Question

A mixture of C2H4 and C3H8 was kept in a 0.820L vessel at 1 atm and 300 K. The weight of the gas mixture in the mixture 0.613g. Calculate the ratio of moles of C3H8 and C2H4.

Answer 1

Explanation:

For O

2

, PV=nRT

or, 1×40=n×0.0821×400

or, n=1.218

C

2

H

6

(g)+

2

7

O

2

(g)→2CO

2

(g)+3H

2

O(l)

C

2

H

4

(g)+3O

2

(g)→2CO

2

(g)+3H

2

O(l)

Let the moles of ethane be a.

Volume of O

2

required by ethane =7/2a

Volume of ethane =(1.218−a)

Volume of O

2

required by ethene=3(1.218−a)

Given,

2

7

a+3(1.218−a)=

32

130

(moles of oxygen)

or

2

7

a+3.654−3a=4.0625

or a=0.817(moles of ethane)

∴1.218−a=1.218−0.817 =0.401(moles of ethene)

Mole fraction of ethane =

1.218

0.817

=0.6707

Mole fraction of ethene=1−0.6707=0.3293