Hydrogen gas is prepared in the laboratory by reacting dilute $HCl$ with granulated zinc. The following reaction takes place: $Zn + 2HCl \to ZnCl_2 + H_2$. Calculate the volume of hydrogen gas liberated at $STP$ when $32.65 \ g$ of zinc reacts with $HCl$. ($1 \ mol$ of a gas occupies $22.7 \ L$ volume at $STP$; atomic mass of $Zn = 65.3 \ u$).

(11.35 L) Given that,mass of $Zn = 32.65 \ g$.
$1 \ mol$ of gas occupies $= 22.7 \ L$ volume at $STP$. Atomic mass of $Zn = 65.3 \ u$.
The given balanced chemical equation is:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2$
From the stoichiometry of the reaction:
$65.3 \ g$ of $Zn$ produces $1 \ mol$ of $H_2$ gas.
$1 \ mol$ of $H_2$ gas occupies $22.7 \ L$ at $STP$.
Therefore,$65.3 \ g$ of $Zn$ produces $22.7 \ L$ of $H_2$ at $STP$.
For $32.65 \ g$ of $Zn$,the volume of $H_2$ produced is:
$\text{Volume} = \frac{22.7 \ L}{65.3 \ g} \times 32.65 \ g = 11.35 \ L$ of $H_2$ at $STP$.