IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - A MIXTURE OF ETHANE ( C2H6) AND ETHENE (C2H4) OCCUPIES 40 LITRES AT 1 ATMOSPHERE AND 400 K

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A MIXTURE OF ETHANE ( C2H6) AND ETHENE (C2H4) OCCUPIES 40 LITRES AT 1 ATMOSPHERE AND 400 K. THE MIXTURE REACTS COMPLETELY WITH 130gm OF O2TO PRODUCE CO2 AND H2O.

ASSUMING IDEAL GAS BEHAVIOR, CALAULATE THE MOLE OF FRACTIONS OF C2H4 AND C2H6 IN THE MIXTURE.

aditya_arora04

Hi,

Look,

By PV = nRT

1*40=n*0.0821*400

So, n = 1.22 moles

Now, let moles of C2H6 be x. so for C2H4 is 1.22-x

Now, wrte the equations and find the total amt of oxygen required. equate it to 130 gms and find x. So, you have found the amt of both and hence can easily find the mole fraction of both.

joyfrancis

i'm getting the answer as 0.11. i,e mole fraction of ethane & ethene both equal to 0.11...is it right?

joyfrancis

isn't ethene supposed to be liquid waise?...we can't apply gas equations to liquids !!..haina?

iitkgp_bipin

Applying PV = nRT gives n=1.22, the no. of moles of mixture of ethane and ethene.

Let the moles of C₂H₆ be x and that of C₂H₄ be 1.22 - x.

C₂H₆ + 3.5O₂ = 2CO₂ + 3H₂O

One mole of ethane requires 3.5 mole of O₂
so x mole of ethane would require 3.5x mole of O₂

C₂H₄ + 3O₂ = 2CO₂ + 2H₂O

One mole of ethene requires 3 mole of O₂
so 1.22-x mole of ethene would require 3(1.22-x) mole of O₂

Total moles of O₂ required = 3.5x + 3(1.22-x)

Total mass of O₂ required = {3.5x + 3(1.22-x)}.32 = 130 gms (given)

this gives x = 0.805

mole fraction of ethane = x/1.22 = 0.805/1.22 = 0.66

so mole fraction of ethene = 1 - 0.66 = 0.34

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