6.5g of $Zn$ is dissolved in excess of $H_2{SO}_4$. The weight of ${ZnSO}_4$ formed and the volume of $H_2$ gas liberated at $STP$ are $(Zn=65)$

 No. of moles of $Zn$ = 6.5/65= 0.1 mole and $H_2{SO}_4$ is in excess

$$\begin{gathered} Zn+H_2{SO}_4\to {ZnSO}_4+H_2 \\ 65g 161g 22.4l \\ 6.5g ? ? \end{gathered}$$

The mass of ZnSO₄ formed=$\frac{6.5}{65}\times 161=16.1\mathrm{g}$

Volume of H₂ liberated=$\frac{6.5}{65}\times 22.4=2.24\mathrm{l}$

Hence option (B) is the correct Answer.