$n$ factor of $F e C_{2} O_{4}$ during its oxidation by acidified $K M n O_{4}$ is:

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Solution

$F e^{+ 2} \to F e^{+ 3}$
n-factor=1
${C_{2} O_{4}}^{- 2} \to C O_{2}$
n-factor= $2 \times + 1$
=2
n-factor= 1+2
=3

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