a
The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.

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

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 \(C l_{2}[\text { or } 11.2 \mathrm{L}]\)