The volume of hydrogen liberated at STP by treating $2.4 \mathrm{g}$ of magnesium with excess of hydrochloric acid is ......$\times {10}^{-2} \mathrm{L}$. Given Molar volume of gas is $22.4 \mathrm{L}$ at STP. Molar mass of magnesium is $24 \mathrm{g} {\mathrm{mol}}^{-1}$
 

The stoichiometric equation of the reaction between magnesium and hydrogen chloride can be written as follows,

One mole magnesium can liberate one mole of hydrogen gas according to the following equation.

 $\underset{\underset{=0.1 \mathrm{mole}}{2.4 \mathrm{g}}}{\mathrm{Mg}}+2\mathrm{HCl}\to {\mathrm{MgCl}}_2+\underset{0.1 \mathrm{mole}}{{\mathrm{H}}_2}$
Volume of $1$ mole ${\mathrm{H}}_2$ at $=22.4 \mathrm{L}$ 
$\therefore 0.1$ mole ${\mathrm{H}}_2$ at STP will occupy $=2.24 \mathrm{L}$