Abinary feed mixture contains 40 mol% hexane (a) and 60 mol% toluene (b) is to be separated continuously into two products d (distillate) and b (bottoms) in a distillation unit. distillate d is 90 mol% hexane and the bottoms b is 90 mol% toluene. using a feed flow rate of 100 lbmoh as basis, compute the flow rates of products b and d in: (a) lbmol/h, and (b) kmol/h.

a) D = 33.44 Lbmol/h

⇒ B = 62.56 Lbmol/h

b) D = 16.848 Kmol/h

⇒ B = 28.152 Kmol/h

Explanation:

global balance:

∴ F = 100 Lbmol/h

balance per component:

A: 0.4*F = 0.9*D + 0.1*B = 0.4*100 = 40 Lbmol/h(2)

B: 0.6*F = 0.1*D + 0.9*B = 0.6*100 = 60 Lbmol/h(3)

from (2):

⇒ 0.9*D = 40 - 0.1*B

⇒ D = ( 40 - 0.1*B ) / 0.9(4)

(4) in (3):

⇒ 0.1*((40-0.1*B)/0.9) + 0.9*B = 60

⇒ B = 62.56 Lbmol/h(5)

(5) in (1):

⇒ D = 100 - B

⇒ D = 37.44 Lbmol/h

∴ Lbmol = 0.45 Kmol

⇒ B = 62.56 Lbmol/h * ( 0.45 Kmol/ Lbmol ) = 28.152 Kmol/h

⇒ D = 37.44 Lbmol/h * ( 0.45 Kmol/h ) = 16.848 Kmol/h

it is stable as it is in group 0 which have full outer shells of electrons