Chemistry - Nomenclature Question with Solution | TestHub

$\because \mathrm{}\mathrm{M}\times \mathrm{V}\left(\text{ml}\right)=\mathrm{m}$mol
10 m mol${\mathrm{H}}_2\text{S}{\text{O}}_4=20$m mol of$\text{N}{\text{H}}_3$
$[{\mathrm{H}}_2\text{S}{\text{O}}_4+\text{2N}{\text{H}}_3\to (\text{N}{\text{H}}_4{)}_2\text{S}{\text{O}}_4]$
1 mol$\text{N}{\text{H}}_3$contains 14 g nitrogen
$20\times 10^{-3}$mol$\text{N}{\text{H}}_3$contains$\text{14}\times 20\times 10^{-3}$nitrogen
0.75 g of sample contains
% Nitrogen$=\frac{14\times 20\times 10^{-3}}{0.75}\times 100=37.33\mathrm{\%}$