`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`. `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` left after reacting with `Sn^(2+)` is

Options

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Answer

B

Explanation

" mol of "`H_(2)O_(2)` intially present `=136gxx(10)/(100)xx(1)/(34)`
`=0.4 " mol of "H_(2)O_(2)`
`" mol of "H_(2)O_(2)` left`=(" mol of "H_(2)O_(2)` initially present `-` " mol of "`H_(2)O_(2)` reduced by `Sn^(2+)`)
`=0.4-0.3=0.1 " mol of "H_(2)O_(2)` left
Equivalent of `H_(2)O_(2)` left `=0.1xx2=(0.2eq)/(100)mL`