When 1 mole of ${\mathbf{N}}_2$ and 1 mole of ${\mathrm{H}}_2$ is enclosed in a $5 \mathrm{L}$ vessel and the reaction is allowed to attain equilibrium. It is found that at equilibrium there is $\mathrm{x}$ mole of ${\mathrm{H}}_2$. The number of moles of ${\mathrm{NH}}_3$ formed would be

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

1        1                0

1-y    1-3y         2y

It is found that at equilibrium there is $\mathrm{x}$ mole of ${\mathrm{H}}_2$. 

$1-3y=x$

$\Rightarrow y=\frac{1-x}{3}$

Therefore,

No. of moles of ${\mathrm{NH}}_3$ formed $=2\mathrm{y}=\frac{2(1-\mathrm{x})}{3}$

Hence the correct option is (C) $\frac{2(1-x)}{3}$.