Solutions/Answers to Exercises: Stoichiometry, Limiting Reactants, and Percent Yield

Solutions to Exercises for Stoichiometry, Mass and Percent Yield

Check your chemical equations to make sure they are all correctly balanced.

Exercise 1. Consider the following unbalanced equation:

Fe2O3 (s) + C → Fe (s) + CO2 (g)

a) Balance the chemical equation.

2 Fe2O3 (s) + 3 C → 4 Fe (s) + 3 CO2 (g)

 

b) How many moles of carbon will react with 35.5 moles of Fe2O3?

\(\displaystyle 35.5\;mol\;Fe_2O_3\times\frac{3\;mol\;C}{2\;mol\;Fe_2O_3}\;=\;\mathbf{53.3\;mol\;C}\)

c) How many grams of carbon will react with 35.5 moles of Fe2O3?


\(\displaystyle 35.5\;mol\;Fe_2O_3\times\frac{3\;mol\;C}{2\;mol\;Fe_2O_3}\times\frac{12.011\;g\;C}{1\;mol\;C}\;=\;\mathbf{640.\;g\;C}\)

 

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Exercise 2. Consider the following unbalanced chemical equation. How many grams of hydrogen are required to react with 243 g of nitrogen?

N2 (g) + H2 (g) → NH3 (g)

First, balance the chemical equation

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Second, write your roadmap

g N2 → mol N2 → mol H2 → g H2

There are three arrows in the roadmap–we need three equivalances

28.0134 g N2 = 1 mole N2. The molar mass of N2 will covert g of N2 to moles of N2
Mole ratio from the balanced chemical equation: 1 mol N2 = 3 mol H2
1 mol H2 = 2.0159 g H2. This will convert moles of H2 to grams of H2

\(\displaystyle 243\;g\;N_2\;\times\;\frac{1\;mol\;N_2}{28.0134\;g\;N_2}\times\frac{3\;mol\;H_2}{1\;mol\;N_2}\times\frac{2.0159\;g\;H_2}{1\;mol\;H_2}\;=\;\mathbf{52.5\;g\;H_2}\)

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Exercise 3. Solid nickel reacts with aqueous hydrochloric acid to produce aqueous nickel(II) chloride and hydrogen gas.

a) Write a balanced chemical equation.

Ni (s) + 2 HCl (aq) → NiCl2 (aq) + H2 (g)

b) How many grams of NiCl2 are produced if 86.6 grams of hydrochloric acid is reacted?

Mm NiCl2 = 129.5994 g/mol
Mm HCl = 36.458 g/mol
Roadmap:   g NiCl2 → mol NiCl2 → mol HCl → g HCl

\(\displaystyle 86.6\;g\;HCl\;\times\frac{1\;mol\;HCl}{36.457\;g\;HCl}\times\frac{1\;mol\;NiCl_2}{2\;mol\;HCl}\times\frac{129.5994\;g\;NiCl_2}{1\;mol\;NiCl_2}\;=\;\mathbf{154\;g\;NiCl_2}\)

c) How many grams of solid nickel is required to react with 86.6 grams of hydrochloric acid?

Mm Ni (s) = 58.6934 g/mol
Roadmap: g HCl → mol HCl → mol Ni → g Ni

\(\displaystyle 86.6\;g\;HCl\;\times\frac{1\;mol\;HCl}{36.457\;g\;HCl}\times\frac{1\;mol\;Ni}{2\;mol\;HCl}\times\frac{58.6934\;g\;Ni}{1\;mol\;Ni}\;=\;\mathbf{69.7\;g\;Ni}\)


 

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Exercise 4. Oxygen gas can be prepared by heating mercury(II) oxide according to the following equation:

\(\displaystyle HgO\;(s)\;\rightarrow\;\;Hg\;(l)\;+\;O_2\;(g)\)

How many grams of oxygen was formed if 535 g of HgO was heated?

First balance the chemical equation

\(\displaystyle 2 HgO\;(s)\;\rightarrow\;\;2 Hg\;(l)\;+\;O_2\;(g)\)

Mm HgO2 = 216.59 g/mol
Mm O2 = 31.999 g/mol
Roadmap:     g HgO → mol HgO → mole O2 → g O2

 

\(\displaystyle 535\;g\;HgO\times\frac{1\;mol\;HgO}{216.59\;g\;HgO}\times\frac{1\;mol\;O_2}{2\;mol\;HgO}\times\frac{31.999\;g\;O_2}{1\;mol\;O_2}\;=\;\mathbf{39.5\;g\;O_2}\)

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Exercise 5. Consider the reaction of aluminum hydroxide with sulfuric acid.

Al(OH)3 (s) + H2SO4 (aq) → Al2(SO4)3 (aq) + H2O (l) (unbalanced)

If 36.52 g of aluminum hydroxide and 40.65 g of sulfuric acid are allowed to react, how many grams of aluminum sulfate will form?

As always, balance the equation.

2 Al(OH)3 (s) + 3 H2SO4 (aq) → Al2(SO4)3 (aq) + 6 H2O (l)

Here, we need to determine which reactant is limiting. Calculate the molar masses of the substances of interest. Then calculate the number of moles of Mm Al(OH)3 formed for each reactant. From that information, we can determine the limiting reactant.

Mm Al(OH)3 = 78.0036 g/mol
Mm H2SO4 = 98.079 g/mol
Mm Al2(SO4)3 = 342.15 g/mol

\(\displaystyle 36.52\;g\;Al(OH)_3\times\frac{1\;mol\;Al(OH)_3}{78.0036\;g\;Al(OH)_3}\times\frac{1\;mol\;Al_2(SO_4)_3}{2\;mol\;Al(OH)_3}\;=\;\mathbf{0.234\;mol\;Al_2(SO_4)_3}\)

\(\displaystyle 44.65\;g\;H_2SO_4\times\frac{1\;mol\;H_2SO_4}{98.079\;g\;H_2SO_4}\times\frac{1\;mol\;Al_2(SO_4)_3}{3\;mol\;H_2SO_4}\;=\;\mathbf{0.152\;mol\;Al_2(SO_4)_3}\)

The limiting reactant is H2SO4, therefore, 52.0 g of Al2(SO4)3 will be produced.

\(\displaystyle 0.152\;mol\;Al_2(SO_4)_3\times\frac{342.15\;g\;Al_2(SO_4)_3}{1\;mol\;Al_2(SO_4)_3}\;=\;\mathbf{52.0\;g\;Al_2(SO_4)_3}\)

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Exercise 6. Benzene, C6H6 reacts with bromine, in the presence of iron, to give bromobenzene, C6H5Br.

C6H6 (l) + Br2 (l) → C6H5Br (l) + HBr (g)

What is the percent yield if 126 g of C6H5Br is recovered when 165 g of C6H6 is reacted with 165 g of Br2?

First, check the chemical equation is balanced. It is already balanced. We need to determine the molar masses of the substances of interest and the limiting reactant.

Mm C6H6 = 78.11 g/mol
Mm Br2 = 159.808 g/mol
Mm C6H5Br = 157.02 g/mol

\(\displaystyle 165\;g\;C_6H_6\times\frac{1\;mol\;C_6H_6}{78.11\;g\;C_6H_6}\times\frac{1\;mol\;C_6H_5Br}{1\;mol\;C_6H_6}\;=\;\mathbf{2.11\;mol\;C_6H_5Br}\)

\(\displaystyle 165\;g\;Br_2\times\frac{1\;mol\;Br_2}{159.808\;g\;Br_2}\times\frac{1\;mol\;C_6H_5Br}{1\;mol\;Br_2}\;=\;\mathbf{1.03\;mol\;C_6H_5Br}\)

The limiting reactant is Br2. Based on this, calculate the grams of C6H5Br formed.

center>\(\displaystyle 1.03\;mol\;C_6H_5Br\times\frac{157.02\;g\;C_6H_5Br}{1\;mol\;C_6H_5Br}\;=\;\mathbf{162\;g\;C_6H_5Br}\)

Now, calculate the percent yield. The actual yield is 126 g of C6H5Br and theoretical yield is 162 g.

\(\displaystyle Percent\;{yield}\;=\;\frac{126\;g}{162\;g}\;\times\;{100}\;=\;\mathbf{77.8\;\%}\)

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Exercise 7. Gaseous hydrogen sulfide, H2S is bubbled into a solution of sodium hydroxide to produce aqueous sodium sulfide and liquid water. Assuming the percent yield of sodium sulfide is 94.0%, how many grams of sodium sulfide are formed if 2.5 g of H2S is bubbled into a solution that contains 4.00 g of NaOH? Make sure you write a balanced chemical equation.

Write a balanced chemical equation:

H2S (g) + 2 NaOH (aq) → Na2S (aq) + 2 H2O (l)

Mm H2S = 34.1 g/mol
Mm NaOH = 39.997 g/mol
Mm Na2S = 78.0452 g/mol

Here, we need to find the actual yield. First, determine the limiting reactant and then calculate the theoretical yield.

\(\displaystyle 2.5\;g\;H_2S\;\times\;\frac{1\;mol\;H_2S}{34.1\;g\;H_2S}\times\frac{1\;mol\;Na_2S}{1\;mol\;H_2S}\;=\;\mathbf{0.0733\;mol\;Na_2S}\)

\(\displaystyle 4.00\;g\;NaOH\;\times\;\frac{1\;mol\;NaOH}{39.997\;g\;NaOH}\times\frac{1\;mol\;Na_2S}{2\;mol\;NaOH}\;=\;\mathbf{0.0500\;mol\;Na_2S}\)

NaOH is our limiting reagent. The theoretical yield of Na2S is 3.90 g.

\(\displaystyle 0.0500\;mol\;Na_2S\times\frac{78.0452\;g\;Na_2S}{1\;mol\;Na_2S}\;=\;\mathbf{3.90\;g\;N_2S}\)

The percent yield is given by:

\(\displaystyle Percent\;{yield}\;=\;\frac{actual\;yield}{theoretical\;yield}\;\times\;{100}\)

Solving for the actual yield:

\(\displaystyle actual\;yield\;=\;\frac{94.0\;\%}{100}\;\times\;{3.90\;g}\;=\;\mathbf{3.67\;g\;Na_2S}\)

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