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
Answers
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