Question

What is the value of F for the B solution to remain 80% KOH if the recycle stream is removed?

(a) 123.1 mole/hr

(b) 666.5 mole/hr

(c) 988.4 mole/hr

(d) 1499.8 mole/hr

I have been asked this question in an online quiz.

Question is taken from Recycle without Chemical Reaction in portion Recycle, Bypass, Purge, Ideal Gases and Real Gases of Basic Chemical Engineering

Answer 1

Right option is (d) 1499.8 mole/hr

Explanation: Second reactor balance, KOH: B*0.8 = C*0.6 + 520.8*[0.9*1 + 0.1*0.6], => 4B = 3C + 2499.8, H2O: B*0.2 = C*0.4 + 520.8*[0.1*0.4], => B = 2C + 104.1, solving both equations we get B = 937.4 mole/hr. First reactor balance, KOH: F*0.5 = 937.4*0.8, => F = 1499.8 mole/hr.