Balance the equation $MnO X_{4} X^{-} + C X_{2} O X_{4} X^{2 -} + H X^{+} Mn X^{2 +} + CO X_{2} + H X_{2} O$ in acidic medium using the smallest whole-number coefficients. Find the sum of the coefficients of all reactants and products. Give the answer as an integer.

Numerical answer type, tolerance 0.1.

Answer: 43

Mn gains 5 electrons per atom; each oxalate loses 2 electrons (2 C from +3 to +4). The ratio

MnO X_{4} X^{-}

C X_{2} O X_{4} X^{2 -}

is 2 : 5, and charge and O balance give

2 MnO X_{4} X^{-} + 5 C X_{2} O X_{4} X^{2 -} + 16 H X^{+} 2 Mn X^{2 +} + 10 CO X_{2} + 8 H X_{2} O

. The coefficients add up to 43.