1.

Calculate the volume of `O_(2)` and volume of air needed for combustion of `1 kg` carbon at `STP`.

Answer» `underset(12 g)(C) + underset(32 g)(O_(2)) rarr CO_(2)`
`12 g` of `C implies g` of `O_(2) = 1 mol` of `O_(2) = 22.4 L` at `STP`
`1000 g` of `C = (22.4 xx 1000)/(12) = 1866.67 L "of" O_(2)`
Air contains `= (20 % "of" O_(2) + 80% "of N_(2))`
Volume of air = 5 times volume of `O_(2) = 5 xx 1866.67`
`= 9333.33 L`


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