Q.

$2 {MnO}_{4} + {bC}_{2} O_{4} {}^{2 -}+ {CH}^{+} \to {xMn}^{2 +} + {yCO}_{2} + {zH}_{2} O$

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

(Round off to the Nearest Integer).

Moderate

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answer is 16.

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Detailed Solution

Writing the half reaction

oxidation half reaction

${M n O}_{4}^{-} \to {M n}^{2 +}$

balancing oxygen

${M n O}_{4}^{-} \to {M n}^{2 +} + 4 H_{2} O$

balancing Hydrogen

$8 H^{+} + {M n O}_{4}^{-} \to {M n}^{2 +} + 4 H_{2} O$

balancing charge

$5 e^{-} + 8 H^{+} + {M n O}_{4}^{-} \to {M n}^{2 +} + 4 H_{2} O$

Reduction half

$C_{2} O_{4} {}^{2 -}\to {C O}_{2}$

Balancing carbon

$C_{2} O_{4} {}^{2 -}\to 2 {C O}_{2}$

Balancing charge

$C_{2} O_{4} {}^{2 -}\to 2 {C O}_{2} + 2 e^{-}$

Net equation

$16 H^{+} + 2 {M n O}_{4}^{-} + 5 C_{2} O_{4} {}^{2 -}\to 10 {C O}_{2} + 2 {M n}^{2 +} + 8 H_{2} O$

$So, C = 16$

Hence the correct answer is 16.

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