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The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. 3B. `1//6`C. 6D. `1//3` |
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Answer» Correct Answer - D `{:(Cr_(2)O_(7)^(2-),rarr,Cr^(3+),Sn^(2+),rarr,Sn^(4+)),(Cr(+6),,Cr(+3),Sn(+2),,Sn(+4)):}` One formula unit of dichromate ion contains two Cr atoms. Thus, total change in oxidation number is of 6 units `(+12` to `+6)` . Thus, 1mole of `Cr_(2)O_(7)^(2-)` contains 6 equivalents. For tin, the oxidation number change by 2 units. Thus, 1 mole of `Sn^(2+)` ion contains 2 equivalents. According to law of equivalents, two equivalentsof `Sn^(2+)` ion will only reduce 2 equivalents of `Cr_(2)O_(7)^(2-)`. 6 equivalents of `Cr_(2)O_(7)^(2-)` come form 1 mole 1 equivalent will come form `(1)/(6)` mole 2 equivalents will come form `(2)/(6)` mole i.. `1//3` mole. |
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