Separation Process Principles- 2n - Seader & Henley - Solutions Manual

90868 781 kg urea remaining in solution 868 781 87 kg

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Unformatted text preview: N = xi D pi DV = 2 3 1/ 3 3 The calculations are made from results of the spreadsheet calculations given above. Surface-mean diameter = D S = Mass-mean diameter = 1 = 0.570 mm 1.7543 DW = 0.633 mm Arithmetic-mean diameter = D N = Volume-mean diameter = DV = 3.4945 = 0.431 mm 8.1012 1 = 0.498 mm 8.10121/ 3 Exercise 17.8 Subject: Dissolution of sparingly soluble solids. Given: Mixing of 1,000 grams of water, 50 grams of Ag2CO3, and 100 grams of AgCl at 25oC. Assumptions: Attainment of equilibrium. Solids have an activity of one. Find: Concentrations of Ag+, Cl-, CO3=, and grams of solids Ag2CO3 and AgCl. Analysis: From Table 17.6, Ag2CO3 and AgCl are both sparingly soluble in water, with the following solubility products: ( ) ( c ) = 6.15 ×10 Ag2CO3 (s) = 2Ag+(aq) + CO3=(aq) with AgCl (s) = Ag+(aq) + Cl-(aq) K c = cAg + By stoichiometry, with K c = cAg + 2 −12 = CO3 ( ) ( c ) = 1.56 ×10 −10 Cl- cAg+ = cCl- + 2cCO= (1) (2) (3) 3 Solving (1), (2), and (3), which constitute a system of nonlinear equations, by Polymath, using the constrained option and noting a high sensitivity to initial guesses, the following results are obtained by assuming initial values for all three concentrations of 1.0 x 10-5 mol/L: cAg+ = 2.311× 10 −4 mol/L cCO= = 1.152 × 10−4 mol/L 3 cCl- = 6.751× 10−7 mol/L Therefore, the concentration of dissolved AgCl = cCl- = 6.751× 10−7 mol/L , and The concentration of dissolved Ag2CO3 = cCO= = 1.152 × 10−4 mol/L 3 Now compute the amount of solids. The molecular weights are 143.32 for AgCl and 275.74 for Ag2CO3. Since we have 1 L of water, the amounts that dissolve are: 6.751× 10−7 (143.32) = 0.000097 g of dissolved AgCl 1.152 ×10 −4 (275.74) = 0.0318 g of dissolved Ag2CO3 Amount of solid AgCl = 100 – 0.000097 = 99.999903 g Amount of solid Ag2CO3 = 50 – 0.0318 = 49.9682 g Exercise 17.9 Subject: Crystallization of (NH4)2SO4 from water by a combination of cooling and evaporation. Given: 5,000 lb/h of a saturated aqueous solution of (NH4)2SO4 at 80oC. Evaporate 50% of the water and cool the solution to 30oC. Assumptions: Equilibrium Find: The lb/h of crystals formed. Analysis: From Table 17.7, the solubility of (NH4)2SO4 at 80oC = 95.3 g/100 g water 95.3 (5, 000) = 2, 440 lb/h of (NH4)2SO4 Therefore, have 95.3 + 100 and 5,000 – 2,440 = 2,560 lb/h of water Evaporate 0.5(2,560) = 1,280 lb/h of water leaving 1,280 lb/h of water in solution From Table 17.7, the solubility of (NH4)2SO4 at 30oC = 78 g/100 g water and the crystals of (NH4)2SO4 have no water of hydration. Therefore, in solution at 30oC, we have 78 (1, 280 ) = 998 lb/h of (NH4)2SO4 100 Therefore, we have 2,440 – 998 = 1,442 lb/h of crystals of (NH4)2SO4 Exercise 17.10 Subject: Crystallization of the hexahydrate of FeCl3 from an aqueous solution by cooling. Given: 7,500 lb/h of a 50 wt% solution of FeCl3 at 100oC. Cooled to 20oC. The solubilities of FeCl3 in g/100 g water are 540 at 100oC and 91.8 at 20oC. Assumptions: Equilibrium Find: The lb/h of FeCl3 6...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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