5 moles of so2 and 5 moles of O2 react in a closed vessesl vessel. At eq60%of the SO2 is consumed the total no of gaseous moles (SO2, O2, and SO3) in the vessel is
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!
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