When 22.4 litres of H_2(g) is mixed with 11.2 litres of Cl_2(g), each at STP, the moles of HCI_(g) formed is equal to

At STP, 1 mole of any gas occupies a volume of 22.4 L.
$\underset{22.4 \mathrm{lt}}{{\mathrm{H}}_2} + \underset{11.2 \mathrm{lt}}{{\mathrm{Cl}}_2} \to 2\mathrm{HCl}$
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.
Thus, when 22.4 liters of H_2​(g) is mixed with 11.2 liters of Cl_2​(g), each at STP, the moles of HCl_(g) formed is equal to 1 mol of HCl_(g).