Question

Balance the equation : $Mn{O}_4-+C_2{O}_{{42}^{-}}+H+\to C{O}_2+Mn+2+H_{2}O$

Answer 1

$Mn+7\to Mn+2;C_2\to C+45e^{-}+Mn+7\to Mn+2(1{)}^{\prime }2;C_2\to 2C+4+2e^{-}(2{)}^{\prime }510e^{-}+2Mn+7\to 2Mn+2;5C_2\to 10C+4+10e^{-}$
Therefore $2Mn{O}^{4-}+5C2O_{4-2}\to 10C{O}^{2-}+2Mn+2$
There must be $8H_{2}O$ on the right to balance O. Therefore there must be 10H+ to balance H.
$2Mn{O}^{4-}+5C2O_{4-2}+16H+\to 10C{O}_2+2Mn+2+8H_{2}O$