Reactions and Gases

Reactions and Gases

Some reactions take place between gases. For these reactions we can work out the volumes of the gases using the fact that volume is proportional to the number of moles.

We can use the following formula:

\begin{align*} V_{\text{A}} & = \frac{a}{b} V_{\text{B}} \end{align*}


\begin{align*} V_{A} & = \text{volume of A}\\ V_{B} & = \text{volume of B}\\ a & = \text{stoichiometric coefficient of A}\\ b & = \text{stoichiometric coefficient of B} \end{align*}


The number in front of a reactant or a product in a balanced chemical equation is called the stoichiometric coefficient or stoichiometric ratio.

Example: Volume and Gases


Hydrogen and oxygen react to form water according to the following equation:

\[2\text{H}_{2} \text{(g)} + \text{O}_{2} \text{(g)} \rightarrow 2\text{H}_{2}\text{O} \text{(g)}\]

If \(\text{3}\) \(\text{dm$^{3}$}\) of oxygen is used, what volume of water is produced?

Step 1: Determine the volume of water produced in the reaction.

We use the equation given above to work out the volume of water needed: \begin{align*} V_{A} & = \frac{a}{b}V_{B} \\ V_{\text{H}_{2}\text{O}} & = \frac{2}{1}V_{\text{O}_{2}} \\ & = 2(\text{3})\\ & = \text{6}\text{ dm$^{3}$} \end{align*}

We can interpret the chemical equation in the worked example above (\(2\text{H}_{2} \text{(g)} + \text{O}_{2} \text{(g)} \rightarrow 2\text{H}_{2}\text{O} \text{(g)}\)) as:

\(\text{2}\) moles of hydrogen react with \(\text{1}\) mole of oxygen to produce \(\text{2}\) moles of water. We can also say that \(\text{2}\) volumes of hydrogen react with \(\text{1}\) volume of oxygen to produce \(\text{2}\) volumes of water.

Example: Gas Phase Calculations


What volume of oxygen at STP is needed for the complete combustion of \(\text{3.3}\) \(\text{dm$^{3}$}\) of propane (\(\text{C}_{3}\text{H}_{8}\))? (Hint: \(\text{CO}_{2}\) and \(\text{H}_{2}\text{O}\) are the products as in all combustion reactions)

Step 1: Write a balanced equation for the reaction.

\(\text{C}_{3}\text{H}_{8}\text{(g)} + 5\text{O}_{2}\text{(g)} → 3\text{CO}_{2}\text{(g)} + 4\text{H}_{2}\text{O (g)}\)

Step 2: Determine the volume of oxygen needed for the reaction.

We use the equation given above to work out the volume of oxygen needed: \begin{align*} V_{A} & = \frac{a}{b}V_{B} \\ V_{\text{O}_{2}} & = \frac{5}{1}V_{\text{C}_{3}\text{H}_{8}} \\ & = 5(\text{3.3})\\ & = \text{16.5}\text{ dm$^{3}$} \end{align*}

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This is a lesson from the tutorial, Quantitative Aspects of Chemical Change and you are encouraged to log in or register, so that you can track your progress.

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