1 N of `H_2O_2=5.6` volume of `O_2`
3 N of `H_2O_2=3xx5.6=16.8` volume of `O_2` (volume strength of `H_2O_2`)
Volume of `O_2` produced by `H_2O_2` at STP
`=` Volume of `O_2xx` volume strength of `H_2O_2`
`=200 mL xx16.8=3360mL`
Same volume will be produced by `KMnO_4=3360mL`
Total volume of `O_2=3360=6720mL=6.72L`