$71 \mathrm{~g} \mathrm{Cl}_{2}$ requires $2 \mathrm{~g} \mathrm{H}_{2}$, than $7.4 \mathrm{~g} \mathrm{Cl}_{2}$ will require, how much gram $\mathrm{H}_{2}$.
$\therefore \mathrm{H}_{2} \text { required }=\frac{2 \times 7.4}{71}=0.208 \mathrm{gm}$
$\therefore \mathrm{Cl}_{2}$ is limiting reagent $\left[\because \mathrm{H}_{2}\right.$ is in excess $]$
$\therefore$ Amount of $\mathrm{HCl}$ formed will depend on $\mathrm{Cl}_{2}$ gas
$\begin{array}{rlr}
& 71 \mathrm{gm} \mathrm{Cl}_{2} & 44.8 \mathrm{~L} \mathrm{HCl} \\
& 7.4 \mathrm{gm} \mathrm{Cl}_{2} & x \mathrm{~L} \mathrm{HCl} \\
\therefore \quad & x=\frac{7.4 \times 44.8}{71} \\
& x=4.67 \mathrm{~L} \text { of } \mathrm{HCl} \text { gas is formed. }
\end{array}$