Stoichiometry Practice Problems - Chemistry Steps (2022)

Stoichiometry

Stoichiometry Practice Problems

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

Practice

1.

Balance the following chemical equations:

a) HCl + O2 → H2O + Cl2

b) Al(NO3)3 + NaOH → Al(OH)3 + NaNO3

c) H2 + N2 → NH3

d) PCl5 + H2O → H3PO4 + HCl

e) Fe + H2SO4 → Fe2(SO4)3 + H2

f) CaCl2 + HNO3 → Ca(NO3)2 + HCl

g) KO2 + H2O → KOH + O2 + H2O2

h) Al + H2O → Al2O3 + H2

i) Fe + Br2 → FeBr3

j) Cu + HNO3 → Cu(NO3)2 + NO2 + H2O

k) Al(OH)3 → Al2O3 + H2O

l) NH3 + O2 → NO + H2O

m) Ca(AlO2)2 + HCl → AlCl3 + CaCl2 + H2O

n) C5H12 + O2 → CO2 + H2O

o) P4O10 + H2O → H3PO4

p) Na2CrO4 + Pb(NO3)2 → PbCrO4 + NaNO3

q) MgCl2 + AgNO3 → AgCl + Mg(NO3)2

r) KClO3 → KClO4 + KCl

s) Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O

a)

answer

2HCl + O2 → H2O + Cl2

b)

answer

Al(NO3)3 + 3NaOH → Al(OH)3 + 3NaNO3

c)

answer

3H2 + N2 → 2NH3

d)

answer

PCl5 + 4H2O → H3PO4 + 5HCl

e)

answer

2Fe + 3H2SO4 → Fe2(SO4)3 + 3H2

f)

answer

CaCl2 + 2HNO3 → Ca(NO3)2 + 2HCl

g)

answer

2KO2 + 2H2O → 2KOH + O2 + H2O2

h)

answer

2Al + 3H2O → Al2O3 + 3H2

i)

answer

2Fe + 3Br2 → 2FeBr3

j)

answer

Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O

k)

answer

2Al(OH)3 → Al2O3 + 3H2O

l)

answer

4NH3 + 5O2 → 4NO + 6H2O

m)

answer

Ca(AlO2)2 + 8HCl → 2AlCl3 + CaCl2 + 4H2O

n)

answer

C5H12 + 8 O2 → 5CO2 + 6H2O

o)

answer

P4O10 + 6H2O → 4H3PO4

p)

answer

Na2CrO4 + Pb(NO3)2 → PbCrO4 + 2NaNO3

q)

answer

MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2

r)

answer

4KClO3 → 3KClO4 + KCl

s)

answer

3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O

2.

Consider the balanced equation:

C5H12 + 8 O2 → 5CO2 + 6H2O

Complete the table showing the appropriate number of moles of reactants and products.

mol C5H12mol O2mol CO2mol H2O
2
2.5
3
5.4

answer

C5H12 + 8 O2 → 5CO2 + 6H2O

mol C5H12mol O2mol CO2mol H2O
2161012
0.31252.51.56251.875
0.64.833.6
0.97.24.55.4

Solution

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:

\[{\rm{n}}\left( {{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.9}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\;{\rm{ \times }}\;\frac{{{\rm{8}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.2}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.4 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\;{\rm{ \times }}\;\frac{{{\rm{5}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.5}}\;{\rm{mol}}\]

3.

How many grams of CO2and H2O are produced from the combustion of 220. g of propane (C3H8)?

C3H8(g) + 5O2(g) → 3CO2(g) +4H2O(g)

answer

660. g CO2

360. g H2O

Solution

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:

Stoichiometry Practice Problems - Chemistry Steps (1)

The moles of C3H8 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 CO2 based on its molar ratio with C3H8:

\[{\rm{n}}\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}}}\;{\rm{ = }}\;{\rm{15}}{\rm{.0}}\;{\rm{mol}}\]

(Video) Step by Step Stoichiometry Practice Problems | How to Pass Chemistry

And finally, the mass of the CO2 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 H2O:

\[{\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{5}}{\rm{.00 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}}}\;{\rm{ = }}\;{\rm{20}}{\rm{.0}}\;{\rm{mol}}\]

\[{\rm{m}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{20}}{\rm{.0 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{18}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{360}}{\rm{.}}\;{\rm{g}}\]

4.

How many grams of CaCl2 can be produced from 65.0 g of Ca(OH)2 according to the following reaction,

Ca(OH)2 + 2HCl → CaCl2 + 2H2O

answer

97.3 g

Solution

Here is the conceptual plan for solving this problem:

Stoichiometry Practice Problems - Chemistry Steps (2)

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 CaCl2 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 CaCl2 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 CaCl2:

\[{\rm{m}}\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.877 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{111}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{97}}{\rm{.3}}\;{\rm{g}}\]

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.

5.

How many moles of oxygen are formed when 75.0 g of Cu(NO3)2 decomposes according to the following reaction?

2Cu(NO3)2→ 2CuO + 4NO2+ O2

answer

0.200 mol

Solution

The amount of product, for a given amount of reactant, is calculated based on their molar/stoichiometric ratio.

So, the conceptual plan is:

Stoichiometry Practice Problems - Chemistry Steps (3)

\[{\rm{n}}\;{\rm{(Cu}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{\;)}}\;{\rm{ = }}\;{\rm{75}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{187}}{\rm{.5}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.400}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{{\rm{O}}_{{\rm{2\;}}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.400}}\;\cancel{{{\rm{mol}}\;{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}{\rm{\;}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{\cancel{{{\rm{2}}\;{\rm{mol}}\;{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}{\rm{\;}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.200}}\;{\rm{mol}}\]

6.

How many grams of MnCl2can be prepared from 52.1 grams of MnO2?

MnO2+ 4HCl → MnCl2+ Cl2+ 2H2O

answer

75.5 g

Solution

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:

Stoichiometry Practice Problems - Chemistry Steps (4)

The moles of MnO2 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 MnCl2 based on its molar ratio with MnO2:

\[{\rm{n}}\;\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.600 }}\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MnC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\]

And finally, the mass of the MnCl2 is determined using its moles and molar mass:

\[{\rm{m}}\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.600 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{125}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{75}}{\rm{.5g}}\]

7.

Determine the mass of oxygen that is formed when an 18.3-g sample of potassium chlorate is decomposed according to the following equation:

2KClO3(s) → 2KCl(s) + 3O2(g).

answer

7.16 g

Solution

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:

Stoichiometry Practice Problems - Chemistry Steps (5)

The moles of MnO2 are calculated using its molar mass:

\[{\rm{n}}\;{\rm{(KCl}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{18}}{\rm{.3}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{122}}{\rm{.6}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.149}}\;{\rm{mol}}\]

Next, we find the moles of O2 based on its molar ratio with KClO3:

\[{\rm{n}}\;\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.149 }}\cancel{{{\rm{mol}}\;{\rm{KCl}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{KCl}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.224}}\;{\rm{mol}}\]

And finally, the mass of the MnCl2 is determined using its moles and molar mass:

\[{\rm{m}}\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.224 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.16}}\;{\rm{g}}\]

8.

How many grams of H2O will be formed when 48.0 grams H2 are mixed with excess hydrogen gas?

2H2+ O2 → 2H2O

answer

432 g

Solution

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:

Stoichiometry Practice Problems - Chemistry Steps (6)

The moles of H2O 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 H2O based on its molar ratio with H2:

\[{\rm{n}}\;\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{24}}{\rm{.0 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{24}}{\rm{.0}}\;{\rm{mol}}\]

And finally, the mass of the H2O is determined using its moles and molar mass:

\[{\rm{m}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)\;{\rm{ = }}\;{\rm{24}}{\rm{.0 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{18}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{432}}\;{\rm{g}}\]

9.

Consider the chlorination reaction of methane (CH4):

CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

What mass of CCl4 will be produced if 0.338 moles of CH4 are used in the reaction?

answer

0.337 mol

Solution

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:

Stoichiometry Practice Problems - Chemistry Steps (7)

The moles of CCl4 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 CH4 based on its molar ratio with CCl4.It is a 1:1 ratio, so there is going to be 0.337 mol CH4:

\[{\rm{n}}\;\left( {{\rm{C}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.337 }}\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{C}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.337}}\;{\rm{mol}}\]

10.

How many grams of Ba(NO3)2 can be produced by reacting 16.5 g of HNO3 with an excess of Ba(OH)2?

answer

34.2 g

Solution

We need to first write the balanced chemical equation to do the calculations:

Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O

The conceptual plan would be:

Stoichiometry Practice Problems - Chemistry Steps (8)

So, let’s start with the moles of HNO3:

\[{\rm{n}}\;{\rm{(HN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{16}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{63}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.262}}\;{\rm{mol}}\]

Once we have the moles, we can determine the moles of Ba(NO3)2 based on the molar ratio, and then convert the moles to the mass:

\[{\rm{n}}\;\left( {{\rm{Ba}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.262 }}\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{HN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.131}}\;{\rm{mol}}\]

\[{\rm{m}}\left( {{\rm{Ba}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.131 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{261}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{34}}{\rm{.2}}\;{\rm{g}}\]

11.

Ethanol can be obtained by fermentation – a complex chemical process breaking down glucose to ethanol and carbon dioxide.

C6H12O6 → 2C2H5OH + 2CO2
glucose ethanol

How many mL of ethanol (d =0.789 g/mL) can be obtained by this process starting with 286 g of glucose?

answer

186 mL

Solution

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:

Stoichiometry Practice Problems - Chemistry Steps (9)

\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{)}}\;{\rm{ = }}\;{\rm{286}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{180}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.59}}\;{\rm{mol}}\]

Now, we can find the moles of C2H5OH:

\[{\rm{n}}\;\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.59 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.18}}\;{\rm{mol}}\]

\[{\rm{m}}\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}} \right)\;{\rm{ = }}\;{\rm{3}}{\rm{.18 }}\cancel{{{\rm{mol}}\;}}\;{\rm{ \times }}\;\frac{{{\rm{46}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;}}}}\;{\rm{ = }}\;{\rm{147}}\;{\rm{g}}\]

The last step is converting the mass to volume:

\[{\rm{v}}\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}} \right)\;{\rm{ = }}\;{\rm{147}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mL}}}}{{{\rm{0}}{\rm{.789}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\;{\rm{186}}\;{\rm{mL}}\]

(Video) Stoichiometry Basic Introduction, Mole to Mole, Grams to Grams, Mole Ratio Practice Problems

12.

36.0 g of butane (C4H10) was burned in an excess of oxygen and the resulting carbon dioxide (CO2) was collected in a sealed vessel.

2C4H10 + 13O2 → 8CO2 + 10H2O

How many grams of LiOH will be necessary to consume all the CO2 from the first reaction?

2LiOH + CO2 → Li2CO3 + H2O

answer

119 g

Solution

There are two reactions here, and the link between them is the CO2 that is formed in the first reaction and used in the second.

So, the plan would be to determine the moles of CO2 in the first reaction, and use it to calculate the moles, and consequently the mass of LiOH.

Stoichiometry Practice Problems - Chemistry Steps (10)

\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{)}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{58}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.620}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.620 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{8}}\;{\rm{mol}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.48}}\;{\rm{mol}}\]

This is the amount of CO2 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.

Limiting Reactant

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.

answer

c) The limiting reactant is the one producing less product than any of the other reactants.

14.

Find the limiting reactant for each initial amount of reactants.

4NH3 + 5O2 → 4NO + 6H2O

a) 2 mol of NH3 and 2 mol of O2

b) 2 mol of NH3 and 3 mol of O2

c) 3 mol of NH3 and 3 mol of O2

d) 3 mol of NH3 and 2 mol of O2

Note:This is not a multiple-choice question. Each row represents a separate question where you need to determine the limiting reactant.

answer

a) 2 mol of NH3 and 2 mol of O2

b) 2 mol of NH3 and 3 mol of O2

c) 3 mol of NH3 and 3 mol of O2

d) 3 mol of NH3 and 2 mol of O2

Solution

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 H2O) and see whether NH3 or O2 will produce less product for the given mole ratio.

Let’s do the calculations based on the amount of NO that can be formed.

a)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{2}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.6}}\;{\rm{mol}}\]

So, for (a), O2 is the limiting reactant.

b)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{2}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4}}\;{\rm{mol}}\]

So, for (b), NO is the limiting reactant.

c)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{3}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4}}\;{\rm{mol}}\]

So, for (c), O2 is the limiting reactant.

d)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{3}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.6}}\;{\rm{mol}}\]

So, for (d), O2 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:

4NH3 + 5O2 → 4NO + 6H2O

a) 2 mol of NH3 and 2 mol of O2

b) 2 mol of NH3 and 3 mol of O2

c) 3 mol of NH3 and 3 mol of O2

d) 3 mol of NH3 and 2 mol of O2

So, for (a), we are comparing 2/4 mol of NH3 with 2/5 mol of O2, and therefore, O2 is the LR.

For (b), we are comparing 2/4 mol of NH3 with 3/5 mol of O2, and therefore, NH3 is the LR.

For (c), we are comparing 3/4 mol of NH3 with 3/5 mol of O2, and therefore, O2 is the LR.

For (d), 3 mol of NH3 and 3/4 mol of NH3 with 2/5 mol of O2, and therefore, O2 is the LR.

15.

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?

N2(g) + 3 H2(g) → 2 NH3(g)

answer

5.0 g

Solution

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:

So, there are:

4.0 – 1.50 = 2.5 mol H2 left out

\[{\rm{m}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.5}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{2}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.0}}\;{\rm{g}}\]

16.

How many grams of PCl3 will be produced if 38.5 g Cl2 is reacted with 56.4 g P4 according to the following equation?

6Cl2(g) + P4(s) → 4PCl3(l)

answer

169 g

Solution

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 Cl2 and P4 are:

\[{\rm{n}}\;{\rm{(C}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{38}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{70}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.84}}\;{\rm{mol}}\]

\[{\rm{n}}\;{\rm{(}}{{\rm{P}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{56}}{\rm{.4}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{123}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.455}}\;{\rm{mol}}\]

Next, we calculate how much PCl3 can be formed from each reactant:

\[{\rm{n}}\;\left( {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.84 }}\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.23}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.455 }}\cancel{{{\rm{mol}}\;{{\rm{P}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{P}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.82}}\;{\rm{mol}}\]

Cl2 gives less PCl3 and therefore, it is the LR and 1.23 mol of PCl3 can be produced in this reaction.

The mass of PCl3 is:

\[{\rm{m}}\;{\rm{(PC}}{{\rm{l}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.23}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{137}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{169}}\;{\rm{g}}\]

17.

How many grams of sulfur can be obtained if 12.6 g H2S is reacted with 14.6 g SO2 according to the following equation?

2H2S(g) + SO2(g) → 3S(s) + 2H2O(g)

answer

17.8 g

Solution

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 H2S and SO2 are:

\[{\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S)}}\;{\rm{ = }}\;{\rm{12}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{34}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.370}}\;{\rm{mol}}\]

(Video) Solution Stoichiometry - Finding Molarity, Mass & Volume

\[{\rm{n}}\;{\rm{(S}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{14}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{64}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.228}}\;{\rm{mol}}\]

Next, we calculate how much PCl3 can be formed from each reactant:

\[{\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.370 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.228 }}\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.684}}\;{\rm{mol}}\]

H2S 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:

\[{\rm{m}}\;{\rm{(S)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{17}}{\rm{.8}}\;{\rm{g}}\]

18.

The following equation represents the combustion of octane, C8H18, a component of gasoline:

2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)

Will 356 g of oxygen be enough for the complete combustion of 954 g of octane?

answer

356 g of oxygen is not enough for the complete combustion of 954 g of octane.

Solution

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 O2 are needed to react with 8.35 mol of C8H18:

\[{\rm{n}}\;\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{8}}{\rm{.35 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{25}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}}}}}\;{\rm{ = }}\;{\rm{104}}\;{\rm{mol}}\]

104 moles of O2 are needed to react with only 8.35 mol of C8H18 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.

19.

When 140.0 g of AgNO3 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 AgNO3 was added?

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

answer

NaCl was the limiting reactant and 35.0 g of it were present in the solution when AgNO3was added.

Solution

We need to find the moles of AgCl which will allow us to calculate how much AgNO3 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 AgNO3 and 0.600 mol of NaCl participating in the reaction.

Now, let’s calculate the moles of 140.0 g of AgNO3and see whether it corresponds to 0.600 mol. If it is more than that, then AgNO3 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 AgNO3 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 AgNO3.

And this confirms that AgNO3 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 AgNO3 was added:

\[{\rm{m}}\;{\rm{(NaCl)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{58}}{\rm{.4}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{35}}{\rm{.0}}\;{\rm{g}}\]

20.

Consider the reaction between MnO2 and HCl:

MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O

What is the theoretical yield of MnCl2 in grams when 165 g of MnO2 is added to a solution containing 94.2 g of HCl?

answer

0.645 mol or 81.1 g MnCl2

Solution

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 MnO2 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 MnCl2 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 MnCl2 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 MnCl2:

\[{\rm{m}}\;{\rm{(MnC}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.645}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{125}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{81}}{\rm{.1}}\;{\rm{g}}\]

21.

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?

answer

79.8 %

Solution

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.

\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{5}}{\rm{.68}}\;{\rm{g}}}}{{{\rm{7}}{\rm{.12}}\;{\rm{g}}}}\; \times \;100\% \; = \;79.8\;\% \]

22.

When 38.45 g CCl4 is reacted with an excess of HF, 21.3 g CCl2F2 is obtained. Calculate the theoretical and percent yields of this reaction.

CCl4 + 2HF → CCl2F2 + 2HCl

answer

70.5 %

Solution

To find the theoretical yield, we need to calculate the moles of CCl4 and, using the mole ratio, determine how much CCl2F2 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 CCl2F2 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 CCl2F2 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:

\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{21}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{30}}{\rm{.2}}\;{\rm{g}}}}\; \times \;100\% \; = \;70.5\;\% \]

23.

Iron(III) oxide reacts with carbon monoxide according to the equation:

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

What is the percent yield of this reaction if 623 g of iron oxide produces 341 g of iron?

answer

78.4%

Solution

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 Fe2O3) converts into a product according to the chemical equation.

So, we are going to calculate the moles of 623 g Fe2O3 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 Fethat 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:

\[{\rm{m}}\;{\rm{(Fe)}}\;{\rm{ = }}\;{\rm{7}}{\rm{.80}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{435}}\;{\rm{g}}\]

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{341}}\;{\rm{g}}}}{{{\rm{435}}\;{\rm{g}}}}\; \times \;100\% \; = \;78.4\;\% \]

24.

Determine the percent yield of the reaction if 77.0 g of CO2are formed from burning 2.00 moles of C5H12in 4.00 moles of O2.

C5H12 + 8 O2 → 5CO2+ 6H2O

answer

70.0 %

Solution

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 CO2.

Now, we are given the moles of the reactants, so we calculate how much CO2 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 CO2 could be formed in this reaction. This is the theoretical yield which is grams is:

\[{\rm{m}}\;{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.50}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{44}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{110}}{\rm{.}}\;{\rm{g}}\]

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{77}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{110}}{\rm{.}}\;{\rm{g}}}}\; \times \;100\% \; = \;70.0\;\% \]

25.

The percent yield for the following reaction was determined to be 84%:

N2(g) + 2H2(g) → N2H4(l)

How many grams of hydrazine (N2H4) can be produced when 38.36 g of nitrogen reacts with 6.68 g of hydrogen?

answer

(Video) Stoichiometry Tutorial: Step by Step Video + review problems explained | Crash Chemistry Academy

36.8 g

Solution

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 N2H4 can each produce:

\[{\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.37 }}\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.37}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{3}}{\rm{.34 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.67}}\;{\rm{mol}}\]

N2 gives less hydrazine so, it is the LR, and therefore the theoretical yield of N2H4 is 1.37 mol. The mass of N2H4 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 N2 by 0.84 and determine the mass of N2H4 based on that.

We would have 1.37 mole x 0.84 = 1.1508 mol N2 which is the moles of nitrogen that actually convert into hydrazine. By the mole ratio, the moles of N2H4 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:

\[{\rm{m}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}_{\rm{2}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.1508}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.8}}\;{\rm{g}}\]

26.

Silver metal can be prepared by reducing its nitrate, AgNO3with copper according to the following equation:

Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 132.5 grams ofAgNO3?

answer

85%

Solution

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 AgNO3) converts into product according to the chemical equation.

So, we are going to calculate the moles of 132.5 gAgNO3 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:

\[{\rm{m}}\;{\rm{(Ag)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.780}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{107}}{\rm{.9}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{84}}{\rm{.2}}\;{\rm{g}}\]

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{71}}{\rm{.5}}\;{\rm{g}}}}{{{\rm{84}}{\rm{.2}}\;{\rm{g}}}}\; \times \;100\% \; = \;85\;\% \]

27.

Industrially, nitric acid is produced from ammonia by the Ostwald process in a series of reactions:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

2NO(g) + O2(g) → 2NO2(g)

2NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq)

Considering that each reaction has an 85% percent yield, how many grams of NH3 must be used to produce 25.0 kg of HNO3 by the above procedure?

answer

2.20 x 104 g

Solution

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 HNO3 and using the mole ratio, find the moles of NO2 which is the link between the 3rd and 2nd reactions.

Before using the mass of HNO3, 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 NO2 that produced 397 mol of HNO3 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 NO2 considering that the yield of the reaction is again 85%.

2NO(g) + O2(g) → 2NO2(g)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{934 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{O}}_{\rm{2}}}}}}}\; \times \;\frac{{100}}{{85}}\;{\rm{ = }}\;{\rm{1098}}\;{\rm{mol}}\]

Next, we do the same for the first reaction and determine how much NH3 was needed to produce 1098 mol NO considering that the yield of the reaction is again 85%.

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

\[{\rm{n}}\;\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{1098 }}\cancel{{{\rm{mol}}\;{\rm{NO}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{NO}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{100}}}}{{{\rm{85}}}}\;{\rm{ = }}\;{\rm{1293}}\;{\rm{mol}}\]

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:

2.20 x 104 g

28.

Aspirin (acetyl salicylic 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:

Stoichiometry Practice Problems - Chemistry Steps (11)

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)Whatmass of aspirin isformed assuming 100% yield (Theoretical yield)?
d)Whatmass of aspirin isformed 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 manykgof salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?

a)

answer

Salicylic acid is the limiting reagent

b)

answer

2.608 x 103g Acetic anhydride

c)

answer

1.305 x 104g Aspirin

d)

answer

9.14 x 103g Aspirin

e)

answer

85.8 %

f)

answer

18.1 kg

Solution

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?

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{11}}{\rm{.2}}\;{\rm{kg}}}}{{{\rm{13}}{\rm{.05}}\;{\rm{kg}}}}\;{\rm{ \times }}\;{\rm{100\% }}\;{\rm{ = }}\;{\rm{85}}{\rm{.8}}\;{\rm{\% }}\]

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 104 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 1.81 kg

(Video) Mole Ratio Practice Problems

FAQs

What are the 5 steps to solving a stoichiometry problem? ›

Stoichiometric Calculations
  1. Balance the equation.
  2. Convert units of a given substance to moles.
  3. Using the mole ratio, calculate the moles of substance yielded by the reaction.
  4. Convert moles of wanted substance to desired units.

What are the 3 steps to solve stoichiometry problems? ›

We can tackle this stoichiometry problem using the following steps:
  • Step 1: Convert known reactant mass to moles. ...
  • Step 2: Use the mole ratio to find moles of other reactant. ...
  • Step 3: Convert moles of other reactant to mass.

Why is learning stoichiometry difficult? ›

Stoichiometry can be difficult because it builds upon a number of individual skills. To be successful you must master the skills and learn how to plan your problem solving strategy. Master each of these skills before moving on: Calculating Molar Mass.

What is the easiest way to learn stoichiometry? ›

So put those numbers out front like you see here here's how it works. You simply solve for X X is

How do you solve stoichiometry step by step? ›

Step by Step Stoichiometry Practice Problems | How to Pass Chemistry

What is stoichiometry formula? ›

The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained. The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation.

How do you solve stoichiometry word problems? ›

How to Solve Stoichiometry Word Problems - YouTube

How do you convert moles to moles in stoichiometry? ›

Chemical Reactions (10 of 11) Stoichiometry: Moles to Moles

How do you find moles using stoichiometry? ›

Stoichiometry Basic Introduction, Mole to Mole, Grams to ... - YouTube

Is stoichiometry the hardest part of chemistry? ›

Stoichiometry is arguably one of the most difficult concepts for students to grasp in a general chemistry class. Stoichiometry requires students to synthesize their knowledge of moles, balanced equations and proportional reasoning to describe a process that is too small to see.

What are the difficulties in stoichiometry? ›

From the findings, six major difficulties identified were lack of understanding of the mole concept, inability to balance chemical equations, use of inconsistent stoichiometric relationships, identifying the limiting reagent, determination of theoretical yields and identification of substances in excess.

What is stoichiometry for dummies? ›

Stoichiometry Tutorial: Step by Step Video + review problems explained

What are the four types of stoichiometry problems? ›

It is important to remember, though, that in every situation you need to start out with a balanced equation.
  • Mole-Mole Problems. ...
  • Mass-Mass Problems (Strategy: Mass g Mole g Mole g Mass) ...
  • Mass-Volume Problems (Strategy: Mass g Mole g Mole g Volume) ...
  • Volume-Volume Problems.
30 Jan 2004

What is a real life example of stoichiometry? ›

Stoichiometry is at the heart of the production of many things you use in your daily life. Soap, tires, fertilizer, gasoline, deodorant, and chocolate bars are just a few commodities you use that are chemically engineered, or produced through chemical reactions.

What is the first step in most stoichiometry problems? ›

the first step in any stoichiometric problem is to always ensure that the chemical reaction you are dealing with is balanced, clarity of the concept of a 'mole' and the relationship between 'amount (grams)' and 'moles'.

What is stoichiometry in chemistry class 11? ›

Stoichiometry is defined as the exact numbers which indicate the actual proportions of the reactant and product. The relative amount of the reactants are important for calculating the exact amount of individual starting material required for the reaction.

How do you calculate stoichiometry with examples? ›

Stoichiometric Formulas based on Chemical Reaction. Formula mass is defined as the sum of the atomic weights of the atoms in the given molecule of the substance. For example, the formula mass of Na₂S is calculated as 2(23) + 1(32) = 78. Avogadro's number is the total number of particles in one mole of a substance.

How do you convert to moles? ›

How to find moles?
  1. Measure the weight of your substance.
  2. Use a periodic table to find its atomic or molecular mass.
  3. Divide the weight by the atomic or molecular mass.
  4. Check your results with Omni Calculator.
8 Sept 2021

What is the mole ratio? ›

Mole Ratio: is a conversion factor between compounds in a chemical reaction, that is derived from the coefficients of the compounds in a balanced equation. The mole ratio is therefore used to convert between quantities of compounds in a chemical reaction.

How do you solve equations in chemistry? ›

Concentration Formula & Calculations | Chemistry | Fuse School - YouTube

How do you convert g mol to Grams? ›

Multiply the given number of moles (2.50 mol) by the molar mass (122.548 g/mol) to get the grams.

How do you find the mole ratio? ›

Determining the Mole Ratio - YouTube

How do I calculate grams? ›

1.2 Calculating mass in grams from amount in mol - YouTube

Why is it called the mole ratio? ›

"In stoichiometry, we shift our unit from molecule to mole. According to this equation, we need two moles of hydrogen to react with one mole of oxygen. This is called the mole ratio. It is defined as the ratio of moles of one substance to the moles of another substance in a balanced equation.

How do you find mass to mass in stoichiometry? ›

Mass-Mass Stoichiometry - YouTube

How do you convert mole ratio to mass ratio? ›

Converting mol ratio to mass ratio - YouTube

How do you solve stoichiometry word problems? ›

How to Solve Stoichiometry Word Problems - YouTube

What is stoichiometry example? ›

Example – Using Stoichiometric Ratio (Moles)

By looking at the coefficients, you can see that for every 1 mole of C6H12O6, 2 moles of CO2 are produced. Using this ratio, you can figure out how many moles of carbon dioxide are made from 2.5 moles of glucose.

What are the laws of stoichiometry? ›

It states that “Mass is neither created nor destroyed in a chemical reaction or The total mass of reactant is equal to the total mass of product in a chemical reaction”.

How do you balance stoichiometry problems? ›

How to Balance Equations in Stoichiometry Problems : Math-Tastic

How do you convert moles to moles in stoichiometry? ›

Chemical Reactions (10 of 11) Stoichiometry: Moles to Moles

How do you find moles using stoichiometry? ›

Stoichiometry Basic Introduction, Mole to Mole, Grams to ... - YouTube

How do you solve a mole problem? ›

How to solve mole problems - YouTube

How do you convert mol to mol? ›

How to Use a Mole to Mole Ratio | How to Pass Chemistry - YouTube

What is the first step in solving stoichiometric problems *? ›

The first and critical step in any stoichiometric calculation is to have a balanced chemical equation.

How do you convert units to moles? ›

To convert formula units to moles you divide the quantity of formula units by Avogadro's number, which is 6.022 x 1023 formula units/mole.

Who invented stoichiometry? ›

Stoichiometry was first discovered by Jeremias Richter, a German chemist. It was Richter who coined the term stoichiometry, a tongue-twisting word that baffles students to this day. Stoichiometry was derived from stoikheion, Greek for "element", and "metron", meaning measure.

What is a mole ratio? ›

Mole Ratio: is a conversion factor between compounds in a chemical reaction, that is derived from the coefficients of the compounds in a balanced equation. The mole ratio is therefore used to convert between quantities of compounds in a chemical reaction.

How is stoichiometry used in real life? ›

Stoichiometry is at the heart of the production of many things you use in your daily life. Soap, tires, fertilizer, gasoline, deodorant, and chocolate bars are just a few commodities you use that are chemically engineered, or produced through chemical reactions.

Videos

1. Stoichiometry - Limiting & Excess Reactant, Theoretical & Percent Yield - Chemistry
(The Organic Chemistry Tutor)
2. Stoichiometry practice problems - Real Chemistry
(Real Chemistry)
3. Stoichiometry Mole to Mole Conversions - Molar Ratio Practice Problems
(The Organic Chemistry Tutor)
4. STOICHIOMETRY PRACTICE- Review & Stoichiometry Extra Help Problems
(sciencepost)
5. Stoichiometry: Percent Yield, Practice Problem 1
(Chemistry Steps)
6. Stoichiometry Equations + Practice Problems & Answers | College Chemistry
(Wize)

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