We think you are located in United States. Is this correct?

# Volume relationships in gaseous reactions

## 8.3 Volume relationships in gaseous reactions (ESBPF)

Using what we have learnt about stoichiometry and about gases we can now apply these principles to reactions involving gases.

We will use explosions as an example.

## Worked example 12: Explosions

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)?

### 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*}

### 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*}

### Work out the volume of oxygen

Recall from earlier in the chapter 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*}

Airbags in cars use a controlled explosion to inflate the bag. When a car hits another car or an object, various sensors trigger the airbag. A chemical reaction then produces a large volume of gas which inflates the airbag.

## Worked example 13: Controlled explosion

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?

### 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*}

### 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*}

### 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*}

# Success in Maths and Science unlocks opportunities

Sign up to get a head start on bursary and career opportunities. Use Siyavula Practice to get the best marks possible.

## Gases 2

Exercise 8.8

What volume of oxygen is needed for the complete combustion of $$\text{5}$$ $$\text{g}$$ of magnesium to form magnesium oxide?

The balanced equation for this reaction is:

$2\text{Mg (s)} + \text{O}_{2}\text{(g)} \rightarrow 2\text{MgO (s)}$

The number of moles of magnesium used is:

\begin{align*} n &= \frac{m}{M}\\ & = \frac{\text{5}}{\text{24}}\\ & = \text{0,2083}\text{ mol} \end{align*}

The mole ratio of $$\text{Mg}$$ 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{Mg}} \times \frac{\text{stoichiometric coefficient O}_{2}}{\text{stoichiometric coefficient Mg}}\\ & = \text{0,2083}\text{ mol} \text{Mg} \times \frac{\text{1}\text{ mol} \text{O}_{2}}{\text{2}\text{ mol} \text{ Mg}}\\ & = \text{0,1042}\text{ mol} \text{O}_{2} \end{align*}

The volume of oxygen is:

\begin{align*} V & = (\text{22,4})n\\ & = (\text{22,4})(\text{0,1042})\\ & = \text{2,33}\text{ dm$^{3}$} \end{align*}

Annalize is making a mini volcano for her science project. She mixes baking soda (mostly $$\text{NaHCO}_{3}$$) and vinegar (mostly $$\text{CH}_{3}\text{COOH}$$) together to make her volcano erupt. The reaction for this equation is: $\text{NaHCO}_{3}\text{(s)} + \text{CH}_{3}\text{COOH (aq)} \rightarrow \text{CH}_{3}\text{COONa (aq)} + \text{H}_{2}\text{O (l)} + \text{CO}_{2}\text{(g)}$ What volume of carbon dioxide is produced if Annalize uses $$\text{50}$$ $$\text{ml}$$ of $$\text{0,2}$$ $$\text{mol·dm^{3}}$$ acetic acid?

The number of moles of acetic acid used is:

\begin{align*} C &= \frac{n}{V}\\ \text{0,2}\text{ mol·dm$^{3}$} & = \frac{n}{\text{0,05}\text{ dm$^{3}$}}\\ n & = \text{0,01}\text{ mol} \end{align*}

The mole ratio of $$\text{CH}_{3}\text{COOH}$$ to $$\text{CO}_{2}$$ is $$1:1$$. So the number of moles of $$\text{CO}_{2}$$ is $$\text{0,01}$$ $$\text{mol}$$.

\begin{align*} V & = (\text{22,4})n\\ & = (\text{22,4})(\text{0,01})\\ & = \text{0,224}\text{ dm$^{3}$} \end{align*}