A mixture of 620 lbmol/h of 36 mol% benzene and the remainder toluene is separated continuously into two products. One product i

Question

alukav5142 [94]

2 years ago

10

A mixture of 620 lbmol/h of 36 mol% benzene and the remainder toluene is separated continuously into two products. One product i

s 98 mol% benzene and the other is 95 mol% toluene. Calculate the flow rates of both products in kmol/h.

Chemistry

1 answer:

Licemer1 [7] · Answer 1 · 2022-09-04T08:52:04+03:00

Answer:

$P_1=206.7lbmol/h\\\\P_2=413.3lbmol/h$

Explanation:

Hello!

In this case, for the separation of the benzene-toluene mixture, we can use the following mole balances including the given mole fraction at each stream per species:

$B:0.36*620=0.98P_1+0.05P_2\\\\T:0.64*620=0.02P_1+0.95P_2$

That can be solved by using a solver for P1 (benzene-rich flow) and P2 (toluene-rich flow):

$P_1=206.7lbmol/h\\\\P_2=413.3lbmol/h$

Best regards!