`H_(2)O_(2)` is marked `22.4` volume. How much of it is required to oxidise `3.5 g H_(2)S` gas?

Answer

7

Explanation

`H_(2)O_(2)` is marked `22.4` volume, which means `1 "litre" H_(2)O_(2)` gives `22.4 "litre"` of `O_(2)` at `NTP`.
`underset(=68 g)underset(2 "mole")(2H_(2)O_(2))rarr2H_(2)O+underset(22400 mL "at" NTP)(O_(2))`
`because 22400 mL` of `O_(2)` is obtained from `=68 g H_(2)O_(2)`
`therefore 22.4 mL` of `O_(2)` is obtained from
`=(68xx22.4)/(22400)=0.068g H_(2)O_(2)`
This amount is present in `1 mL H_(2)O_(2)` solution
`underset(68 g)(H_(2)O_(2))+underset(34 g)(H_(2)S)rarr2H_(2)O+S`
`because 34 g H_(2)S` is oxidised by `=68g H_(2)O_(2)`
`therefore 3.5 gH_(2)S` is oxidised by `=(68xx3.5)/(34)=7gH_(2)O_(2)`