$6.5 g$ 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 (Atomic number of $Zn=65$ )

$$\begin{gathered} \mathrm{Zn}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{ZnSO}}_4+ {\mathrm{H}}_2 \\ 1\mathrm{mole} 1\mathrm{mole} 1\mathrm{mole} \\ 6.5\mathrm{gm} \to ? ? \end{gathered}$$
 

The no. of moles of zinc involved is $=6.5/65=0.1$ moles. The no. of moles of zinc sulphate and hydrogen formed is equal to no. moles of zinc decomposed.

Therefore no. of moles of zinc sulphate formed $=$ no. of moles of hydrogen formed $=0.1$

At STP any gas will occupy $22.4$lit., so the volume of Hydrogen liberated =

$0.1\times 22.4=2.24ltr$

Weight of zinc sulphate $=$ no. of moles of zinc sulphate $\times MW$ of zinc sulphate $=0.1\times (65+32+64)=0.1\times 161=16.1 g$

Hence the correct option is (B).