What quantity of ammonium sulphate is necessary for the production of ${\mathrm{NH}}_3$ gas sufficient to neutralize a solution containing 292 g of HCl ? $\left[\mathrm{HCl}=36.5;{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4=132;{\mathrm{NH}}_3=17\right]$

 $\begin{array}{l}\left[\mathrm{HCl}=36.5;{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4=132;{\mathrm{NH}}_3=17\right]\\ \underset{132}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4}\equiv 2{\mathrm{NH}}_3\equiv \underset{2\left(36.5\right)=73\mathrm{g}}{2\mathrm{HCl}}\\ \mathrm{Since} 1 \mathrm{mole} \mathrm{of} {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\to 2 {\mathrm{NH}}_3 \mathrm{which} \mathrm{consumes} 2\mathrm{moles} \mathrm{ofHCl} \mathrm{for} \mathrm{neutralization}\\ \\ 73\mathrm{g}\mathrm{HCl}\equiv 132\mathrm{g}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\\ 292\mathrm{g}\mathrm{HCl}\equiv 528\mathrm{g}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\end{array}$