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

Answer

B

Explanation

Initial no. of GEW.s of `H_2O_2 = 13.6/17 = 0.8`
no. of GEW.s of `Sn^(+2) = 6xx 0.1 =0.6`
no. of GEW.s of `H_2O_2` lift over = 0.2
no.of GEW.s of `H_2O_2`consumed = 0.6
no. of `H_2O_2` left =`0.2 xx 1000/100 = 2N`
Vol. strength = 11.2