Question

A definite amount of gaseous hydrocarbon was burnt with 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 the similar conditions. The molecular formula of the compound is

Answer 1

Answer:

Explanation:

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