A mixture of 50.0 mL of NH₃ and 60.0 mL of O₂ gas reacts as

$4{\mathrm{NH}}_3\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)⟶4\mathrm{NO}+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

If all the gases are at the same temperature and pressure and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced?

By Avogadro's law, equal volumes of gases contain equal number of moles.

$$\begin{gathered} 4{\mathrm{NH}}_3\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)⟶4\mathrm{NO}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \\ 4\mathrm{mL}{\mathrm{NH}}_3=6{\mathrm{mLH}}_2\mathrm{O} \end{gathered}$$

$$\begin{gathered} \therefore 50\mathrm{mL}{\mathrm{NH}}_3=75{\mathrm{mLH}}_2\mathrm{O} \\ 5{\mathrm{mLO}}_2=6{\mathrm{mLH}}_2\mathrm{O} \\ \therefore 60{\mathrm{mLO}}_2=72{\mathrm{mLH}}_2\mathrm{O} \\ \mathrm{stoichiometrically}, {\mathrm{O}}_2 \mathrm{is} \mathrm{consumed} \mathrm{completely} \\ \mathrm{Thus}, 72 {\mathrm{mLH}}_2\mathrm{O} \left(\mathrm{g}\right). \end{gathered}$$