Question

A cylinder containing CH4, C2H6, and N2 has to be prepared in which the ratio of the moles of CH4 to C2H6 is 1.3 to 1. Available are (1) a cylinder containing a mixture of 70% N2 and 30% CH4, (2) a cylinder containing a mixture of 90% N2 and 10% C2H6, and a cylinder of pure N2. Determine the proportions in which the respective gases from each cylinder should be used

Answer 1

Answer:

The proportions in which the respective gases from each cylinder should be used; C1 : C2 : C3 = 1 : 2.308 : C3

Explanation:

Given that;

CH4 to C2H6 =  1.3 to 1

Cylinder 1 =  70% N2 and 30% CH3

Cylinder 2 = 90% N2 and 10% C2H6

Cylinder 3 = ( pure N2 ) = 100% N2

CH4 to C2H6 =  (1.3 to 1)

we have CH4 and C2H6 in cylinder 1 and 2 only

we only have balance material in c1 and c2

Now, Let x be from cylinder and ( 1-x ) be from cylinder 2

so, 30x / 10( 1-x ) = 1.3 / 1

30x = (10 - 10x)1.3

30x = 13 - 13x

30x + 13x = 13

x = 13/43

x = 0.3023

Also, ( 1-x ) = ( 1 - 0.3023) = 0.6977

so

C1/C2 = 0.3023/0.6977 = 1 / 2.308   while c3 can be anything;

Therefore, C1 : C2 : C3 = 1 : 2.308 : C3