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Equivalent weight of FeS₂ in the half reaction, FeS₂→ Fe₂O₃ + SO₂ is :

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M/6

M/11

M/1

M/10

answer is B.

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Detailed Solution

${FeS}_{2} + O_{2} \to {Fe}_{2} O_{3} + {SO}_{2}$

Balanced reaction is

$4 {FeS}_{2} + 11 O_{2} \to 2 {Fe}_{2} O_{3} + 8 {SO}_{2}$

1 mole of ${FeS}_{2}$ =11/4 moles of O₂

Each O− atom takes 2e⁻for reduction

∴1 O₂molecule takes 4e−

NO. of electrons gained by O₂ = $4 \times \frac{11}{4}$ =11

∴ Number of e⁻lost by FeS₂=11

∴ Equivalent wt of FeS₂ in this reaction= $\frac{M}{11}$

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