When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P, the moles of HCl (g)formed is equal to

$1 mole \equiv 22.4 litres at S.T.P.$

$n_{H_{2}} = \frac{22.4}{22.4} = 1 mol, n_{{Cl}_{2}} = \frac{11.2}{22.4} = 0.5 mol$

reaction is as

$H_{2 (g)} + {Cl}_{2 (g)} ⟶ 2 {HCl}_{(g)}$

$\begin{array}{l} Initial & 1 mol & 0.5 mol & 0 \\ Final & (1 - 0.5) = 0.5 & (0.5 - 0.5) = 0 & 2 \times 0.5 = 1 \end{array}$

Here, ${Cl}_{2}$ is limiting reagent. So, 1 mole of HCl(g) is formed.

When 22.4 litres of $H_{2} (g)$ is mixed with 11.2 litres of ${Cl}_{2} (g)$ , each at S.T.P, the moles of HCl (g)formed is equal to