When 22.4 litres of ${\mathrm{H}}_2\left(\mathrm{g}\right)$ is mixed with 11.2 litres of ${\mathrm{Cl}}_2\left(\mathrm{g}\right)$, each at S.T.P, the moles of HCl (g)formed is equal to

$1\text{ mole }\equiv 22.4\text{ litres at S.T.P. }$

${\mathrm{n}}_{{\mathrm{H}}_2}=\frac{22.4}{22.4}=1 \mathrm{mol},{\mathrm{n}}_{{\mathrm{Cl}}_2}=\frac{11.2}{22.4}=0.5 \mathrm{mol}$

reaction is as

                       ${\mathrm{H}}_{2\left(\mathrm{g}\right)} + {\mathrm{Cl}}_{2\left(\mathrm{g}\right)} ⟶ 2{\mathrm{HCl}}_{\left(\mathrm{g}\right)}$

$\begin{array}{llll}\text{ Initial }& 1 \mathrm{mol}& 0.5 \mathrm{mol}& 0\\ \text{ Final }& (1-0.5)=0.5 & (0.5-0.5)=0& 2\times 0.5=1\end{array}$

Here, ${\mathrm{Cl}}_2$ is limiting reagent. So, 1 mole of HCl(g) is formed.