1.4 g of an organic compound was digested according to Kjeldahl's method and the ammonia evolved was…

Total millimoles of sulphuric acid =$60\times \frac{1}{10}=6\mathrm{ }\mathrm{m}\mathrm{m}\mathrm{ }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$

Excess ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\equiv 20\times \frac{1}{10}\times \frac{1}{2}=1\mathrm{ }\mathrm{m}\mathrm{m}\mathrm{ }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$

${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ used in reaction with ammonia $=6-1=5\mathrm{ }\mathrm{m}\mathrm{m}$

Based on the reaction stoichiometry, $2\mathrm{N}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4⟶{\left(\mathrm{N}{\mathrm{H}}_4\right)}_2\mathrm{S}{\mathrm{O}}_4$

mm of $\mathrm{N}{\mathrm{H}}_3=10\mathrm{ }\mathrm{m}\mathrm{m}$

Mass of $\mathrm{N}=10\times {10}^{-3}\times 14\left(\frac{\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}\right)=0.140\mathrm{g}$

$\mathrm{\%}\mathrm{ }{\mathrm{N}}_2=\frac{0.140}{1.4}\times 100=10\mathrm{\%}$