Total volume occupied by a mixture containing 1 equivalent each of N₂, H₂ and NH₃ based on the following equation at STP is

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)⟶2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

${\mathrm{N}}_2\equiv 3{\mathrm{H}}_2\equiv 6\mathrm{H}$

$\text{ Thus, equivalent mass of }{\mathrm{N}}_2=\frac{\text{ Molar mass }}{6}$

Volume occupied by I equivalent N₂ at STP =$\frac{22.4}{6}L$

NH₂ $\equiv$ 3H

Thus, equivalent mass of NH₃ at STP =$\frac{22.4}{3}L$

H₂ $\equiv$ 2H

$\text{ Thus, equivalent mass of }{\mathrm{H}}_2=\frac{\text{ Molar mass }}{2}$

$\text{ Volume occupied by }1\text{ equivalent }{\mathrm{H}}_2\text{ gas at STP }=\frac{22.4}{2}L$

$\text{ Thus, total volume }=\mathrm{V}\left({\mathrm{N}}_2\right)+\mathrm{V}\left({\mathrm{NH}}_3\right)+\mathrm{V}\left({\mathrm{H}}_2\right)$

$\begin{array}{l}=\frac{22.4}{6}+\frac{22.4}{3}+\frac{22.4}{2}\\ =\frac{22.4}{6}+\frac{2\times 22.4}{6}+\frac{3\times 22.4}{6}=22.4\mathrm{L}\end{array}$