Question

A mixture of ethane and ethene occupied 41 L at 1 atm and 500 K. The mixture reacts completely with
10
3
mol of O2
to produce CO2
and H2O. The number of moles of ethane in the mixture is

Answer 1

Answer:

the mole fraction of ethane is 0.66 & ethene is 0.33 in the mixture.

Explanation:

formula of ethane = C2 H6 C2 H6

formula of ethene = C2 H4 C2 H4

total mole of (C2 H6 + C2 H4) = 1. 2195

PV = nRT

1 × 41 =n×0.082×500

n1+n2=1---------- 1( eq.)

C2 H6 + 7/2 C2 H6 __2 CO2+ 3 H2O

C2 H4 + 3O2__2CO2+2H2O

7/2n1+3n2= 10/3--------2(eq.)

on solving 1 and 2

n1=2/3and n2=1/3

n1=2/3

=0.67 (approx.)

n2=1/3

=0.33(approx.)

therefore, the mole fraction of ethane and ethene in the mixture is 0.67 and 0.33 respectively.

hope this helps!!!!!

thanks