Solution Manual For Intro to Chemical Processes Principles, Analysis, Synthesis 2e Regina Murphy Chapters 1-9
Chapter 1 Solutions Warm-ups P1.1 Paper bag — agricultural materials Plastic soda bottle — fossil fuels Wine glass — minerals P1.2 minerals, fossil fuels, air, water, agricultural materials N2 - air Fe - minerals C - fossil fuels or agricultural materials H - water, fossil fuels, or agricultural materials P1.3
1C3H6
Set
P1.4
2NH3 1=
3O2
4C3H3N
1
C balance:
4
1
0
4
N balance:
2
4
0
2
H balance: O balance:
6 2
C3H6
NH3
1(NH2)2CO
1
2
3
3
2 5
1.5 O2 2O2
4
2
(NH2)2 CO 1NH2CONH2
H2O
5
0 3
C3H3N 3H2O
2NH3
2C2H6O2
3H2O 4NH3
5CO2
4 2 5 1 3 1 2 0
CO2 3C3H4O3
1 1 5
0
let 1 1 N balance: 21 4 0 C balance: 1 5 0 H balance: 41 23 3 4 0 O balance: 1 22 3 25 0
P1.5
5H2O
4NH3
1.5
3
let 1 1 N balance: 2 C balance:
1
2
1
H balance:
4
4
0
2
3
6
1
O balance:
1SiO2
2( 1)
3(1)
C2H6O2
C3H4O3
3SiC
4CO
1
O balance:
2
C balance:
1
1
6
2
4
2
3
SiC
1
3 4
0
4
2 3
2
2 CO
Use C, H, O balances to get 25 O2 8CO2 9H2O 2
1(CH3)3Ga
2NH3
Ga:
1
3
1
H:
9
1
N:
0
3
C:
3
(CH3)3Ga
Ca:
1
Cl:
2
4
4
2
4
0
3
NH3
1Ca(OH)2
2
GaN
0 3
1 3
0
2
1
check
2NH4Cl 3
4CH4
1
0
4
2
3GaN
1
3
0
3
1
check
0
0
4
2
choose
P1.9
1,
3
0
3
0
3
3C
C8H18
P1.8
3
3
3
1
Si balance:
P1.7
H simultaneously 2
2
0
4
1
1
SiO2
3
3
2
2C
let
4
2
1
NH2CONH2 2NH3 P1.6
0
3
2
Solve C
2
4
3 CH4 3CaCl2
4NH3
5H2O
2
N:
choose
2
0
4
O:
2
1
H:
2
1
0
5
4
2
3
2
4
0
5
1, solve Ca, Cl, N, O balances: 1, 2 2, 4 2, 5 3 check H balance 1
Ca(OH)2 P1.10
2 NH4Cl
CaCl2
2NH3
3O2
1C3H6
choose 1 C balance: N balance: H balance: O balance: 2 C3H6 P1.11
1
3
1
3
2
O:
9
2 H2O
4C3H3N
5H2O
3
1 2
1
2
3 4 2
4
2
0
5
5
0
3
3
2 C3H3N
6 H2O
3H2O
4N2
2CO2
1
1
5
2
2
3
2
0 0 0
2
5O2
3 3 5 2 4 3 2
3
2
5
0 5 1 4
5 3 1 C3H5(NO3 )3 3CO2 H2O N2 O2 2 2 4
P1.12
2
4
0
3 O2
1
2 2
0
4
4
6 2
2 NH3
1
2 NH3
2 3
1C3H5(NO3)3
choose C: 3 H: 5 N: 3
2
1(NH4)2PtCl6 2Pt 3NH4Cl N balance: 2 1 2 0 3 4 H balance: 8 1 4 3 0 5 Pt balance: 0 1 2 Cl balance: 6 1 0 3 5 Lots of choices for basis: I’ll choose 5 6 solve H and Cl balance simultaneously 1 9 8
4N2
N5HCl
6
use
Pt balance Cl balance
2 9 8 3 6 8
N balance
4 6 8
9 9 6 6 (NH4 )2 PtCl6 Pt NH4Cl N2 6HCl 8 8 8 8
or could clear fractions 9 (NH4)2PtCl6 P1.13
9 Pt
6 NH4Cl
S
O2 SO2 1 SO2 O2 SO3 2 SO3 H2O H2SO4 net j ij
6 N2
48 HCl
(R1) (R2) (R3)
let 1 2 3 1 net SO2 1SO2 1 2SO2 2 (1)(1) (1)( 1) 0 net SO3 2SO3 2 3SO3 3 (1)(1) (1)( 1) 0
check check
P1.14 (a) reactant - negative (b) intermediate - zero (c) product - positive (d) byproduct - positive k is counter for reaction. k has only one subscript because it applies to the entire reaction. ik has subscripts i and k because it is a stoichiometric coefficient for a specific compound in a specific reaction. P1.15 N 14 g/gmole, C 12, O 16, H 1 (NH2)2CO:
2(14)
1(12)
1(16)
4(1)
60 g/gmol
2 gmolN 1 gmol urea 14 g N 46.6 gN 100g 100 g urea gmol urea 60 g urea gmol N
P1.16
N2
3H2
2NH3
1 billion lb NH3
1 lbmol 5.88 107 lbmolNH3 17 lb
1lbmol N2 28 lbN2 5.88 107 lbmol NH3 8.24 108 lb N2 2lbmol NH3 lbmol N2
3lbmol H2 2 lb H2 5.88 107 1.76 108 lb H2 2lbmol NH3 lbmol H2
8.24 108
check: P1.17
1.76 108
1 109 lbs
18 lb H2O kg lbmol H2O 2.2 lb
4.4 lbmol H2O
79.2 lb H2O 36 kg H2O 0.036 metric tons H2O
P1.18
P1.19
1 lbmol (8.35 lb) 0.464 lbmol H2O 18lb 454 gmol 210.6 gmol H2O lbmol NH2 CH(R) COOH with
N
C
R
CH2 CH2 S CH3
S
O
H
(1 14) (5 12) (1 32) (2 16) (11 1) 149 g/gmol P1.20
CH3 [C10H8O4]n OH, n
50
C
H
O
1 (10 50) 12 4 8 50 1 1 4 50 16 9632 g/gmole P1.21
45 106 tons H2SO4, Mw 98 g/gmole 2000 lb 1 kg 45 106 tons 4.09 1010 kg H2SO4 tons 2.2 lb (45 106)(2000) 9 1010 lb H2SO4 (45 106)(2000)(454) 4.1 1013 grams H2SO4
(45 × 106)(2000)(454)( 1 98 )
4.09 1010 kg 5.2 kg 9 7.8 10 people person
P1.22
342 g 342 lb gmol lbmol
342 g 1 lb 0.749 lb gmol 453.9 g gmol
4.17 1011 gmol H2SO4