2Cl 2 b) answer Al (NO 3) 3 + 3NaOH → Al (OH) 3 + 3NaNO 3
Chemistry: Stoichiometry - Problem Sheet 2 KEY 9) 2 24 2 2 23 2 2 2 2 4.63 x 10molecules I 1 mol I 6.02 x 10 moleculesI 1 mol Cl 1mol 71 g Cl Cl x 546 g Cl 10) 292 g Ag 1 mol Ag 108 g Ag 1 mol Cu 1 mol Ag 63.5 g Cu
Stoichiometry example problem 1 Stoichiometry Limiting reactant example problem 1 edited Specific gravity Stoichiometry questions Google Classroom One type of anaerobic respiration converts glucose ( C_6 H_ {12} O_6 C 6H 12O6) to ethanol ( C_2 H_5 OH C 2H 5OH) and carbon dioxide.
Step 1: grams of A is converted to moles by multiplying by the inverse of the molar mass. Step 2: moles of A is converted to moles of B by multiplying by the molar ratio. Step 3: moles of B is converted to grams of B by the molar mass. To illustrate this procedure, consider the combustion of glucose.
Balance the following reaction: __ C 2 H 2 + __ O 2 → __ CO 2 + __ H 2 O The main idea when balancing equations is that there should be the same number of each element on both sides of the reaction. You can balance the carbons and the hydrogens first, then move onto the oxygen. The balanced equation should look like this:
Homework Set 1: 2C3H7OH + 9O2 --> 6CO2 + 8H2O How many moles of O2 are required to react with 58 moles of C3H7OH? How many moles of CO2 are required to react with 17 moles of O2? How many grams of CO2 would be required to react with 7.8 moles of H2O? How many grams of C3H7OH are needed to produce 0.45 moles of water?
Answer a Answer b Click here to see a video of the solution PROBLEM 5.2.4 What mass of silver oxide, Ag 2 O, is required to produce 25.0 g of silver sulfadiazine, AgC 10 H 9 N 4 SO 2, from the reaction of silver oxide and sulfadiazine? 2C 10H 10N 4SO 2 + Ag 2O → 2AgC 10H 9N 4SO 2 + H 2O Answer
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements.
Problem Set ST7: Mass-to-Mass Stoichiometry 2 To use molar mass values and a balanced chemical equation to relate the mass of reactants to the mass of products. Includes 5 problems. Problem Set ST8: Mass-to-Mass Stoichiometry 3 To use molar mass values and a balanced chemical equation to relate the mass of reactants to the mass of products.
Solution. Step 1. First use stoichiometry to solve for the number of moles of CO 2 produced. (2molC 2H 6)( 4molCO 22molC 2H 6) = 4molCO 2. So 4 moles of Carbon Dioxide are produced if we react 2 moles of ethane gas. Step 2. Now we simply need to manipulate the ideal gas equation to solve for the variable of interest.
There are 4 major categories of stiochiometry problems. It is important to remember, though, that in every situation you need to start out with a balanced equation. 1. Mole-Mole Problems Problem: How many moles of HCl are needed to react with 0.87 moles of Al? Step 1: Balance The Equation & Calculate the Ratios
Problem Set MS13: Solution Stoichiometry 2. Apply stoichiometric principles to reactions between two aqueous-state reactants to relate the volumes and molarities of the reactants to the mass of product. The net ionic reaction is given. All problems are limiting reactant problems; there is significant scaffolding for each problem.
This volume makes intuitive sense for two reasons: (1) the number of moles of Pb(NO 3) 2 required is half of the number of moles of NaCl, based off of the stoichiometry in the balanced reaction (Equation 13.8.1 ); (2) the concentration of Pb(NO 3) 2 solution is 50% greater than the NaCl solution, so less volume is needed. Example 13.8.1
A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following: a. 9.36 × 10 24 formula units of NaCl.
1 Balancing the Chemical Equation 2 Converting Between Grams and Moles 3 Converting Between Liters of Gas and Moles + Show 1 more... Other Sections Questions & Answers Video References Article Summary Co-authored by Bess Ruff, MA Last Updated: December 2, 2022 References
Check your understanding and truly master stoichiometry with these practice problems! In this video, we go over how to convert grams of one compound to grams...
# Steps 2 # Steps 3 # Steps 3Stoichiometry Map Problems #2 SOLUTIONS 1. moles of FeCl3 to grams of MgCl2 2FeCl3 + 3MgO Fe2O3+ 3MgCl2 How manygrams of MgCl 2are produced from the reaction of.84 moles of FeCl3? 2. grams of O2 to grams of SO2 2ZnS + 3O2 2ZnO + 2SO 2 How manygrams of SO2are produced from2.8 grams of oxygen gas? 3.