KEY2CHEM

Stoichiometry Calculations

Stoichiometry relates measured laboratory quantities (such as mass) to microscopic quantities (such as number of atoms/molecules) through the use of amount (moles) and coefficients in balanced chemical equations. These relationships can be extended to other laboratory quantities, such as pressures or volumes or gases and volumes of solutions. Additionally, the limiting reactant of a reaction (one reactant that runs out first and limits the quantity of product that can form) can be determined by stoichiometric calculation.


Example 1.

If \(3.0\text{ mol}\) \(\require{mhchem}\ce{H2}\) react with \(2.0\text { mol}\) \(\require{mhchem}\ce{N2}\) by the following reaction, how many moles of ammonia (\(\require{mhchem}\ce{NH3}\)) can be formed? 

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

 

A. \(2.0\text{ mol}\)

B. \(3.0\text{ mol}\)

C. \(4.0\text{ mol}\)

 

 

 

 

 

 

 

Solution 

A. \(2.0\text{ mol}\)

The limiting reactant is the reactant that runs out first and therefore limits the amount of product that can form. It is not necessarily the reactant present in the smaller starting quantity. Determine the yield of product from each given quantity of reactant.

\(3.0\text{ mol H}_2 \times \frac{2 \text{ mol NH}_3}{3\text{ mol H}_2} = 2.0\text{ mol NH}_3\)

\(2.0\text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1\text{ mol N}_2} = 4.0\text{ mol NH}_3\)


Example 2.

How many grams of \(\require{mhchem}\ce{Fe2O3}\) are formed when \(16.9\text{ g Fe}\) reacts with excess \(\require{mhchem}\ce{O2}\) by the following balanced chemical equation?

        \(\require{mhchem}\ce{4 Fe(s) + 3 O2(g)-> 2Fe2O3(s) }\)

 

 

A. \(24.2\text{ g}\)

B. \(48.4\text{ g}\)   

C. \(12.1\text{ g}\)

 

 

 

Solution

 

A. \(24.2\text{ g}\)

\(16.9\text{ g Fe} \times \frac{1\text{ mol Fe}}{55.85\text{ Fe}} \times \frac{2\text{ mol Fe}_2\text{O}_3}{4 \text{ mol Fe}} \times \frac{159.7\text{ g Fe}_2\text{O}_3}{1\text{ mol Fe}_2\text{O}_3} = 24.2\text{ g Fe}_2\text{O}_3 \)


Example 3.

How many liters of \(\require{mhchem}\ce{N2(g)}\) can form at  \(1.0\text{ atm}\) and \(298\text{ K}\) when \(44.3\text{ g }\require{mhchem}\ce{NaN3}\) react by the following balanced equation?

             \(\require{mhchem}\ce{2 NaN3(s) -> 2 Na(s) + 3 N2(g)}\)

 

A. \(2527\text{ L}\)

B. \(24.9\text{ L}\)

C. \(1.02\text{ L}\)

 

 

 

Solution

B. \(24.9\text{ L}\)

 \(44.3\text{ g NaN}_3 \times \frac{1\text{ mol NaN}_3}{65.02\text{ NaN}_3} \times \frac{3\text{ mol N}_2}{2 \text{ mol NaN}_3} = 1.02\text{ mol N}_2\)

\(\begin{align}PV &= nRT \\ V &= \frac{nRT}{P} = \frac{(1.02\text{ mol})(0.08206\frac{L\cdot atm}{mol \cdot K})(298\text{ K})}{1.0\text{ atm}} = 24.9\text{ L}\end{align}\)