$10\, \mathrm{m} \,\mathrm{mol} \,\mathrm{H}_{2} \mathrm{SO}_{4}=20\, \mathrm{m}\, \mathrm{mol}$ of $\mathrm{NH}_{3}$

$\left[\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NH}_{3} \longrightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\right]$

$1\, \mathrm{mol}\; \mathrm{NH}_{3}$ contains $14 \,\mathrm{g}$ nitrogen $20 \times 10^{-3}\, \mathrm{mol}\; \mathrm{NH}_{3}$ contains $14 \times 20 \times 10^{-3}$ nitrogen

$0.75\, \mathrm{g}$ of sample contains

$\%$ Nitrogen $=\frac{14 \times 20 \times 10^{-3}}{0.75} \times 100=37.33 \%$