Equivalent mass of $F e S_{2}$ in the following reaction ( $M :$ Mol Mass of $F e S_{2}$ ) is:
$F e S_{2} + O_{2} \to F e S O_{4} + S O_{2}$

\frac{M}{12}

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\frac{M}{11}

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\frac{M}{6}

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\frac{M}{4}

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Solution

The correct option is D $\frac{M}{11}$
$F e S_{2} + O_{2} ⟶ F e_{2} O_{3} + S O_{2}$
Balanced reaction is
$4 F e S_{2} + 11 O_{2} ⟶ 2 F e_{2} O_{3} + 8 S O_{2}$
$1$ mole of $F e S_{2} = 11 / 4$ moles of $O_{2}$
Each $O -$ atom takes $2 e^{-}$ for reduction
$∴ 1 O_{2}$ molecule takes $4 e^{-}$
$∴$ Number of $e^{-}$ gained by $O_{2} = \frac{11}{4} \times 4 = 11$
$∴$ Number of $e^{-}$ lost by $F e S_{2} = 11$
$∴$ Equivalent weight of $F e S_{2}$ in this reaction= $\frac{M}{11}$

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