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Equivalent weight of $F e S_{2} [M o l . w t . = M]$ in the following reaction is $F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}$

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$\frac{M}{14}$

$\frac{M}{4}$

$\frac{M}{11}$

$\frac{M}{2}$

answer is C.

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Detailed Solution

${Fe}^{+ 2} S_{2}^{- 2} \to {Fe}^{+ 3} O^{- 2} + S^{+ 4} O^{- 2}$

Total change in Oxidation number of FeS₂

$= 1 (+ 1) + 2 (+ 5) = + 11$

Equivalent weight of FeS₂ $= \frac{M}{11}$

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