Ethane burns in oxygen according to the following equation:
`2C_(2)H_(6)(g) + 7O_(2)(g) rarr 4CO_(2)(g) + 6H_(2)O(I)`
`2.5 L` of ethane are burnal in excess of oxygen at `27^(@)C` and 1 bar pressure. Calculate how many litre of `CO_(2)` are formed at `50^(@)C` and `1.5` bar.

Explanation

The equation for the combustion reaction is :
`underset(2vol("Litre"))(2C_(2)H_(6)(g)) + 7O_(2)(g) + underset(4vol("Litre"))(4CO_(2(g)) + 6H_(2)O(l)`
Under the given conditions i.e., at `27^(@)C` and 1 bar pressure
`2L` of ethane evolve `CO_(2)(g) = 4L`
`2.5 L` of ethan evolve `CO_(2)(g) = (4L) xx ((2.5 L)/(2L)) = 5.0L`
The volume of `CO_(2)(g)` at `50^(@)C` and `1.5` bar pressure can be pressure can be calculated with the help of gas equation.
`V_(1) = 5.0L , V_(2) = ?`
`P_(1) = 1` bar , `P_(2) = 1.5` bar
`T_(1) = (27+273) = 300 K` , `T_(2) = 50+273 = 323 K`
According to gas equation : `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(P_(2)T_(1))`
Substituting the values : `V_(2) = ((1 "atm" ) xx (5.0 L) xx (323 K))/((1.5 "atm") xx (3.00 K)) = 3.59 L`