12 litre of $H_2$ and 11.2 litre of ${\mathrm{Cl}}_2$ are mixed and exploded. The amount of remaining hydrogen is (all volumes are measured at STP) _____?

${\mathrm{H}}_2+{\mathrm{Cl}}_2\to 2\mathrm{HCl}$

At STP, one mole of gas occupies 22.4 L of volume.

Moles of gases present in given volume:

$$\begin{gathered} \mathrm{Moles} \mathrm{present} \mathrm{in} 12 \mathrm{L} \mathrm{of} {\mathrm{H}}_2=\frac{12 \mathrm{L}}{22.4 \mathrm{L}}=0.54 \mathrm{mol} \\ \mathrm{Moles} \mathrm{present} \mathrm{in} 11.2 \mathrm{L} \mathrm{of} {\mathrm{Cl}}_2=\frac{11.2 \mathrm{L}}{22.4 \mathrm{L}}=0.5 \mathrm{mol} \end{gathered}$$

${\mathrm{Cl}}_2$ is limiting reagent. Unreacted moles of H₂:

$\mathrm{Unreacted} \mathrm{Moles}=\frac{12 \mathrm{L}}{22.4 \mathrm{L}}-\frac{11.2 \mathrm{L}}{22.4 \mathrm{L}}=0.035 \mathrm{moles}$

Volume of H₂ present in 0.035 moles is:

$\mathrm{Volume}:0.035\times 22.4 \mathrm{L}=0.8 \mathrm{L}$