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

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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