`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+)` is

Options

A. `0.2`
B. `0.3`
C. `0.4`
D. `0.6`

Answer

D

Explanation

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`