$2{\mathrm{MnO}}_4+{\mathrm{bC}}_2{\mathrm{O}}_4{}^{2-}+{\mathrm{CH}}^{+}\to {\mathrm{xMn}}^{2+}+{\mathrm{yCO}}_2+{\mathrm{zH}}_2\mathrm{O}$

If the above equation is balanced with integer coefficients, the value of C is ?

(Round off to the Nearest Integer).

Writing the half reaction

oxidation half reaction

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

balancing oxygen

${MnO}_4^{-}\to {Mn}^{2+}+4H_{2}O$

balancing Hydrogen

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

balancing charge

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

Reduction half

$C_2{O}_4{}^{2-}\to {CO}_2$

Balancing carbon

$C_2{O}_4{}^{2-}\to 2{CO}_2$

Balancing charge

$C_2{O}_4{}^{2-}\to 2{CO}_2+2e^{-}$

Net equation

$16H^{+}+2{MnO}_4^{-}+5C_2{O}_4{}^{2-}\to 10{CO}_2+2{Mn}^{2+}+8H_{2}O$

$\mathrm{So}, \mathrm{C}=16$

Hence the correct answer is 16.