4-3
For each of the following reactions, calculate the amount of the compounds asked for: (Note: The equations are not balanced.) a) Benzene + Oxygen → Carbon dioxide + water
per 120 kg of benzene burned, how many kilograms CO2 and kilograms H2O are produced? How much oxygen is needed?
b) NaCl + H2O + SO3 → Na2SO4 + HC1
Per 300 kg NaC1, how much Na2SO4 and HC1 are formed?
Solution:
a) C6H6 + 7.5O2 → 6CO2 + 3H2O Basis: 120 kg benzene
kg CO2= 120 kg C6H6% 6MWCOMWC6H62 = 120 % 6(44.01) kg
Basis: 300 kg NaCl
kg Na2SO4 = 300 kg NaCl % MWNa2SO4 kg
2MWNaCl = 300 % 142.04
2(58.44) = 364.6
(310.18) = 112.4 tons tons P= 100 tons Ca3(PO4)2% 2MWP
MWCa3(PO4)2 = 100 % 2(30.97)
(310.18) = 19.97 tons tons CO2= 100 tons Ca3(PO4)2% (5/2)MWCO2
MWCa3(PO4)2 = 100 % (5/2)(44.01)
(310.18) = 35.47 tons
d) C2H6 + (7/2)O2 → 2CO2 + 3H2O
DRAFT
DRAFT
4-6
Basis: 1,000 kg of fertilizer
MW (NaNO3)= 85.0 kg N2 = 0.02(1000) = 20 kg MW Ca3(PO4)2 =310.2 kg P2O5 = 0.12(1000) = 120 kg MW KCl = 74.55
kg K2O = 0.06(1000) = 60 kg MW P2O5 = 141.94 MW K2O = 94.18 kg 98% NaNO3 needed =
20 % 2(MW NaNO3)
MW N2 % 1 kg 98% NaNO3
0.98 kg NaNO3 = 20 % 2(85) 28 % 1
0.98 = 123.8 kg kg Ca3(PO4)2 = 120 % MW Ca3(PO4)2
MW(P2O5) = 120 % 310.2141.94 = 262.5 kg kg 97% KCl =
60 % 2(MW KCl)
MW K2O % 1 kg 97% KCl
0.97 kg KCl = 60 % 2(74.55) 94.18 % 1
0.97 = 97.93 kg Overall Mass Balance:
kg inerts = kg mixed fertilizer - (kg 98% NaNO3 + kg 97% KCl + kg Ca3(PO4)2) kg inerts = 1,000 - (123.8 + 262.5 + 97.93) = 515.8 kg
DRAFT
4-7
Basis: 1,000 kg of fertilizer 2% N2, 12% P2O5, 6% K2O Calcium nitrate, pure
Phosphoric acid KNO3
Inert fillers.
kg N2 = 0.02(1000) = 20 kg MW Ca(NO3)2 = 226.1 kg P2O5 = 0.12(1000) = 120 kg MW H3PO4 = 98
kg K2O = 0.06(1000) = 60 kg MW KNO3 = 101.1
MW P2O5 = 141.94 MW K2O = 94.18 kg KNO3 needed = 60 % 2(MW KNO3)
MW K2O = 60 %2(101.1)
94.18 = 128.8 kg
N2 from KNO3 = 128.8 kg KNO3% MWN2 17.84
2MW KNO3 %= 128.8 % 28 2(101.1) = N2 balance
N2 from Ca(NO3)3 = total N2 - N2 from KNO3 = 20 - 17.84 = 2.16
g Ca(NO3)3 = 2.16 kg N2% 2MW Ca(NO3)3
3MW(N2) = 2.16 % 2(226.1)
3(28) = 11.63 kg kg 97% H3PO4 =
120 % 2(MW H3PO4)
MW P2O5 = 1202(98)
141.9 = 165.8 kg Overall Mass Balance:
kg inerts = kg mixed fertilizer - ( kg KNO3 + kg Ca (NO3)3 KCl + kg H3PO4)2) kg inerts = 1,000 - (128.8 + 11.63 + 165.8) = 693.8 kg
DRAFT
Basis: 5,000 kg mixed fertilizer
MW (NH4)3PO4 = 149
DRAFT
4-10 Solution:
Basis: 1,000 kg of solution at 30°C
We have no tie component and we have to use the algebraic method Let x = Na2SO4⋅10H2O produced
1,000 - x = kg solution at 15°C (Using an overall mass balance) Na2SO4 balance:
Na2SO4 in feed = Na2SO4 in Na2SO4⋅10H2O + Na2SO4 in final solution
1000 % 40 kg Na2SO4
(100 + 40)kg solution = x MW Na2SO4
MW Na2SO4$10H2O + (1000 − x) %
13.5 kg Na2SO4
(100 + 13.5) kg solution 1000 % 40140 = x 142
322.2 + (1000 − x)13.5 113.5 Solving for x,
= 518.3 kg Na2SO4⋅10H2O x=
40000
140 − 13500 113.5 322.2 −142 13.5
113.5
DRAFT
4-13 For the following reactions, find the composition the final product. If the second reactant used is 10% in excess of the theoretical amount or complete reaction. (The first is the limiting reactant) The reaction is 100% complete and the reactants are pure.
a. 2KCl + H2SO4 → K2SO4 + 2HCl
b. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2
Solution:
a. 2KCl (s) + H2SO4(l) → K2SO4(s) + 2HCl(g) MW Cl: 35.45 H: 1 S: 32 O: 16 K: 39 basis: 100 kg KCl
kg theoretical H2SO4 needed = 100 kg KCl % MWH2SO4 kg
2MWKCl = 98
2 % 74.45 = 65.82 excess H2SO4 needed = 0.1( 65.82 ) = 6.58 kg
kg K2SO4 produced = 100 kg KCl % MWK2SO4 kg
2MWKCl = 174
2 % 74.45 = 116.9 kg HCL produced = 100 kg KCl % MWHClMWKCl = 36.45 kg
74.45 = 48.96
product composition: HCl is a gas and normally will not mix with the other substances. The product mixture consist of H2SO4 and K2SO4.
% H2SO4= 6.58
6.58+ 116.9 % 100 = 5.33%
% K2SO4= 116.9
6.58+ 116.9 % 100 = 94.67%
b. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2
Basis: 100 kg Na2CO3
kg theoretical Br2 needed = 100 kg Na2CO3 % MWBr2
MWNa2CO3 = 159.8106 = 150.8 kg excess Br2 needed = 0.1(150.8) = 15.08 kg
kg NaBr produced = 100 kg Na2CO3 % 5MWNaBr3MWNa2CO3 = 5(102.8)
3(106) = 32.33 kg kg NaBrO3 produced = 100 kg Na2CO3 % MWNaBrO3MWNa2CO33 = 342.63(106) = 107.7 kg kg product = 15.08 + 32.33 + 107.7 = 155.1 kg
% Br2 = 15.08155.1 % 100 = 9.722%
% NaBr = 32.33155.1 % 100 = 20.84%
% NaBrO3 = 107.7155.1 % 100 = 69.44%
CO2 is a gas and will not be in the solid and liquid substances.
DRAFT
4.14a For the following, the reaction goes only to 95% completion. Pure reactants and theoretical amounts are used. For every 500 kg of the impure products, how much of each of the reactants are used?
a. KCl +NaNO3 → KNO3 + NaCl Basis: 100 kg KCl
kg NaNO3 needed = 100 kg KCl % MWNaNOMWKCl3 = (100)(84.995)
74.551 = 114.009 kg KNO3 produced = 100 kg KCl % MWKNOMWKCl % 0.95 =3 (100)(101.103)(0.95
74.551 = 128.835 kg CaCl2 produced = 100 kg KCl % MWNaClMWKCl % 0.95 = (100)(58.442)(0.95)
74.551 = 74.473
The product consists of KNO3, NaCl2, and unreacted KCl and NaNO3. unreacted KCl = 5 kg unreacted NaNO3 = 0.05(114.009) = 5.7 kg kg product = 5 + 5.7 + 128.835 + 74.473 = 214.008 kg
By ratio and proportion, we can obtain the reactants needed to form 500 kg product.
kg KCl : 500 = 100 : 214.008 kg KCl = 500 % 100
214.008 = 233.636 kg kg NaNO3 : 500 = 114.009 : 214.008 kg NaNO3 = 500 % 114.009214.008 = 266.366 kg
4.14b For the following, the reactions go only to 95% completion. Pure reactants and theoretical amounts are used. For every 500 kg of the impure products, how much of each of the reactants are used?
MgO + CaCl2 + CO2 → MgCl2 +CaCO3
Basis: 100 kg MgO
kg CO2 needed = 100 kg MgO % MWCOMWMgO =2 (100)(44.01)
(40.3) = 109.2 kg
kg CaCl2 needed = 100 kg MgO % MWCaClMWMgO =2 (100)(111)
(40.3) = 275.4 kg
kg MgCl2 produced = 100 kg MgO%MWMgCl2
MWMgO % 0.95 = (100)(95.21)(0.95)
40.3 = 224.4 kg kg CaCO3 produced = 100 kg MgO %MWCaCO3
MWMgO % 0.95 = (100)(100.1)(0.95)
40.3 = 236 kg
The product consists of MgCl2, CaCO3, and unreacted MgO and CaCl2
unreacted MgO = 5 kg unreacted CaCl2 = 0.05(275.4) = 13.77kg kg product = 5 + 13.77 + 224.4 + 236 = 479.2 kg
By ratio and proportion, we can obtain the reactants needed to form 500 kg product.
DRAFT
kg MgO : 500 = 100 : 479.2 kg MgO = 479.2 % 100 = 104.3 kg500 kg CaCl2 : 500 = 275.4 : 479.2 kg CaCl2 = 479.2 % 275.4 = 287.4 kg500 kg CO2 : 500 = 109.2 : 479.2 kg CO2 = 479.2 % 109.2 = 113.9 kg500
DRAFT
4-18 100 kg of pure carbon is burned. For each of the following cases, calculate the composition of the combustion gases:
a. Theoretical amount of pure O2 used; complete combustion b. Theoretical amount of air used; complete combustion
c. 30% excess air is used: 90% of the C burns to CO2 and 10% to CO
d. 20% excess air is used: 90% of the C burns to CO2 while the 10% goes out as unburned solid.
Basis: 100 kg pure carbon
a. Theoretical pure O2 used; complete combustion Reaction: C + O2 → CO2
katom C = 100/12 = 8.333 katom kmol CO2 formed = 8.333 kmol O2 needed = 8.333 kmol combustion gas = 100% CO2
b. Theoretical air used; complete combustion kmol CO2 formed = 8.333 kmol
Theo O2 needed = 8.333 kmol
N2 from air = 8.333(79/21) = 31.348 kmol
Total mols combustion gas = 8.333 + 31.348 = 39.681 kmol
% CO2 = 8.333(100)/39.681 = 21% CO formed = 0.1(8.333) = 0.833 kmol
O2 left due to incomplete reaction to CO = 0.833/2 = 0.416 kmol O2 in combustion gas = 2.5 + 0.416 = 2.92 kmol
O2 left due to incomplete reaction = 0.833 kmol O2 in combustion gas = 1.667 + 0.833 = 2.5 kmol
DRAFT
100 47.62
Total
79 37.62
N2
5.25 2.5
O2
15.75 7.5
CO2
% kmol
Comp
DRAFT
4.19e The following pure compound (CO) is burned with 25% excess of air required over the theoretical.
Basis: 100 kmol CO
Reaction: CO + 1/2 O2 → CO2
Theo O2 required = 100/2 = 50 kmol Excess O2 required = 0.25(50) = 12.5 kmol Total O2 required = 50+12.5 = 62.5 kmol N2 from air = 62.5(79/21) = 235.1 CO2 formed = 100 kmol
100.01 347.6
total
67.64 235.1
N2
3.6 12.5
O2
28.77 100
CO2
% kmol
Component
DRAFT
4-34 500 m3 per min of a fuel gas at 30°C and 800 mm Hg consisting of 15% CH4, 40% C2H6, 15% C3H8, 10%
C6H6, and 20% N2 is burned with 20% excess air. All the C burns to CO2 and H2 to H2O. The air enters at a temperature of 60°C and a total pressure of 760 mm Hg wing PH2O = 12 mm Hg. Calculate:
a. the flue gas analysis (dry basis)
b. the partial pressure of water in the flue gas c. m3 humid air used/m3 fuel at the given conditions PT of flue gas = 755 mm Hg.
Given:
Required: a. flue gas analysis
b. partial pressure of water in the flue gas c. m3 humid air used/m3
Basis: 1 minute operation fuel input = 500 mmin %3 0+ 273 Air required = 81.37 + 306.09 = 387.46 kmol H2O in air = 387.46 % 12
760− 12 (a) Products of combustion
(b) Partial pressure of the flue gas total H2O in flue gas = 50.85 + 6.22 = 57.07 kmol
DRAFT
SOLUTION TO SOME PROBLEMS IN CHAPTER 5
5-2 One kg of iron is initially at 35°C. It absorbs 500 joules of heat? What will be its final temperature?
Solution:
Basis: 1 kg of iron at 35°C that absorbs 500 joules of heat Required: Final temperature of the iron
Let Tf as the final temperature of the iron The sensible heat gain of the iron is 500 J.
m
¶
TTifCpdT= 500m = 1 kg Ti = 273 + 35 = 315 K
Cp = 4.13 + 0.00635T Cp, cal/(mol)(K) T, K (1000)
¶
315Tf (4.13 + 0.00635T)dT = 500 (1000) 4.13T + 0.00635 T22 315Tf
= (1000) 4.13(Tf− 315) + 0.00635Tf2− 3152
2 = (500)(4.18)
4.13Tf−(4.13)(315) +(0.00635/2)Tf2−(0.00635/2)(315)2= 4.18/2 Simplifying,
Tf = 315.3 K = 42.3 °C
DRAFT
SOME SOLUTION TO PROBLEMS IN CHAPTER 7
7-1 Find the degrees of freedom for the following systems in equilibrium:
Formula: F = C - P + 2
a. salt and sugar (with undissolved quantities) in water.
F = 2 - 2 + 2 = 2
b. hexane, heptane and octane in equilibrium with the vapor.
F = 3 - 2 + 2 = 3
c. oil and water in a closed vessel at 100°C.
F = 2 - 2 + 2 - 1= 1 d. carbon, iron and silicon.
F = 3 - 1 + 2 = 4 e. coffee, milk, and sugar.
F = 3 - 1 + 2 = 4 7-2 Calculate the boiling point of
a. benzene at 30 in Hg.
30 in Hg %760 mm Hg
29.92 in Hg = 762.0 mm Hg From Cox Chart, BP at 762.0 mm Hg = 81°C From Antoine Equation,
t= 1211.033
6.90565− log 762.0 − 220.790 = 80.2°C b. water at 10 mm Hg.
From Cox Chart, BP at 10 mm Hg = 12°C c. toluene at 10 atm.
10 atm %760 mm Hg
1 atm = 7600 mm Hg
From Cox Chart, BP at 7600 mm Hg = 220°C From Antoine equation,
t= 1344.800
6.95464− log 7600 − 219.482 = 218°C
d. propane at 50 psig.
50 psig %760 mm Hg
14.7 psi + 760 mm Hg = 3345 mm Hg From Cox Chart, BP at 3345 mm Hg = −4°C From Antoine equation,
t= 813.20
6.82973− log 3345 − 248.00 = −2.0°C
DRAFT
7-3 What is the vapor pressure of a. water at 80°F?
(80-32)/1.8 = 26.7°C
From Cox Chart, Pv at 26.7°C = 27 mm Hg b. n-octane at 250°C?
From Cox Chart, Pv at 250°C = 9000 mm Hg c. n-pentane at 125°C?
From Cox Chart, Pv at 125°C = 7400 mm Hg From Antoine equation,
= 7414 mm Hg Pv= anti log 6.85221 − 1064.63
232.000+ 125 d. mercury at 200°C?
From Cox Chart, Pv at 200°C = 16 mm Hg
7-4 Given that glycerol at mm Hg boils at 167.2°C and at 760 mm Hg, 290°C, calculate the constants A and B in the equation
Log PV= A - B T+ C
where PV is vapor pressure in mm Hg T, temperature in K, and C = 230
What is the vapor pressure of glycerol at 200°C?
(1) Pv = anti log 13.1456− 8140.01
200+ 273 + 230 = 36.9 mm Hg
7-5 A flue gas on a wet basis contains 10% CO2 , 1% CO, 5% O2, 5% H2O and 79% N2. It is at 200°C and 758 mm Hg. To what temperature should it be cooled to make it saturated with water vapor, keeping the total pressure constant at 758 mm Hg?
Given: Flue gas composition on a wet basis: 10% CO2, 1% CO, 5% O2, 79% N2, and 5%H2O T = 200°C P = 758 mm Hg
Required: Temperature to which it should be cooled to make it saturated with water vapor PH2O = 0.05(758) = 37.9 mm Hg
DRAFT
The equilibrium temperature corresponding to the an equilibrium vapor pressure of 37.9 mm Hg is what we want. We can find this from Appendix 9 to Appendix 7.
From Appendix 9 (Cox Chart) the temperature corresponding to P = 37.9 mm Hg for water is 33°C. From the Appendix 7 (Steam Table), the pressure is 5.053 kPa. By interpolation, the corresponding equilibrium temperature is
T = 32 + 2 (5.053 - 4.759)/(5.324 - 4.759) = 32 + 0.98 = 33°C
7-6 In a room measuring 4m × 5m × 4m was left an open tank containing ethyl alcohol. Considering that the room is sealed from the outside, calculate the amount of alcohol that has evaporated into the room if the temperature is 30°C and the atmospheric pressure is 755 mm Hg. Assume that the room becomes completely saturated with alcohol.
PvEtOH, 30oC= anti log 8.16290 − 1623.22228.98+ 30 = 78.552 mm Hg YEtOH,sat.= PvEtOH
PT− PvEtOH = 78.522
755− 78.552 = 0.11612 kmol EtOH kmol dry air
7-7 The atmospheric pressure varies with elevation according to the equation P = 29.92 − 0.00111h
where P is pressure in in Hg and h is the altitude in ft.
What is the elevation where water boils at 95°C?
At 95°C, from Cox chart, Pvwater = 650 mm Hg = 25.59 in. Hg h= 29.92 − 25.590.00111 = 3900 ft.
7-8 A liquid mixture at 0°C consists of 30% ethane, 40% propane, and 30% n-butane. Find the composition of the vapor mixture in equilibrium with this liquid. What is the total equilibrium pressure?
From Antoine equation,
Pvethane= anti log 6.80266 − 656.40256.00 = 17322 mm Hg Pvpropane= anti log 6.82973 − 813.20248.00 = 3553.8 mm Hg Pvbu tan e= anti log 6.83029 − 945.90240.00 = 774.53 mm Hg
PT = 0.30(17322) + 0.40(3553.8) + 0.30(774.53) = 6850 mm Hg yethane= 0.30(17322)
6850 = 0.7586 ypropane= 0.40(3553.8)
6850 = 0.2075
DRAFT
ybu tan e= 0.30(774.53)
6850 = 0.0339
7-9 A gaseous mixture consists of 61% n-hexane, 26% n-octane and 13% n-decane at a temperature of 50°C. Find the equilibrium pressure and the composition of the liquid in equilibrium with this mixture.
At 50°C, from Cox chart, Pvhexane = 390 mm Hg
7-10 Find the dew point of a gaseous mixture of 40% ethyl alcohol and 60% water at a total pressure of 2 atm. What is the composition of the first liquid that appears?
PT= 2 atm = 1520 mm Hg
DRAFT
7-11 Find the bubble point of a mixture of 50% n-pentane and 50% n-butane at a pressure of 4 atm.
What is the composition of the first vapors formed?
PT = 4 atm = 3040 mm Hg
Using Antoine equation:
(1) Pvpen tan e= anti log 6.85221 − 1064.63232.000+ T
Pvbu tan e= anti log 6.83029 − 945.90240.00+ T (2) 0.50 (3)
3040 Pvpen tan e+ 0.503040 Pvbu tan e= 1
Solving simultaneously (1), (2) and (3), bubble point = 58.2°C
DRAFT
7-12 The composition of a liquefied petroleum gas (LPG) is 0.5% ethane, 0.1% ethylene, 16.4%
propane, 2.1% propylene, 74.0% n-butane, and 6.9% n-butene. If the room temperature is 30°C, what is the pressure inside the tank and the composition of the gas that first issues from the mixture?
At T = 30°C
Pvethane= anti log 6.80266− 656.40
256.00+ 30 = 32177.7 mm Hg Pvethylene= anti log 6.74756 − 585.00
255.00+ 30 = 49536.9 mm Hg Pvpropane= anti log 6.82973 − 813.20
248.00+ 30 = 8026.94 mm Hg Pvpropylene= anti log 6.81960 − 785.00
247.00+ 30 = 9675.31 mm Hg Pvn−bu tan e= anti log 6.83029 − 945.90
240.00+ 30 = 2123.03 mm Hg Pvn−butene= anti log 6.84290 − 926.10
240.00+ 30 = 2587.62 mm Hg
PT= 0.005(32177.7) + 0.001(49536.9) + 0.164(8026.94) + 0.021(9675.31) + 0.74(2123.03) +0.069(2587.62) = 3479.62
yethane= 0.005(32177.7)
3479.62 = 0.04624 yethylene= 0.001(49536.9)
3479.62 = 0.01424 ypropane= 0.164(8026.94)
3479.62 = 0.37832 ypropylene= 0.021(9675.31)
3479.62 = 0.05839 yn−bu tan e= 0.74(2123.03)
3479.62 = 0.45150 yn−butene= 0.069(2587.62)
3479.62 = 0.05131
7-16 A helium-toluene mixture contains 30% mole toluene at 100°C and 760 mm Hg. Calculate the percentage saturation, relative saturation, dew point and the molar humidity of the mixture.
T = 100°C PT = 760 mm Hg
Ptoluene = 0.30(760) = 228 mm Hg
Pvtoluene= anti log 6.95464 − 1344.800
219.482+ 100 = 556.3 mm Hg
DRAFT
7-17 200 grams of benzene is allowed to evaporate in a 10 m3 tank containing CO2 gas at 40°C and 750 mm Hg. If the mixture is brought to 60°C and 1.2 atm, what is the percentage saturation and the relative saturation of the mixture?
At 60°C
Pvbenzene= anti log 6.90565 − 1211.033220.790+ 60 = 391.5 mm Hg PT= 1.2 atm %760 mm Hg
1 atm = 912 mm Hg 10 m3CO2% kmol22.4 m3 % 750760 % 273 K
273+ 40 K = 0.384 kmol CO2 0.2 kg benzene % kmol78.12 kg = 0.00256 kmol benzene
Pbenzene= ybenzene$PT= 0.00256
0.00256+ 0.38425 (912) = 6.04 percent saturation =
relative saturation = Pbenzene
Pvbenzene % 100= 6.04391.5 % 100 = 1.54 %
7-18 A nitrogen-acetone mixture containing 20% acetone and 80% nitrogen is at a temperature of 30°C and a pressure of 755 mm Hg. To what temperature should the mixture be chilled so that when heated back to 30°C at the same pressure, the percentage saturation will be 40%?
At 30°C, Pvacetone= anti log 7.23157 − 1277.03237.23+ 30 = 283.7 mm Hg
At chilling temperature, Pacetone = Pvacetone
t = 1277.03
7.23157− log 146.5 − 237.23 = 14.86°C
DRAFT
7-19 A mixture of benzene-nitrogen contains 15% benzene and is at 40°C and 780 mm Hg. What will happen if the mixture is compressed until the final pressure becomes 1500 mm Hg with temperature kept at 40°C? What is the final percentage saturation and concentration, (kg benzene)/(kg
benzene-free nitrogen)?
Pvbenzene= anti log 6.90565 − 1211.033220.790+ 40 = 182.8 mm Hg
percent saturation =
Pbenzene
PT− Pbenzene
Pvbenzene
PT− Pvbenzene
% 100=
0.15(780) 1500− 0.15(780)
182.8 1500− 182.8
% 100=