`
$2 \mathrm{MnO}_{4}^{-}+\mathrm{b} \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{c} \mathrm{H}^{+} \rightarrow \mathrm{X} \mathrm{Mn}^{2+}+\mathrm{y} \mathrm{CO}_{2}$
$+\mathrm{z} \mathrm{H}_{2} \mathrm{O}$
If the above equation is balanced with integer coefficients, the value of c is _______. (Round off to the Nearest Integer).
Writting the half reaction oxidation half reaction
$\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}$\
balancing oxygen
$\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$
balancing Hydrogen
$8 \mathrm{H}^{+}+\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$
balancing charge
$5 \mathrm{e}^{-}+8 \mathrm{H}^{+}+\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$
Reduction half
$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{CO}_{2}$
Balancing carbon
$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{CO}_{2}$
Balancing charge
$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{CO}_{2}+2 \mathrm{e}^{-}$
Net equation
$16 \mathrm{H}^{+}+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}$
So c = 16