Question

a definite amount of gaseous hydrocarbon was burnt in just sufficient amount of O2. The volume of all reactants was 600 ml. after the explosion the volume of the products co2 (g) and h2o (g) was found to be 700 ml under similar conditions. the molecular formula of compound is
a) c3h8
b) c3h6
c) c3h4
d) c4h10

Answer 1

The general standard chemical equation for Hydrocarbon combustion is :

CxHy +(x+y/4)CO2--->XCO2 +y/2 H2O

we need to find the values of x and y in the above equation.
Let us use the ideal gas equation:PV=nRT
let us say n1  be the total number of moles of gas present on reactant side .PV1=n1RT---------(1)
let us say n2  be the total number of moles of gas present on product side .PV2=n2RT---------(2)
Dividing equation 1 by 2
 we getPV1/pV2=n1RT/n2RT
Therefore,V1/V2=n1/n2n1
= totla number of moles present on reactant side
=1+(x+y/4)n2
= total number of moles present on product side
=x+y/21+(x+y/4)/x+y/2   = 600/700
1+x+y/4=6x=5-y/4x+y/2=7
5-y/4+y/2=7
5-(y-2y/4)=7
5+y/4=7y/4=2y=8
then x=5-8/4=20-8/4=12/4=3
Therefore Molecular formula of Unkown Hydrocabon is C3H8 --Propane
Option A is correct Answer.