When 1 mole of $N_2$ and 1 mole of $H_2$ is enclosed in a 5L vessel and the reaction is allowed to attain equilibrium. It is found that at equilibrium there is $x$ mole of $H_2$ . The number of moles of $N{H}_3$ formed would be

$N_2+3H_2\rightleftharpoons 2N{H}_3$

$110$

$\underset{\_}{\alpha 3\alpha 2\alpha }$

$\left(1-\alpha \right)\left(1-3\alpha \right)2\alpha$

          $Given1-3\alpha =x$

                       $3\alpha =1-x$

                         $\alpha =\frac{1-x}{3}$      

$\therefore$ Number of moles$N{H}_3=2\alpha =2\left(\frac{1-x}{3}\right)$