3 moles of $N{H}_3$  are allowed to dissociate in a 5 litre vessel and the equilibrium concentration of $N_2$  is 0.2 mole / lit. Then the total number of moles at equilibrium is

$2N{H}_3\rightleftharpoons N_2+3H_2$ 

$300$

$\underset{\_}{2xx3x}$

$\left(3-2x\right)x3x$

        $Given\frac{x}{5}=0.2M$

                      $x=1M$

$Totalnumberofmolesatequilibrium=$

                                                      $\left(3-2x\right)+x+3x$

                                                  $=3+1+1=5$