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

\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 $\frac{M o l . w t}{1}$
Here the Oxidation state of $F e$ is changed from $+ 2$ to $+ 3$ .
Therefore, net change is $1$ .
Hence, Equivalent weight $= \frac{M o l . w t}{1}$

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F e S_{2}

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

F e S_{2}

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

K M n O_{4}

K M n O_{4}

S O_{2}

S O_{2}

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(F e = 56, S = 32)

K I O_{3}

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

{C u}_{2} S

(m o l . w t . = M)

{C u}_{2} S + 4 {M n O}_{4} ⟶ {C u}^{2} + {S O}_{2} + {M n}^{2 +}

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