A mixture of 2 moles of ${\mathrm{N}}_2$ and 8 moles of ${\mathrm{H}}_2$ are heated in a 2 lit vessel, till equilibrium is established. At equilibrium, 0.4 moles of ${\mathrm{N}}_2$  was present. The equilibrium concentration of  ${\mathrm{H}}_2$ will be

${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$

Initial :          2         8        0 

Acc.to eqn    x        3x     2x Equilibrium: 2 - x   8-3x   2x

At equilibrium; 0.4 moles of ${\mathrm{N}}_2$ are present.

2 - x = 0.4 

x =1.6moles.

Equilibrium: $$\begin{gathered} \underset{\downarrow }{0.4} \underset{\downarrow }{3.2} \underset{\downarrow }{3.2} \\ {N}_2 H_2 N{H}_3 \end{gathered}$$

So equilibrium ionic of ${\mathrm{H}}_2=\frac{{\mathrm{n}}_{{\mathrm{H}}_2}}{\mathrm{V}\left(\mathrm{H}\right)}=\frac{3.2}{2}=1.6 \mathrm{mole}/\mathrm{lit}.$