A binary 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 = D + B ... (1)

∴ 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