A mixture of 5 ml of gaseous hydrocarbon with 12 oxygen was placed in a eudiometer and detonated.

A mixture of 5 ml of gaseous hydrocarbon with 12 oxygen was placed in a eudiometer and detonated. after bringing the conditions back to the original, the volume of the gas mixture was 7 ml. and after it was passed through the alkali solution, it decreased to 2 ml, with the remaining gas supporting combustion.

Given:
V (CxHy) = 5 ml
V (O2) = 12 ml
V mixture2 = 7 ml
V mixture3 = 2 ml

To find:
CxHy -?

Solution:
1) (a) CxHy + (4x + a * y / 4) O2 => (a * x) CO2 + (a * y / 2) H2O;
CO2 + 2KOH => K2CO3 + H2O;
2) V rest. (O2) = V mixture3 = 2 ml;
3) V (CO2) = V mixture2 – V mixture3 = 7 – 2 = 5 ml;
4) V react. (O2) = V (O2) – V rest. (O2) = 12 – 2 = 10 ml;
5) (a): (4x + a * y / 4): (a * x) = V (CxHy): V (O2): V (CO2) = 5: 10: 5 = 1: 2: 1;
a = 1;
a * x = 1;
x = 1;
((4x + a * y) / 4) = 2;
((4 * 1 + 1 * y) / 4) = 2;
y = 4;
6) Unknown substance – CH4.

Answer: Unknown substance – CH4.