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.
Surfacemean diameter = D S =
Massmean diameter = 1
= 0.570 mm
1.7543 DW = 0.633 mm Arithmeticmean diameter = D N =
Volumemean 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 105 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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 Spring '11
 Levicky
 The Land

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