4 moles of each SO2 and O2 gases are allowed to react to form So, in a closed vessel. At
equilibrium 50% of SO2 is used up. The moles of O2 present at equilibrium are:
A.2
B. 2.5
C. 3.0
D. 3.5
E.1
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Explanation:
Solution:-
SO
2
(g)
+
2
1
O
2
(g)
⇌SO
3
(g)
From the above reaction-
No. of moles of SO
3
formed = No. of moles of SO
2
consumed.
No. of moles of O
2
consumed = half of no. of moles of SO
2
consumed.
Given that 60% of 5 moles of SO
2
is consumed.
No. of moles of SO
3
formed =
100
60
×5=3 mol
No. of moles of O
2
consumed =
2
1
×3=1.5 mol
Therefore, at equilibrium,
No. of moles of SO
2
present =5−3=2
No. of moles of O
2
present =5−1.5=3.5
No. of moles of SO
3
present =3
∴ Total no. of gaseous moles in the vessel =2+3.5+3=8.5
Hence the total no. of gaseous moles in the vessel is 8.5.
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