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200 " mL of " acidified 3 N `H_2O_2` is reacted with `KMnO_4` solution till there is a light tinge of purple colour. Calculate the volume of `O_2` produced at STP.

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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`

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