Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

$\begin{matrix} 4\text{NH}$3 & + & 5\text{O}2 & \longrightarrow & 4\text{NO} & + & 6\text{H}_2\text{O} \\ 4 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4 \text{ vol.} \\ \end{matrix}

[By Gay Lussac's law]

9 litres of reactants produces = 4 litres of NO

So, 27 litres of reactants will produces

$=\dfrac{4}{9} \times 27 \\[0.5em] = 12 \text{ litres}$

Hence, volume of nitrogen monoxide produced = 12 litres