The volume of ${\mathrm{H}}_2$ liberated at STP when one gram equivalent of ‘$\mathrm{Al}$’ is treated with excess of aqueous $\mathrm{NaOH}$ solution is

$\mathrm{Al}+\mathrm{NaOH}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NaAlO}}_2+\frac{3}{2}{\mathrm{H}}_2$

mole-mole analysis we get,

1 mole of $\mathrm{Al}\Rightarrow \frac{3}{2}$ mole of ${\mathrm{H}}_2$

Since, one gram equivalent of ‘$\mathrm{Al}$’ is equal to 27 grams of aluminium.

$\mathrm{n}\left(\mathrm{Al}\right)=\frac{27}{27}=1$

$\therefore \mathrm{n}\left({\mathrm{H}}_2\right)\text{ formed }=\frac{3}{2}$

Molar volume of gas at $\mathrm{STP}=22.4$ Litres

$\frac{3}{2}$ moles will have a volume of $=\frac{3}{2}\times 22.4$

$=33.6$litres