Question

15 ml of mixture of C2H4 & CH4 on complete combustion with excess oxygen gives 20 ml CO2. The volume of C2H4 in the original mixture is

Answer 1

Answer:

53.33% CH

4

and 46.67% C

2

H

4

Explanation:

we have,

CH

4

+O

2

→CO

2

+2H

2

O

and, C

2

H

4

+O

2

→2CO

2

+2H

2

O

if they are producing equal volume of CO

2

gas, that means moles of CH

4

is double the moles of C

2

H

4

let, moles of CH

4

= 1, and mass becomes = 16 g

and moles of C

2

H

4

= 0.5 and mass becomes = 14 g

so, percentage composition of C

2

H

4

=

30

14

∗100 = 46.6 %

percentage composition of CH

4

=

30

16

∗100 = 53.33%