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. \(1.25\text{ mol}\)

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