3 moles of NH3  are allowed to dissociate in a 5 litre vessel and the equilibrium concentration of N2  is 0.2 mole / lit. Then the total number of moles at equilibrium is

2NH3⇌N2+3H2            3                0          0           2x              x         3x_     (3−2x)         x        3x        Given    x5=0.2 M                      x= 1 MTotal number of moles at equilibrium =                                                      (3−2x)+x+3x                                                  =3+1+1=5