When 22.4 litres of $H_{2\left(g\right)}$  is mixed with 11.2 litres of  $C{l}_{2\left(g\right)},$ each at S.T.P, the moles of  $HC{l}_{\left(g\right)}$ formed is equal to 

1 mole = 22.4 litres at S.T.P
$n_{H_2}=\frac{22.4}{22.4}=1mol;n_{C{l}_2}=\frac{11.2}{22.4}=0.5mol$
Reaction is as,
$\begin{array}{l}{H}_{2\left(g\right)}+C{l}_{2\left(g\right)}\to 2HC{l}_{\left(g\right)}\\ Initial1mole0.5mol0\\ Final(1-0.5)(0.5-0.5)2\times 0.5\\ =0.5mol=0mol1mol\end{array}$
Here, $C{l}_2$  is limiting reagent. So, 1 mole of $HC{l}_{\left(g\right)}$  is formed.