Q.79..

Q79. An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is :

   (a) 0.6                           (b) 0.4                     (c) 0.5                         (d) 0.8 
 

Dear Student

As given that the mixture of ethane and ethene occupies 29 litre at 1 atm and 273 K. 
And the mixture reacts completely with 128 g of O2 to produce CO2 and H2O. 
Total amount of mixture of ethane and ethene = n = PVRT= 1 ×280.0831 × 273= 1.234 mol

Amount of oxygen used in complete combustion = 12832= 4 mol

The reaction of the ethane with oxygen:

C2H6 +72O2 2CO2 + 3H2OAccording to stoichiometry 1 mole requires 7/2 mole of oxygen or 112 g of oxygen 

The reaction of the ethene with oxygen:

C2H4 +3O2 2CO2 + 2H2OAccording to stoichiometry 1 mole of ethene requires 3 mole of oxygen or 96 g of oxygen 

So if n1 and n2 are the amount of C2H4 and C2H6 n1 +n2 = 1.234 mol ........(1)3 n1 +7/2 n2 = 4 mol  .......(2)Multiply eq (1)by 3 we get 3n1 +3n2 = 3.702 mol  ..........(3)Subtracting eq 3 from 2 we get 0.5 n2 = 0.298n2 = 0.596 molSolving for n1 we get :n1 = 1.234 - 0.596       = 0.638 mol 
So, the mole fraction of C2H6 : 

= 0.6380.638 + 0.596= 0.6381.234= 0.517

So, the correct answer is (c) 

Regards

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