KEY2CHEM

Stoichiometry

Stoichiometry is a useful tool that allows the conversion between masses (such as can be weighed out in a laboratory) and number of atoms or molecules (which are very small and cannot be weighed easily in a laboratory). The coefficients in the balanced chemical equation, often referred to as the mole ratio, provide an essential conversion factor between moles of different substances in a chemical equation.

Example 1.

If $$3$$ moles of $$\require{mhchem}\ce{H2}$$ react with excess $$\require{mhchem}\ce{N2}$$ by the following reaction, how many moles of ammonia ($$\require{mhchem}\ce{NH3}$$) can be formed by the following balanced equation?

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

A. 2 moles

B. 3 moles

C. 4 moles

Solution

A. 2 moles

$$3\text{ mol H}_2 \times \frac{2 \text{ mol NH}_3}{3\text{ mol H}_2} = 2\text{ mol NH}_3$$

Example 2.

In the following reaction, $$2.0$$ moles of $$\require{mhchem}\ce{Al}$$ reacts with excess $$\require{mhchem}\ce{O2}$$ by the following balanced chemical equation. If the reaction yields $$0.65$$ moles of $$\require{mhchem}\ce{Al2O3}$$, what is the percent yield of the reaction?

$$\require{mhchem}\ce{4 Al(s) + 3 O2(g)-> 2Al2O3(s) }$$

A. $$65\%$$

B. $$33\%$$

C. $$16\%$$

Solution

A. $$65\%$$

Theoretical yield: $$2.0\text{ mol Al} \times \frac{2\text{ mol Al}_2\text{O}_3}{4 \text{ mol Al}} = 1.0\text{ mol Al}_2\text{O}_3$$

Percent yield: $$\frac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 = \frac{0.65\text{ mol}}{1.0\text{ mol}}\times 100 = 65\%$$

Example 3.

If $$2.50\text{ mol}$$ of product $$\require{mhchem}\ce{AB3}$$ are desired from the following reaction, how many moles of $$\require{mhchem}\ce{A2}$$ are required? Assume there is excess $$\require{mhchem}\ce{B2}$$ present.

$$\require{mhchem}\ce{A2(g) + 3 B2(g)->2 AB3(g)}$$

A. $$2.50\text{ mol}$$

B. $$1.25\text{ mol}$$

C. $$5.00\text{ mol}$$

B. $$1.25\text{ mol}$$

$$2.50\text{ mol AB}_3 \times \frac{1 \text{ mol A}_2}{2\text{ mol AB}_3} = 1.25\text{ mol A}_2$$