A liquid mixture containing 40.0 wt% (by mass) benzene (B) and 60.0 wt% toluene (T) is fed to a distillation column. A product stream leaving the top of the column (the overhead product) contains 90.0 wt% B, and 95% of B fed to the column is recovered in the overhead. The volumetric flow rate of the feed stream is 2000 L/h and the specific gravity of the feed mixture is 0.872. Determine the flowrates of all components in all streams using our 9-step process.

Flow rate of benzene = 706. 22kg/h

Flow rate of toluene = 1003 kgT/h

Explanation:

m1 = (2000 ltr/h) × (0.872 kg/ltr) = 1744 kg/h

Since 95% is recovered in the overhead, it means that the bottom stream contained 5% of benzene. Thus;

M(b3) = (0.05) (0.4m1) = 0.05 x 0.4 x 1744 = 34.88 kgB/h

Now in overhead product;

Basis : 100 kmol overhead

90 kmol B, 10 kmol T

Molar masd of benzene is approximately 78g/mol

Benzene = 90 kmolB × 78 kgB/kmolB = 7020 kgB

Molar mass of toluene = 92g/mol.hence,

Toluene = 5 × 92 = 461 kg T

Y(b2) = 7020/(7020 + 461) = 0.9384

Benzene balance:

0.4 m1 = m2 (yB2) + mB3

0.4 × 1744 = m2 (0.9384) + 34.88

⇒ m2 = 706. 22kg/h

Toluene balance:

0.6 m1 = (1-yB2) m2 +mT3

0.6 × 1744 = {(1-0.9384) × 706. 22} +mT3

So, mT3 = 1003 kgT/h

high-sugar diet and high cholesterol