Hi Aspirant!

Solution for the query is:

SO2 (g) + 1/2 O2(g)  <=> SO3 (g)

This is the equilibrium reaction which is taking place and from this reaction,

Number of moles of SO3 formed = number of moles of SO2 consumed

Number of moles of O2 consumed =  half of number of moles of SO2 consumed

It is given that 60% of 5 moles of SO2 are consumed, so number of moles of SO3 formed are = (60/100)* 5 = 3 mol

Number of moles of O2 consumed = (1/2)*3 = 1.5 mol

Thus, at equilibrium,

Number of moles of SO2 present = 5-3 = 2

Number of moles of O2 present = 5-1.5 = 3.5

Number of moles of SO3 present = 3

Therefore, total number of gaesous moles present in the vessel = 2 + 3.5 + 3 = 8.5

Hence, the total number of gaseous moles present in vessel are 8.5 moles.

Thankyou!