Study the following reactions :
$$\begin{gathered} 4\mathrm{Zn} + 7\mathrm{NaOH} + {\mathrm{NaNO}}_3 \to 4{\mathrm{Na}}_2 {\mathrm{ZnO}}_2 + {\mathrm{NH}}_3 + 2{\mathrm{H}}_2\mathrm{O} \\ 2{\mathrm{NH}}_3 + {\mathrm{H}}_2{\mathrm{SO}}_4 \to {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4 \end{gathered}$$

The correct ratio of number of $\mathrm{g}-$equivalents of $\mathrm{Zn}$ and ${\mathrm{H}}_2{\mathrm{SO}}_4$ reacted, if all ${\mathrm{NH}}_3$ produced from ${\mathrm{NaNO}}_3$ is reacted with ${\mathrm{H}}_2{\mathrm{SO}}_4$, is :

$$\begin{gathered} \mathrm{For} 1\mathrm{st} \mathrm{reactions} :– \\ \mathrm{Equivalents} \mathrm{of} \mathrm{Zn} = \mathrm{equivalents} \mathrm{of} {\mathrm{NH}}_3 \\ \mathrm{Moles} \mathrm{of} \mathrm{Zn} \times \mathrm{n}-\mathrm{factor} \mathrm{of} \mathrm{Zn} = \mathrm{equivalents} \mathrm{of} {\mathrm{NH}}_3 \\ 4 \times 2 = \mathrm{equivalents} \mathrm{of} {\mathrm{NH}}_3 \\ 8 = \mathrm{equivalents} \mathrm{of} {\mathrm{NH}}_3 \\ 8 = \mathrm{moles} \mathrm{of} {\mathrm{NH}}_2 \times 8 \\ 1 = \mathrm{moles} \mathrm{of} {\mathrm{NH}}_3 \\ \mathrm{For} 2\mathrm{nd} \mathrm{reactions} :– 1 \mathrm{mole} {\mathrm{NH}}_3 \mathrm{reacts} \mathrm{with} \frac{1}{2} \mathrm{mol} {\mathrm{H}}_2{\mathrm{SO}}_4 \\ \mathrm{therefore} \mathrm{equivalent} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4 = \frac{1}{2} \times 2 = 1 \\ \frac{\mathrm{equivalents} \mathrm{of} \mathrm{Zn}}{\mathrm{equivalents} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4} =\frac{8}{1} = 8 : 1 \mathrm{Ans}. \end{gathered}$$