Number of moles of electrons produced per mole of
$F e C r_{2} O_{4}$ in the following redox reaction.
$F e C r_{2} O_{4} + K O H + O_{2} \to K_{2} C r O_{4} + F e_{2} O_{3}$

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Solution

$F e$ and $C r$ both are getting oxidised.
$F e$ oxidation state is changing from $+ 2$ (in $F e C r_{2} O_{4}$ ) to $+ 3$ (in $F e_{2} O_{3}$ ) and
$C r$ from $+ 3$ (in $F e C r_{2} O_{4}$ ) to $+ 6$ (in $K_{2} C r O_{4}$ )

$F e^{2 +} \to F e^{3 +} + e^{-}$
$C r_{2}^{6 +} \to 2 C r^{3 +} + 6 e^{-}$
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$F e C r_{2} O_{4} \to F e_{2} O_{3} + K_{2} C r O_{4} + 7 e^{-}$

So 7 moles of electrons are produced per mole of $F e C r_{2} O_{4}$

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