2 MnO _ 4 ^ - + bC _ 2 O _ 4 ^ 2 - + cH ^ + → xMn ^ 2 + + yCO _ 2 +...

Writing the half reaction

Reduction half reaction

${MnO}_{4}^{-} \to {Mn}^{2 +}$

balancing oxygen

${MnO}_{4}^{-} \to {Mn}^{2 +} + 4{H}_{2}O$

balancing Hydrogen

$8{H}^{+} + {MnO}_{4}^{-} \to {Mn}^{2 +} + 4{H}_{2}O$

balancing charge

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

Oxidation half

${C}_{2}{O}_{4}^{2 -} \to {CO}_{2}$
Balancing carbon

${C}_{2}{O}_{4}^{2 -} \to 2{CO}_{2}$

Balancing charge

${C}_{2}{O}_{4}^{2 -} \to 2{CO}_{2} + 2{e}^{-}$

Net equation

$16{H}^{+} + 2{MnO}_{4}^{-} + 5{C}_{2}{O}_{4}^{2 -} \to 10{CO}_{2} + 2{Mn}^{2 +} + 8{H}_{2}O$

So $c = 16$