Chemistry question?

Question

25cm3 of a solution of X2O5 of concentration 0.100M is reduced by SO2 to a lower oxidation state.To reoxidise X to its original oxidation number requires 50cm3 of 0.02M KMnO4 solution.To what oxidation number was X reduced by SO2?

Answer 1

the answer is +4

Answer 2

MnO4- + 5e- + 8H+ -> Mn2+ + 4H2O

Xn+ -> X5+ + (5-n)e-

Millimoles of KMnO4 = 50x0.02 = 1

Thus millimoles of X = 5/(5-n)

Actual millimoles = 2.5*2 = 5

Thus 5 = 5/(5-n)
or 5-n = 1
or n = 4

Thus X was reduced from +5 to +4.