Chapter 3 — Stoichiometry

Section 3.4 — Mass Relationships and Limiting Reagents

Problems

1

Calculate the molar mass of ammonium sulfate, $\ce{(NH4)2SO4}$. Report your answer to four significant figures.

2

How many moles are present in $\pu{45.0 g}$ of glucose, $\ce{C6H12O6}$?

3

Balance the following equation:

$\ce{C3H8 + O2 -> CO2 + H2O}$

4

How many molecules of $\ce{CO2}$ are produced when $\pu{2.50 mol}$ of propane burns completely?

5

Iron(III) oxide reacts with carbon monoxide:

$\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$

What mass of iron is produced from $\pu{125 g}$ of $\ce{Fe2O3}$?

6

$\pu{10.0 g}$ of $\ce{H2}$ reacts with $\pu{64.0 g}$ of $\ce{O2}$. Which is the limiting reagent?

7

$\pu{25.0 g}$ of $\ce{N2}$ reacts with $\pu{5.00 g}$ of $\ce{H2}$ to form ammonia:

$\ce{N2 + 3H2 -> 2NH3}$

Determine the limiting reagent and the theoretical yield of $\ce{NH3}$ in grams.

8

A reaction with a theoretical yield of $\pu{18.6 g}$ produces $\pu{15.2 g}$ of product. Calculate the percent yield.

9

A sample of $\ce{CaCO3}$ has a mass of $\pu{0.4820 g}$. How many formula units does it contain? Report to the correct number of significant figures.

10

$\pu{15.0 g}$ of $\ce{Al}$ reacts with $\pu{75.0 g}$ of $\ce{Fe2O3}$ in a thermite reaction. If $\pu{22.1 g}$ of iron is recovered, what is the percent yield?

Solutions

1

Answer: $\pu{132.1 g/mol}$

Solution:

$2(\pu{14.01}) + 8(\pu{1.008}) + \pu{32.07} + 4(\pu{16.00}) = \pu{132.14 g/mol}$

To four significant figures, $\pu{132.1 g/mol}$.

3

Answer: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$

Solution:

Balance carbon first ($3\ce{CO2}$), then hydrogen ($4\ce{H2O}$), which fixes oxygen at $6 + 4 = 10$ atoms, so $5\ce{O2}$.

5

Answer: $\pu{87.4 g}$ of $\ce{Fe}$

Solution:

$n(\ce{Fe2O3}) = \frac{\pu{125 g}}{\pu{159.69 g/mol}} = \pu{0.7828 mol}$

The equation gives $\pu{2 mol}$ $\ce{Fe}$ per $\pu{1 mol}$ $\ce{Fe2O3}$:

$m(\ce{Fe}) = 2(\pu{0.7828 mol})(\pu{55.85 g/mol}) = \pu{87.4 g}$

7

Answer: $\ce{H2}$ is limiting; theoretical yield $= \pu{28.1 g}$ of $\ce{NH3}$

Solution:

$n(\ce{N2}) = \pu{25.0 g} / \pu{28.02 g/mol} = \pu{0.892 mol}$

$n(\ce{H2}) = \pu{5.00 g} / \pu{2.016 g/mol} = \pu{2.48 mol}$

$\ce{N2}$ would need $3(\pu{0.892}) = \pu{2.68 mol}$ of $\ce{H2}$, but only $\pu{2.48 mol}$ is available, so $\ce{H2}$ limits.

$m(\ce{NH3}) = \frac{2}{3}(\pu{2.48 mol})(\pu{17.03 g/mol}) = \pu{28.1 g}$

9

Answer: $\pu{2.900e21}$ formula units

Solution:

$n = \frac{\pu{0.4820 g}}{\pu{100.09 g/mol}} = \pu{4.816e-3 mol}$

$N = (\pu{4.816e-3 mol})(\pu{6.022e23 mol-1}) = \pu{2.900e21}$

Four significant figures, matching the mass given.