5 moles of SO2 and 5 moles of O2 are allowed to react to form SO3 in a closed vessel. At the equilibrium stage 60% of SO2 is used up. The total number of moles of SO2, O2 and SO3 in the vessel now is

2 2 3 2SO O 2SO No. of moles of 2 SO =5 No. of moles of 2 O =5 At equilibrium, 60% sulfur dioxide is used up 5×60 = = 3 100 moles. And that of 2 O will react 2.5×60 = = 1.5 100 moles. And no. of moles of 3 SO formed = 3 moles as much as 2 SO reacted. No. of moles of 2 SO left 5 3 2 moles No. of moles of 2 O left 5 1.5 3.5 moles. Total no. of moles left at equilibrium 3 2 3.5 8.5 moles.