Question

HW3 Differential Distillation of Benzene-Toluene. A mixture containing 70 mol % benzene and 30 mol % toluene is distilled under differential conditions at 101.32 kPa (1 atm). A total of one third of the moles in the feed is vaporized. Calculate the average composition of the distillate and the composition of the remaining liquid. Use equilibrium data in Table 11.1-1.

Answer 1

Sobuollona Let Loo mol feed. F= 100mol, D= {xF = F x100mles = 100 moh. # F=D+W 7 W=FFD = (100-109 ) mol = 290 mol >p = 0.70 (70mol % Berzene) from eqfurelibolun data we can find belatne kolability (w foom egelton Table 11:1-1 1 0.456 = x x 0.258 1+ 6-1)0.298 0.456 + 0.117648 C -0.117648= 0.2580 0.78 0.900 0.581 0.777 0-1403520C = 0.338352 #[ QC = 2.411 Abfros 0.411 0-258 0.130 0.632 0-456 0.261 # We Know log ( 7 ) = log ( 12 ) 1-ZE) i 100x (1-0.70) 7 longe than a loyat xw22422 Taking Antslog both sides.

2W 0.146 12.411 2.411 Palu 1,05 = (0.45)2-411 (2-Xw)2.441 - 4-w)2.411 By hit & total method #xw = 0.635] Affoox Ghana #compositron of remaining leqpuid. # Applying Componente Balance, FZq = DXg+ Waw 100x 0.4 = 1ou xxy + 290 x 0635 4,00 * = 70- 127. = 210-127 = 89 x = 0.83 comportementet dhewerty xp = 0.83] Afwer composttion of dostallate.

I have used actual table 11.1-1 which is for benzene and toluene at 1 atm.

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