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Reaction with Metals

Equivalent we...

Question

Equivalent weight of ${F e}_{2} S_{3}$ in the reaction is :
(M=molecular weight of ${F e}_{2} S_{3}$ )
${F e}_{3} S_{3} + O_{2} ⟶ {F e S O}_{4} + {S O}_{2}$

\frac{M}{2}

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\frac{M}{14}

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\frac{M}{18}

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\frac{M}{20}

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Solution

The correct option is A $\frac{M}{2}$
${F e}_{2} S_{3} + 5 O_{2} ⟶ 2 F e S O_{4} + S O_{2}$
Oxidation state of $F e$ in ${F e}_{2} S_{3} = + 3$
Oxidation state of $F e$ in $F e S O_{4} = + 2$
Valence factor $= 2 \times ((+ 3) - (+ 2)) = 2 \times 1 = 2$
Given that the molecular weight of ${F e}_{2} S_{3}$ be M
$∴ Equivalent weight = \frac{mol. wt.}{valence factor} = \frac{M}{2}$

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