Question:

\(2\mathrm{MnO_4^{-}} + b\,\mathrm{C_2O_4^{2-}} + c\,\mathrm{H^{+}} \rightarrow x\,\mathrm{Mn^{2+}} + y\,\mathrm{CO_2} + z\,\mathrm{H_2O}\)
If the above equation is balanced with integer coefficients, the value of c is ___________. (Round off to the Nearest Integer).

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Balance by the ion-electron (or oxidation number) method: Mn goes from +7 to +2 and C from +3 to +4; then balance charge using H+.
Updated On: Nov 4, 2026
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Correct Answer: 16

Solution and Explanation

Mn: +7 to +2, gain of 5 e- per Mn, so 2 MnO4- gain 10 e-.
C in C2O42-: +3 to +4, loss of 1 e- per C, i.e. 2 e- per oxalate. To lose 10 e-: b = 5.
So x = 2, y = 10 (from 5 oxalate ions).
Charge balance: left = 2(-1) + 5(-2) + c = c - 12; right = 2(+2) = +4. So c - 12 = 4, c = 16.
Check H and O: H: 16 on left gives z = 8; O: 8 + 20 = 28 on left, 20 + 8 = 28 on right. Balanced.
Answer: 16
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