A process stream contains 90 mole % of desired liquid chemical A and 10 mole % o

Question

A process stream contains 90 mole % of desired liquid chemical A and 10 mole % of a contaminant C. It is proposed to remove most of contaminant C from liquid A using liquid-liquid extraction as shown in the figure below. The process stream is mixed in an extractor with a liquid B (stream 2) that is immiscible (doesn't mix) with liquid ?. However, contaminant C is soluble in both liquids ? and B. In the extractor, some of the contaminant C moves from liquid A to liquid B and exits in stream 3. The remainder exits with liquid ? in stream 4. Since ? and B don't mix, there is no A in stream 3 and no B in stream 4. Assuming equilibrium is reached in the extractor, the relation between the mole fractions of contaminant C in the two exit streams is given by: x_c, 3/x_c, 4 = K where, for this particular set of chemical species, K = 4. If the flow rate of the process stream is 100 moles/hr and it is desired to remove 80% of the contaminant C present in that stream, determine the required flow rate (moles/hr) of liquid B in stream 2.

Explanation / Answer

Stream-4 contains all the A and stream-3 contains all the B. Let flow rate of B x mole/hr

All the B will be in stream-3. Stream-3 contains x moles/hr of stream B. Stream-4 contains 100*0.9= 90 moles/hr of A. C in feed =10 moles/hr. It is desired to remove 80% of C.   C that needs to be removed = 8 moles/hr. C in stream-3 = 8 moles/hr and C in stream-4 = 2 moles/hr

Mole fraction of C in stream-3 = 8/(x+8) and mole fraction of C in stream-4= 2/(2+90)=0.0217

Hence 8/(x+8)/0.0217= 4

8/(x+8)= 4*0.0217, 8= 0.0217*(x+8), 8= 0.0217*x+0.0217*8,   8*(1-0.0217)= 0.0217x, x=361moles/hr

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