67.2 litres of H₂ combines with 44.8 litres of N₂ to form NH₃ :

N₂(g) + 3H₂(g) ⟶ 2NH₃(g).

Calculate the vol. of NH₃ produced. What is the substance, if any, that remains in the resultant mixture ?

[By Lussac's law]

$\begin{matrix} \text{N}$2 & + & 3\text{H}2 & \longrightarrow & 2 \text{NH}_3 \ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}

To calculate the volume of ammonia gas formed.

$\begin{matrix} \text{H}$2 & : & \text{NH}3 \ 3 & : & 2 \ 67.2 & : & x \end{matrix}

$\therefore \dfrac{2}{3} \times 67.2 = 44.8 \text{ lit}$

Hence, volume of NH₃ formed is 44.8 lit.

We know,

$\begin{matrix}\text{H}$2 & : & \text{N}2 \ 3 & : & 1 \ 67.2 & : & x \end{matrix}

$\therefore \dfrac{1}{3} \times 67.2 = 22.4 \text{ lit}$

∴ nitrogen left = 44.8 - 22.4 = 22.4 lit.

Hence, 22.4 lit of nitrogen remains in the resultant mixture.