Volume Relationships in Gaseous Reactions

Ammonium nitrate is used as an explosive in mining. The following reaction occurs when ammonium nitrate is heated:

\[2\text{NH}_{4}\text{NO}_{3}\text{(s)} → 2\text{N}_{2}\text{(g)} + 4\text{H}_{2}\text{O (g)} + \text{O}_{2} \text{(g)}\]

If \(\text{750}\) \(\text{g}\) of ammonium nitrate is used, what volume of oxygen gas would we expect to produce (at STP)?

Step 1: Work out the number of moles of ammonium nitrate

The number of moles of ammonium nitrate used is: \begin{align*} n &= \frac{m}{M}\\ & = \frac{750}{80}\\ & = \text{9.375}\text{ mol} \end{align*}

Step 2: Work out the amount of oxygen

The mole ratio of \(\text{NH}_{4}\text{NO}_{3}\) to \(\text{O}_{2}\) is \(2:1\). So the number of moles of \(\text{O}_{2}\) is: \begin{align*} n_{\text{O}_{2}} & = n_{\text{NH}_{4}\text{NO}_{3}} \times \frac{\text{stoichiometric coefficient O}_{2}}{\text{stoichiometric coefficient NH}_{4}\text{NO}_{3}}\\ & = \text{9.375}\text{ mol} \text{NH}_{4}\text{NO}_{3} \times \frac{\text{1}\text{ mol} \text{O}_{2}}{\text{2}\text{ mol} \text{ NH}_{4}\text{NO}_{3}}\\ & = \text{4.6875}\text{ mol} \end{align*}

Step 3: Work out the volume of oxygen

Recall from earlier in the section that we said that one mole of any gas occupies \(\text{22.4}\) \(\text{dm$^{3}$}\) at STP.

\begin{align*} V & = (\text{22.4})n\\ & = (\text{22.4})(\text{4.6875})\\ & = \text{105}\text{ dm$^{3}$} \end{align*}

Sodium azide is sometimes used in airbags. When triggered, it has the following reaction:

\[2\text{NaN}_{3}\text{(s)} → 2\text{Na(s)} + 3\text{N}_{2}\text{(g)}\]

If \(\text{55}\) grams of sodium azide is used, what volume of nitrogen gas would we expect to produce?

Step 1: Work out the number of moles of sodium azide

The number of moles of sodium azide used is: \begin{align*} n &= \frac{m}{M}\\ & = \frac{\text{55}}{\text{65}}\\ & = \text{0.85}\text{ mol} \end{align*}

Step 2: Work out the amount of nitrogen

The mole ratio of \(\text{NaN}_{3}\) to \(\text{N}_{2}\) is \(2:3\). So the number of moles of \(\text{N}_{2}\) is: \begin{align*} n_{\text{N}_{2}} & = n_{\text{NaN}_{3}} \times \frac{\text{stoichiometric coefficient N}_{2}}{\text{stoichiometric coefficient NaN}_{3}}\\ & = \text{0.85}\text{ mol} \text{NaN}_{3} \times \frac{\text{3}\text{ mol} \text{N}_{2}}{\text{2}\text{ mol} \text{ NaN}_{3}}\\ & = \text{1.27}\text{ mol} \text{ N}_{2} \end{align*}

Step 3: Work out the volume of nitrogen

\begin{align*} V & = (\text{22.4})n\\ & = (\text{22.4})(\text{1.27})\\ & = \text{28.4}\text{ dm$^{3}$} \end{align*}