`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`. `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)`
Calculate the volume of `O_2` produced at `27^@C` and `1` atm after `H_2O_2` is reacted with `Sn^(2+) and the mixture is allowed to stand.

Options

A. 2.46 L
B. 4.92 L
C. 1.23 L
D. 7.38 L

Answer

A

Explanation

`underset(2 mol)(2H_(2)O_(2))to2H_(2))+underset(1 mol)(O_(2))`
`2 " mol of "H_(2)O_(2) gives 1 " mol of "O_(2)`
`0.1 " mol of "H_(2)O_(2) gives 0.1 " mol of "O_(2)`
`V=(nRT)/(P)=(0.1xx0.082xx300)/(1)=2.46L`