Question

`H_2O_2` is reduced rapidly by `Sn^(2+). H_2O_2` is decomposed slowly at room temperature to yeild `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+)` is
A. `0.2`
B. `0.3`
C. `0.4`
D. `0.6`

Answer 1

Correct Answer - D
The problem can be solved simply by mole concept since `n` factor for both `H_(2)O_(2)` and `Sn^(2+)is 2`.
`2H^(o+)+underset( "1 mole")(H_(2)O_(2))+underset(1 mol)(Sn^(2+))toSn^(4+)+2H_(2)O` ltMbrgt mmol of `H_(2)O_(2)`= mmol of `Sn^(2+)` ltbRgt `=100 mLxx3M=300 mmol`
`=300mmol of H_(2)O_(2)`
`=600 m Eq=0.6 Eq`