Combustion of carbon may be given as,
`underset(12g)underset(1"mol")(C(s))+underset(1"mol")underset(1"mol")(O_(2)(g)) to CO_(2)(g)`
`therefore "12 g of carbon required 1 mole" O_(2) "for complete combustion"`
`therefore "1000g of carbon will requires" (1)/(12)xx1000 "mole" O_(2)"for combustion i.e. 83.33 mole" O_(2)`
Volume of `O_(2)` at NTP=`83.33xx22.4"litre"=18.66.592"litre"`
`therefore21 "litre" O_(2)` is present in 100 litre air
`therefore 18.66.592 "litre" O_(2)` will be present in `(100)/(21)xx1866.592 "litre" O_(2) =8888.5"litre=8.8885xx10^(5)xx10^(3)"litre"`
