2 MnO 4 - + bC 2 O 4 2 - + cH + → xMn 2 + + yCO 2 + zH 2 O If the above equation is balanced with…

Writing the half reaction

Reduction half reaction 

${\mathrm{MnO}}_4^{-}\to {\mathrm{Mn}}^{2+}$

balancing oxygen 

${\mathrm{MnO}}_4^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

balancing Hydrogen 

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

balancing charge 

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

Oxidation half 

${\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to {\mathrm{CO}}_2$
Balancing carbon

${\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to 2{\mathrm{CO}}_2$

Balancing charge

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

Net equation

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

So $\mathrm{c}=16$