When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at STP, the moles of HCl(g) formed is equal to:

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

Therefore, 22.4 L of H2(g) represents 1 mole of H2(g).
And 11.2 L of Cl2(g) represents 11.2 L / 22.4 L/mol = 0.5 moles of Cl2(g).

The balanced chemical equation for the reaction is:
H2(g) + Cl2(g) → 2HCl(g)

From the stoichiometry of the reaction, 1 mole of H2(g) reacts with 1 mole of Cl2(g) to produce 2 moles of HCl(g).
Since we have 1 mole of H2(g) and 0.5 moles of Cl2(g), Cl2(g) is the limiting reactant.

Therefore, 0.5 moles of Cl2(g) will react with 0.5 moles of H2(g) to produce 2 * 0.5 = 1 mole of HCl(g).