Elementary Principles 222 - 6.75 (cont'd) Analysis of feed:...

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6- 56 6.75 (cont’d) Analysis of feed: 2KOH H SO K SO 2H O 24 2 +→ + x 0 0 427 = = 22.4 mL H SO l 1 L 0.85 mol H SO 2 mol KOH 56.11 g KOH 5 g feed soln 10 mL L 1 mol H SO 1 mol KOH g KOH g feed 3 bg . 60% recovery: ( )( ) 875 0.427 0.60 224.2 kg KOH h = () 2 3 22 224.2 kg KOH 92.15 kg KOH 2H O 368.2 kg KOH 2H O h 143.8 kg H O h h 56.11 kg KOH m == KOH balance: 0.427 875 224.2 1.03 145.1 kg h mm =+ = Total mass balance: ( ) 11 2 875 368.2 2.03 145.1 212kg H O h evaporated + = 6.76 a. C R C CRC R A A AA g A dissolved mL solution Plot vs. = = 03 0 4 5 00 2 0 3 0 0 150 .. / b. Mass of solution: 500 mol 1.10 g ml g = 550 (160 g A, 390 g S) The initial solution is saturated at 10.2 ° C. Solubility @ 10.2 ° C = ° 160 0 410 410 g A 390 g S g A g S g A 100 g S @ 10.2 C At 0 ° C, R = 17 5 . C A g A 1 mL soln mL soln g soln g A g soln 17 5 150 110 0106 . . . Thus 1 g of solution saturated at 0 ° C contains 0.106 g A & 0.894 g S.
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