(Solute) A 90 10 E, E2 Ks 80 20 2 70 30 R2 60 R 40 Plaît 50 point 50 Feed F = 250 kg Solvent S = 100 kg 40 1 60 30 70 XF

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Solute A 90 10 E E2 Ks 80 20 2 70 30 R2 60 R 40 Plait 50 Point 50 Feed F 250 Kg Solvent S 100 Kg 40 1 60 30 70 Xf 1 (135 KiB) Viewed 108 times

(Solute) A 90 10 E, E2 Ks 80 20 2 70 30 R2 60 R 40 Plaît 50 point 50 Feed F = 250 kg Solvent S = 100 kg 40 1 60 30 70 XF, A = 0.24 XF,C = 0.76 XF,s = 0.00 XS, A = 0.0 XS,C = 0.0 Xs, s = 1.0 20 80 10 90 (Solvent) C(Carrier) 90 80 70 60 50 40 30 20 10 In the counter current extraction process of the second stage as shown in the figure, we want to operate so that the concentration (XR2,A) of solute A in the raffinate of the second stage is less than 0.05. (Construct the triangle on the diagram upper, and explain the reason in the order of construction.) 1. What is the concentration (YE1,A) of solute A in the extract (El) in the first stage? 2. What is the concentration (Y£2,A) of solute A in the extract (E2) in the second stage?