Chemistry - Stoichiometry Question with Solution | TestHub

${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\stackrel{40\%}{\to }{\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{NH}}_3$

${\mathrm{NH}}_3+\mathrm{HCL}\stackrel{80\%}{\to }{\mathrm{NH}}_4\mathrm{Cl}$

(B) Moles of ${\mathrm{NH}}_4\mathrm{Cl}=\frac{10.7}{53.5}=0.2$

Moles of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ required

$=\left(\frac{10.7}{53.5}\right)\times \frac{1\times 1}{2\times 1}\times \left(\frac{100}{40}\times \frac{100}{80}\right)=\frac{10}{32}$

Mass of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4=\frac{10}{32}\times 132=10.25\mathrm{gm}$

(C) Moles of NaOH required

$=0.2\times \left(\frac{2\times 1}{2\times 1}\right)\times \left(\frac{100}{40}\times \frac{100}{80}\right)$

$=\frac{20}{32}=\frac{5}{8}$

Weight of NaOH required $=\frac{5}{8}\times 40=25\mathrm{gm}$

(D) Moles of NaOH taken $=\frac{4}{40}=0.1$ mole

Moles of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ formed $=\left(0.1\times \frac{40}{100}\right)\times \frac{1}{2}$

Moles of ${\mathrm{NH}}_4\mathrm{Cl}$ formed $=0.1\times \left(\frac{2\times 1}{2\times 1}\right)\times \left(\frac{40}{100}\times \frac{80}{100}\right)$ $\frac{\text{ moles of }{\mathrm{NH}}_4\mathrm{Cl}}{\text{ moles of }\mathrm{NaOH}}=\frac{0.1\times \frac{40}{100}\times \frac{80}{100}}{0.1\times \frac{40}{100}\times \frac{1}{2}}=\frac{160}{100}=1.6$ times