`H_(2)O_(2)` is reduced rapidlly by `Sn^(2+)` to give `H_(2)O and Sn^(+)H_(2)O_(2)` is decomposed slowly at room temperature so yield `O_(2) and H_(2)O`. 136g of % by mass of `H_(2)O` in water treated with 100 mL `3 M 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+) to Sn^(2+)+2H_(2)O: 2H_(2)O+O_(2)`
The valume strength of `H_(2)O_(2)` left after reacting with `Sn^(2+)` is

Options

A. `1.12V`
B. `11.2V`
C. `2.24V`
D. `22.4V`

Answer

B

Explanation

Intial no.of GEW.s of `H_(2)O_(2)=(13.6)/(17)=0.8`
no.of GEW.s of `Sn^(+2)=6xx0.1=0.6`
no.of GEW.s of `H_(2)O_(2)` left over `=0.2`
no.of GEW.s of `H_(2)O_(2)` consumed `=0.6`
no.of `H_(2)O_(2)` left `=0.2xx(1000)/(100)2N`
Vol strength `=11.2`