1000 kmol of an equimolar mixture of benzene and toluene is distilled in a multi

Question

1000 kmol of an equimolar mixture of benzene and toluene is distilled in a multistage baicil distillation unit, 90 % of the benzene is in the top product (distillate). The top product has a benzene mole fraction of 0.95. Describe the chemical process using a labeled flowchart, and demonstrate that the stream is completely labeled? Carry out the degree-of-freedom analysis and demonstrate that you have enough information to calculate all unknown variables on the completely labeled flow chart. 1. 2. 3. In case you have all the required information, caleulate the quantities of top and bottom products and the composition of the bottom product. In case you don't have enough information, suggest a reasonable calculation basis.

Explanation / Answer

In btch distillation, material is added one time and and then process is carried out and again and again batch cycle repeated. So in these cases total mass entering in the system is equal to the output of the system. They follow the law of conservation of mass.

Input + Generation = Output + Consumption.

Flowchart:

1000 kmol Benzene + Toulene ---------------I-

                                                            I                                                           Top Product Benzene 90%

                                                                                                                       Bottom Product Toulene

Now according to degree of freedom analysis,

We have total volume of mixture, Mole fraction of one component and its percentage in top product. We have to calculate the top and bottom product quantities and percentages in bottom product. We have enough information to solve this equation. So degree of freedom is 0.

Firstly we need to calculate the molecular weight of benzene (C6H6)

Molecular weight = 12 X 6 + 1 X 6 = 72 + 6 = 78 g

1 kmol = 78 Kg

As we know 1000 Kmol solution is present i.e, 500 kmol of benzene and 500 kmol of toulene

1 kmol = 78 Kg

500 kmol = 78 X 500 = 39000 Kg

Now Mole fraction of benzene = Moles of benzene / total moles

0.95 = 39000 / total moles

Total moles = 39000/ 0.95 = 41052.63 Kg

In top product, 90% of benzene is present so

Amount of benzene = 41052.63 X 90 /100 = 36947.37 Kg of benzene in top product

And amount of toulene in bottom = 41052.63 - 36947.37 = 4105.26 Kg

Distillation

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.