5 moles of ${\mathrm{SO}}_2$ and  5 moles of $O_2$  react in a closed vessel.  At equilibrium 60% of the ${\mathrm{SO}}_2$ is consumed.  The total number of gaseous moles $({\mathrm{SO}}_2,{\mathrm{O}}_2 \text{and }{\mathrm{SO}}_3)$ in the vessel is 

$2{\mathrm{SO}}_{2_{\left(\mathrm{g}\right)}}+{\mathrm{O}}_{2_{\left(\mathrm{g}\right)}}\to 2{\mathrm{SO}}_{3_{\left(\mathrm{g}\right)}}$        At equilibrium  60%of the SO₂ is consumed

 5                  5                0

5-2x           5-x              2x

5-3           5-1.5            3

2               3.5               3

 Total no. of moles = 2 + 3.5 + 3 =8.5