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*}
where:
\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*}
Tip:
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
Question
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
Question
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*}
This lesson is part of:
Quantitative Aspects of Chemical Change