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

The balanced equation is:
$${\text{H}}_2(g)+{\text{Cl}}_2(g)\to 2\text{HCl}(g)$$
Moles of ${\text{H}}_2=\frac{22.4}{22.4}=1\text{mol}$ Moles of ${\text{Cl}}_2=\frac{11.2}{22.4}=0.5\text{mol}$ According to the stoichiometry, $1\text{mol}$ of ${\text{H}}_2$ requires $1\text{mol}$ of ${\text{Cl}}_2$. Since we only have $0.5\text{mol}$ of ${\text{Cl}}_2$, ${\text{Cl}}_2$ is the limiting reagent.Moles of $\text{HCl}$ produced $=2\times \text{moles of limiting reagent}$ Moles of $\text{HCl}=2\times 0.5=1\text{mole}$.