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Equivalent weight of `FeS_(2)` in the half reaction `FeS rarr Fe_(2)O_(3) + SO_(2)` is :A. `M//10`B. `M//11`C. `M//8`D. `M//7` |
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Answer» Correct Answer - D `S^(2-) rarr S^(4+) + 6e` `2Fe^(2+) rarr (Fe^(3+))_(2) + 2e` `FeS Fe_(2)O_(3) + SO_(2)` involves a change of `7` electrons per mole. |
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