InterviewSolution
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InterviewSolution
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`H_2O_2` is reduced rapidly by `Sn^(2+). H_2O_2` is decomposed slowly at room temperature to yield `O_2` and `H_2O_2`. `136g` of `10%` by mass of `H_2O_2` in water is treated with `100mL` of `3M Sn^(2+)` and then a mixture is allowed to stand until no further reaction occurs. The reactions involved are: `2H^(o+)+H_(2)O_(2)+Sn^(2+)toSn^(4+)+2H_(2)` `2H_(2)O_(2)to2H_(2)O+O_(2)` The equivalent of `H_2O_2` reacted with `Sn^(2+)` isA. `0.2`B. `0.3`C. `0.4`D. `0.6` |
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Answer» Correct Answer - D The problem can be solved simply by mol concept since n factor for both `H_(2)O_(2)` and `Sn^(2+)` is 2. `2H^(o+)+underset(1 mol)(H_(2)O_(2))+underset(1mol)(Sn^(2+))toSn^(4+)+2H_(2)O` m" mol of "`H_(2)O_(2)=m" mol of "Sn^(2+)` `=100mLxx3M=300mmol` `=300m" mol of "H_(2)O_(2)` `=300xx2` (n factor) `mEq H_(2)O_(2)` `=600mEq=0.6eq` |
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