$\quad \quad \quad \quad H_{2}(g)+C l_{2}(g) \rightarrow 2 H C l(g)$

Initial vol. $22.4 L \;\;11. 2 L\;\;\;\;\;\;\;\;\;2 m o l$

$\therefore 22.4 \mathrm{L}$ volume at $\mathrm{STP}$ is occupied by

$C l_{2}=1 \text { mole }$

$\therefore 11.2 \mathrm{L}$ volume will be occupled by

$C l_{2}=\frac{1 \times 11.2}{22.4} \mathrm{mol}=0.5 \mathrm{mol}$

$22.4 \mathrm{L}$ volume at $\mathrm{STP}$ is occupied by $H_{2}=1 \mathrm{mol}$

Thus, $H_{2}(g)+C l_{2}(g) \rightarrow 2 H C l(g)$

$\quad \quad 1 \mathrm{mol} \quad \quad 1\mathrm{mol}\quad \quad 0.5 \mathrm{mol}$

since, $C l_{2}$ possesses minimum number of moles,

thus it is the limiting reagent.

As per equation,

$1\; mole$ of $C l_{2}=2$ moles of $\mathrm{HCl}$

$\therefore 0.5$ mole of $C l_{2}=2 \times 0.5$ mole of $\mathrm{HCl}$

$=1.0$ mole of $\mathrm{HCl}$

Hence, $1.0 \;mole$ of $\mathrm{HCl}(\mathrm{g})$ is produced by $0.5\; mole$ of $C l_{2}[\text { or } 11.2 \mathrm{L}]$