1. We need to balance the redox reaction:

$2\text{MnO}_4^- + b\text{C}_2\text{O}_4^{2-} + c\text{H}^+ \to x\text{Mn}^{2+} + y\text{CO}_2 + z\text{H}_2\text{O}$

We are asked to find the coefficient $c$ of $\text{H}^+$.

2. Use the ion-electron method in acidic medium.

Oxidation half-reaction

Oxalate is oxidized to carbon dioxide:

$\text{C}_2\text{O}_4^{2-} \to 2\text{CO}_2 + 2e^-$

Reduction half-reaction

Permanganate is reduced to $\text{Mn}^{2+}$ in acidic medium:

$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$

3. Equalize electrons.

• Oxidation half gives $2e^-$
• Reduction half needs $5e^-$

LCM of 2 and 5 is 10.

So multiply:

• oxidation half by 5
• reduction half by 2

Thus,

$5\text{C}_2\text{O}_4^{2-} \to 10\text{CO}_2 + 10e^-$

$2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \to 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

4. Add the two half-reactions:

$2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \to 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$

5. Compare with the given form:

$2\text{MnO}_4^- + b\text{C}_2\text{O}_4^{2-} + c\text{H}^+ \to x\text{Mn}^{2+} + y\text{CO}_2 + z\text{H}_2\text{O}$

So,

$c = 16$

6. Since this is already an integer, rounding to nearest integer still gives:

$\boxed{16}$