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

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