20 ml of a mixture of methane and a gaseous compound of acetylene series were mixed with 100 mol of oxygen and exploded to complete combustion. The volume of the products after colling to original room temperature and pressure, was 80 ml and on treatment with potash Sol. a further contraction of 40 ml was observed. Calculate (a) the molecular formula of the hydrocarbon, (b) the percentage composition of the mixture.

20 ml of a mixture of methane and a gaseous compound of acetylene seri

1.	20 ml of a mixture of methane and a gaseous compound of acetylene series were mixed with 100 mol of oxygen and exploded to complete combustion. The volume of the products after colling to original room temperature and pressure, was 80 ml and on treatment with potash Sol. a further contraction of 40 ml was observed. Calculate (a) the molecular formula of the hydrocarbon, (b) the percentage composition of the mixture.
Answer» SOLUTION :Oxygen is PRESENT in excess. This is because the combustion has been done completely (which means methane and the other gas are consumed completely) and on `CO_(2)` absorption 40 ml of a gas is still left over. This let over gas is oxygen. `therefore` Oxygen used up =100-40=60 ml `CO_(2)` produced =40 ml If the molecular formula of the gas of acetylene series is `C_(N)H_(2n-2)` and its volume is a ml, then the volume of `CH_(4)` is (20-a) ml. On combustion `CH_(4)(g) +2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l)` `C_(n)H_(2n-2)(g)+((3n-1))/(2)O_(2)(g) rarr nCO_(2)+(n-1)H_(2)O(l)` Oxygen used up `=2(20-a) +((3n-1))/(2)a=60` (from `CH_(4)`) (from `C_(n)H_(2n-2)`) `CO_(2)` produced =(20-a)+na =40 (from `CH_(4)`) (from `C_(n) H_(2n-2)`) Solving a=10 ml and n=3 `therefore` Formula of the gas is `C_(3)H_(4)` % of `C_(3)H_(4)=(10)/(20)xx100=50%` % of `CH_(4)=50%`