When $22.4$ liters of ${\mathrm{H}}_2 \left(\mathrm{g}\right)$ is mixed with 11.2 liters of $C{l}_2\left(g\right)$, each at STP, the moles of $HCl\left(g\right)$ formed is equal to

At STP, 1 mole of any gas occupies a volume of 22.4 L.

$\underset{\underset{22.4\mathrm{l}}{22.4\mathrm{l}}}{{\mathrm{H}}_2}+\underset{\underset{11.2\mathrm{l}}{22.4\mathrm{l}}}{{\mathrm{Cl}}_2}\to \underset{\underset{?}{2\times 22.4}\mathrm{l}}{2\mathrm{HCl}}$
Limiting reagent is Cl₂​.
1 mole of chlorine forms 2 moles of HCl.
Hence, 0.5 moles of chlorine will form 1 mole of HCl.