The equivalent weight of $F e S_{2}$ in the following reaction is $F e S_{2} + O_{2} ⟶ {F e}^{+ 3} + S O_{2}$

\frac{M o l . w t}{1}

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\frac{M o l . w t}{7}

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\frac{M o l . w t}{11}

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\frac{M o l . w t}{9}

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Solution

The correct option is A $\frac{M o l . w t}{1}$
$F e S + O_{2} ⟶ F e^{+ 3} + S O_{2}$
Here, as $S_{2}$ has oxidation state $2 -$
$F e$ has oxidation state $2 +$
After oxidation $F e$ has oxidation state $3 +$
$∴$ No. of $e^{-}$ transferred= $3 - 2 = I e^{-}$
We know,
Equivalent weight= $\frac{M o l e c u l a r w e i g h t}{n f a c t o r}$
Here $n$ factor is no. of $e^{-}$ transferred as this is a redox reactor.
$∴$ Equivalent weight= $\frac{M o l . w t}{1}$

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Similar questions

Q. Equivalent weight of

F e S_{2}

[mol.wt = M] in the following reaction is?

F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}

Q. The equivalent weight of

K I O_{3}

in the reaction

2 C r (O H)_{3} + 4 O H + K I O_{3} \to 2 C r O_{4}^{2 -} + 5 H_{2} O + K I

Q. In acidic medium, equivalent weight of

K_{2} {C r}_{2} O_{7} (M o l . w t . = M)

is:

$K M n O_{4}$ acts as an oxidising agent in the neutral medium and gets reduced to $M n O_{2}$ . The equivalent weight of $K M n O_{4}$ in neutral medium is ______ .

Q. One litre of acidified

K M n O_{4}

containing 15.8 g

K M n O_{4}

(mol.wt=158) is decolourised by passing sufficient

S O_{2}

. If

S O_{2}

is produced by

F e S_{2}

, then which of the following options is/are correct?

(F e = 56, S = 32)

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The equivalent weight of FeS2 in the following reaction is FeS2+O2⟶Fe+3+SO2

The equivalent weight of $F e S_{2}$ in the following reaction is $F e S_{2} + O_{2} ⟶ {F e}^{+ 3} + S O_{2}$