28 g N₂ and 6 g H₂ were mixed. At equilibrium 17 g NH₃ was formed. The mass of N₂ and H₂ of equilibrium are respectively:

$\underset{1-\frac{1}{2}}{\underset{\frac{28}{28}=1}{{\mathrm{N}}_2}}+\underset{3-\frac{3}{2}}{\underset{\frac{6}{2}=3}{3{\mathrm{H}}_2}}\rightleftharpoons \underset{\frac{17}{17}=1}{\underset{0}{2{\mathrm{NH}}_3}} \underset{\mathrm{mole} \mathrm{after} \mathrm{reaction}}{\underset{\mathrm{mole} \mathrm{before} \mathrm{reaction}}{{ }_{ }}}$
 $\therefore$ Mole of ${\mathrm{N}}_2=\frac{1}{2}$
$\therefore$ mass of N₂ = 14 g
Mole of H₂ $=\frac{3}{2}$
$\therefore$ mass of ${\mathrm{H}}_2=\frac{3}{2}\times 2=3\mathrm{g}$