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Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many moles of KI are required for one mole of `K_(2)Cr_(2)O_(7)` ? |
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Answer» Correct Answer - `(6)` `[Cr_(2)O_(7)^(2-)+14H^(o+)+6e^(Θ)rarr 2Cr^(3+)+7H_(2)O]` `2I^(Θ)rarr I_(2)+2e^(Θ)]xx3` `ulbar(Cr_(2)O_(7)^(2-)+14H^(o+)+61^(Θ)rarr 2Cr^(3+)+7H_(2)O+3I_(2)` 1 mol of `K_(2)Cr_(2)O_(7)` reacts with 6 moles of KI. |
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