Correct Answer - A
Volume of `O_(2)` at STP`=100mLxx11.2` volume strength ltBrgt `=1120 " mL of " O_(2) at STP`
(Since `1N=5.6` volume strength, `2N=11.2` volume strength of `H_(2)O_(2))`
Volume of `O_(2)` produced by `H_(2)O_(2)=1120mL`
Same volume of `O_(2)` will be produced by `KMnO_(4)=1120mL`
Total volume of `O_(2)=2240mL=2.24L`