1.2 moles of ${\mathrm{SO}}_3$are allowed to dissociate in a 2 litre vessel the reaction is  $2{\mathrm{SO}}_{3\left(\mathrm{g}\right)}\iff 2{\mathrm{SO}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}$and the concentration of oxygen at equilibrium is 0.1 mole per litre. The total number of moles at equilibrium will be

The balanced equation is

$2{\mathrm{SO}}_3\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

Initial(mole):     1.2        0       0 

Change (mole): -2x        2x     x 

Equilib(mole): 1.2 -2x    2x     x

             $=1.2-2\times 0.1 2\times 0.1 0.1$

$\left[{\mathrm{O}}_2\right]=0.1 mole.L^{\_1}$

$=\frac{{\mathrm{n}}_{{\mathrm{O}}_2}}{\mathrm{V} \mathrm{in} \mathrm{Litres}}=0.1 \mathrm{mol} .{\mathrm{L}}^{-1}$

$={\mathrm{N}}_{{\mathrm{O}}_2}=0.1\times 2$

=0.2 mole

So total no. of moles = 0.8 +0.4 +0.2

=1.4