2 mole of zinc is dissolved in $\mathrm{HCl}$ at 25° C. The work done in open vessel is :

For the given process,

$\mathrm{Zn}+2\mathrm{HCl}⟶{\mathrm{ZnCl}}_2+{\mathrm{H}}_2\left(\mathrm{g}\right)$

$\begin{array}{l}\mathrm{W}=-{\mathrm{\Delta n}}_8\mathrm{RT}\\ =-1\times 8.314\times 298\end{array}$

For 2 moles $=\frac{-1\times 8.314\times 298\times 2}{1000}$

$=-4.955 \mathrm{kJ}$.