$F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O$
The $n -$ factor for $F e C_{2} O_{4}$ in the above reaction is:

2

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3

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2.5

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1

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Solution

The correct option is B $3$
The balanced chemical equation is:
$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 20 C O_{2} + 6 M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O$
$n-factor of an ion in case of redox change = Change in oxidation state \times Number of atoms undergoing this change per molecule$
$F e^{2 +} \to F e^{3 +} + e^{-}$
$n$ factor for $F e^{2 +}$ to $F e^{3 +} = 3 - 2 = 1$
$C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-}$
$n$ factor for $C_{2} O_{4}^{2 -}$ to $C O_{2}$
$= | 4 - 3 | \times 2 = 2$
Overall $n$ factor is $2 + 1 = 3$ .

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Similar questions

F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O

K M n O_{4} + H 2 S O_{4} + F e S O 4 \to K 2 S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O

K M n O_{4} + H_{2} S O_{4} + F e S O_{4} \to K_{2} S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O

F e S O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O + K_{2} S O_{4}

K M n O_{4} + H_{2} S O_{4} + K_{2} C_{3} O_{4} \to M n S O_{4} + C O_{2} + K_{2} S O_{4}

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