16cm cube of a gaseous hydrocarbon were mixed with 90cm cube of oxygen. After explosion and cooling, the volume was 66cm cube. It was reduced (to 34cm cube of residual oxygen) by absorption by KOH solution. What is the molecular formula of the hydrocarbon?

Answer 1

CxHy + O2 ==> CO2 + H2O
16 cc....90 cc.............................
90 cc = O2 initially
34 cc = excess O2
---------
32 cc = CO2 formed
90 O2 initially - 34 xs O2 = 56 cc O2 used so
CxHy + O2 ==> CO2 + H2O
16 cc....56 cc......32..............

56/16 = 3.5 so coefficient for O2 is 7/2 or 3.5.
32 CO2/16 CxHy = 2 so coefficient for CO2 is 2. Let's update
CxHy + 3.5O2 ==> 2CO2 + H2O
H2O is left. If we have 3.5 O2 and we use 2 O2 from 2CO2, that leaves 1.5 CO2 for the water. 1.5 O2 = 3 O so the coefficient for H2O is 3 and a new update is
CxHy + 7/2 O2 ==> 2CO2 + 3H2O
Now finish balancing. x must be 2 and y must be 6 so CxHy must be C2H6 or ethane.