A gas mixture of 3.0 L of propane and butane on complete combustion at `27^(@)C` produced 10.0 L of `CO_(2)`. Find out the composition of the gas mixture.

Explanation

`C_(3)H_(8)+5O_(2)rarr 3CO_(2)+4H_(2)O`
`C_(4)H_(10)+6(1)/(2)O_(2)rarr 4CO_(2)+5H_(2)O`
Suppose propane = xL. Then butane = (3-x) L
`1 L C_(3)H_(8)` produces 3 L of `CO_(2)` and 1 L of `C_(4)H_(10)` produced 4 L of `CO_(2)`
Hence, `CO_(2)` produced `=3x +4(3-x)=12-x=10 L`
or x = 2 L
`therefore` Propane = 2 L and butane = 3 - 2 = 1 L.