Mass Balance Problems PDF
Mass Balance Problems PDF
Mass Balance Problems PDF
PROBLEM 1:
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
1 mile = 5280 ft
PROBLEM 2:
SOLUTION:
Product, P=?
Feed, F=1 ton DRYER
Water, W=0.70 Water,W=0.25
Solids, S=0.30 Solids,S=0.75
1.00 1.00
Water,W=?
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
MATERIAL BALANCE:
Overall Balance:
(1)
COMPONENT BALANCE:
Water Balance
By eq. 1 &eq.2
P=F–W
0.7 = (0.25)(1-W) + 1W
0.7 = 0.25-0.25W+1W
0.45 = 0.75 W
F = 2000 lb
P = 800 lb
PROBLEM 3:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
SOLUTION:
Dilute acid,
D=100Kg
Product,P=?
H2SO4=0.28 H2SO4=0.50
H2O= 0.72 H2O=0.50
1.00 1.00
H2SO4=0.96
H2O=0.04
1.00
Conc. Acid,C=?
Overall Balance:
(1)
COMPONENT BALANCE:
H2SO4 Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
PROBLEM 4:
SOLUTION:
Dry Nitrocellulose,D= ?
Overall Balance:
(1)
COMPONENT BALANCE:
Nitrocellulose Balance:
(0.055) F + 1 D = (0.08) P
Water Balance:
(0.945) F = (0.02) P
F = 973.5lb
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
From eq. 1
973.5 + D = 1000
D = 26.5lb
PROBLEM 5:
SOLUTION:
No 100% conversion
Waste Product,
Feed, F=6.22Kgmol
W(Kgmol)=?
REACTOR
XH2=0.580 XHCl =?
XSiHCl3 = 0.420 XH2=0.223
1.00 XSiHCl3=?
Product, P(Kgmol)= ?
Si = 1.00
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
nSiHCl = F
3 SiHCl3 = (6.22) (0.420) = 2.61 moles
nHCl= W HCl
Overall Balance:
(1)
Atomic Balance:
Hydrogen Balance:
Chlorine Balance:
Silicon Balance:
2.6124 = P + nSiHCl3(4)
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
W= (7)
9.8276 = + nH2
W= = 8.03Kgmol
From eq.4
P = 2.6124 – 0.79
P = 1.82 Kg moles/hr
hr 60 min 1 Kg mol Si
P = 16.9 Kg Si
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
PROBLEM 6:
SOLUTION:
Water,W=?
Product, P=?
Feed, F=1 Kg DRYER
H2O= 0.201 H2O= 0.086
Wet Timber, T=0.799 Wet Timber, T=0.914
1.00 1.00
BASIS: 1 Kg of timber
MATERIAL BALANCE:
Overall Balance:
(1)
COMPONENT BALANCE:
(0.799) F = (0.914) P
P = 0.87 Kg
Water Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
PROBLEM 7:
In the manufacture of vinyl acetate, some unreacted acetic acid and other compounds are
discharged to the sewer. Your company has been cited as discharging more than the
specified limit of several pollutants. It is not possible to measure the discharge directly
because no flow measuring device are in place, but you can take samples of liquid at
different places in the sewer line, and measure the concentration of potassium chloride. At
one manhole, the concentration is 0.105%. You introduced a solution of 400 g of KCl in 1L
at a steady rate of 1L per minute over ½ hr at a manhole 500 ft downstream, and at 200 ft
downstream measure the average steady state concentration of KCl as 0.281%. What is the
flow rate of fluid in the sewer in kg/min?
SOLUTION:
A=1Kg/min
K=0.400
Fluid=0.600
1.00
Product, P=?
Feed, F=?
K = 0.00281
K= 0.00105 Fluid= 0.99719
Fluid= 0.9989 1.00
1.00
BASIS: 1 Kg/min
MATERIAL BALANCE:
Overall Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
(1)
COMPONENT BALANCE:
K Balance:
Fluid Balance:
From eq.2
( )
P=
Put in eq.1
( )
F+1=
F = 225.676 Kg/min
PROBLEM 8:
Ammonia is a gas for which reliable analytical methods are available to determine its
concentrations in other gases. To measure the flow in a natural gas pipeline, pure ammonia
gas is injected into the pipeline at a constant rate of 72.3 kg/min for 12 min. Five miles
downstream from the injection point; the steady state ammonia concentration is found to
be 0.382 weight percent. The gas upstream from the point of ammonia injection contains no
measureable ammonia. How many kilograms of natural gas are flowing through the
pipeline per hour?
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
SOLUTION:
T=72.3 Kg/min
NG=0
NH3=1
Product, P=?
Feed, F=? DRYER
Natural Gas, NG= 1.00 NG= 0.618
Ammonia, NH3=0 NH3= 0.382
1.00
Overall Balance:
(1)
COMPONENT BALANCE:
NG Balance:
F = (0.618) P (2)
Ammonia Balance:
T = P (0.382)
72.3 = P (0.382)
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
PROBLEM 9:
Water pollution in the Hudson River has claimed considerable recent attention, especially
pollution from sewage outlets and industrial wastes. To determine accurately how much
effluent enters the river is quite difficult because to catch and weigh the material is
impossible, weirs are hard to construct, and so on. One suggestion that has been offered is
to add a tracer of Br ions to a given sewage stream, let it mix well, and sample of sewage
stream after it mixes. On one test of the proposal you add ten pounds of NaBr per hour for
24 hours to a sewage stream with essentially no Br in it. Somewhat downstream of the
introduction point a sampling of the sewage stream shows 0.012% NaBr. The sewage
density is 60.3 lb/ft3 and river water density if 62.4 lb/ft3. What is the flow rate of the
sewage in lb/min?
SOLUTION:
NaBr=1.00
NaBr=10 lb/hr
Sewage=0.00
Product, P=?
Sewage=60.3 lb/ft3 Feed, F=?
NaBr= 0.00012
Br= 0.00
S= 0.99988
Sewage, S=1.00
1.00
BASIS:10 lb/hr
MATERIAL BALANCE:
Overall Balance:
(1)
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
COMPONENT BALANCE:
P = 10/0.00012
PROBLEM 10:
Acetone is used in the manufacture of many chemicals & as a solvent. Many restrictions
are placed on the release of acetone vapors to the environment. You are asked to design an
acetone recovery system having flow sheet illustrated in figure. All the concentrations
shown in figure of both gases & liquid are specified in weight %.Calculate A, F, W, B & D
per hour?
Water=100% Air
W (Kg) Air= 0.995
c A (Kg) Water=0.005
1.00 Distillate
b D (Kg)
Condenser Acetone=0.99
a Water= 0.01
1.00
Absorber Column
Distillation
Entering Gas=1400 Kg/hr Column
Air= 0.95
F (Kg)
Water= 0.02
Acetone=0.03 G (Kg)
Acetone=0.19
1.00 B(Kg)
Water=0.81
1.00 Acetone=0.04
Water= 0.96
1.00
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
SOLUTION:
Overall Balance:
(1)
COMPONENT BALANCE:
Air Balance:
(0.95) G = (0.995) A
A = 1336.68 Kg/hr
Water Balance:
W = 157.69 Kg/hr
Acetone Balance:
(0.03) G = (0.19) F
F = 221.05 Kg/hr
Overall Balance:
(2)
COMPONENT BALANCE:
Water Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Acetone Balance:
From eq. 3
B= (5)
41.99 = + (0.99) D
40.31 = + 0.9504 D
40.31 = 0.95 D
D = 34.89 Kg/hr
From eq. 5
B=
B = 186.095 Kg/hr
PROBLEM 11:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
SOLUTION:
Q4 = 40 Kg/hr Q5 = 30 Kg/hr
d
A=0.9 A=0.6
B= 0.1 B=0.4
1.00 1.00
a c
Feed, F=100 Kg/hr b 3 Q3
Unit 1 1 2 Unit 2
A=0.5 X3A
B=0.5 1-X3B
1.00
Q1 Q2
X1A X2A
1-X1B 1-X2B
Q=30 Kg/hr
A=0.30
B=0.70
1.00
Overall Balance:
F + Q = Q3 + Q4 + Q5
100 + 30 = Q3 + 40 + 30
Q3 = 60 Kg/hr
COMPONENT BALANCE:
Balance For A:
Overall Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
F = Q4 + Q1
100 = 40 + Q1
Q1 = 60 Kg/hr
COMPONENT BALANCE:
Balance For A:
Q1 + Q= Q2
Q2 = 90 Kg/hr
COMPONENT BALANCE:
Balance For A:
PROBLEM 12:
A liquid mixture containing 30 mole% benzene, 25% toluene & 45% xylene is fed at rate of
1275 Kmol/hr to a distillation unit consisting of two columns. The bottom product of first
column is to contain 99 mole% xylene & no benzene. And 98% of xylene in feed is to be
removed in this stream. The overhead product from the first column is fed into the second
column.The overhead product from the second column contains 99% benzene & no xylene.
The benzene recovered in this stream represents 96% of benzene in feed. Calculate molar
flow rates & component mole fraction in each stream.
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
SOLUTION:
a b c
F=1275Kmol/hr O1 =? O2=?
COLUMN 1 COLUMN 2
B=0.30 B=? B=0.99
T=0.25 T=? T=0.1
X=0.45 X=? X=0.0
1.00 1.00
B=0.0 B=?
T=0.1 T=?
X=0.99 X=?
1.00
P1=? P2 =?
By Statement:
562.275 = P1(0.99)
P1 = 567.95Kmoles/hr
By Statement:
O2 = 370.91 Kmoles/hr
Overall Balance:
F = P1+ P2 + O2
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
P2 = 336.151 Kmoles/hr
Overall Balance:
F = P1+ O1
1275 = O1 + 567.95
O1 = 707.05 Kmoles/hr
COMPONENT BALANCE:
Benzene Balance:
0.30 × F = 0 × P1 + O1 × BO1
BO1 = 0.54
Xylene Balance:
XO1= 11.479/707.05
XO1 = 0.016
TO1 = 0.44
COMPONENT BALANCE:
Benzene Balance:
BP2 = 0.043
Toluene Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
TP2 = 0.82
XP2 = 0.137
PROBLEM 13:
Sea water is to be desalinized by reverse osmosis using the scheme indicated in fig. Use the
(b) The rate of desalinized water (called potable water) production (D)
(c) The fraction of the brine leaving the reverse osmosis cell that is recycled.
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
b Recycle, R=?
BASIS: 100lb/hr
R = 5.25% salt = 0.0525
Overall Balance:
F=D+B
100 = D + B (1)
COMPONENT BALANCE:
Salt Balance:
From eq.1
D = 100 – B
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
B = 39.4 lb/hr
D = 100-39.4
D = 60.6 lb/hr
Overall Balance:
F1 = D + P1
F1 = 60.6 + P1(3)
COMPONENT BALANCE:
Salt Balance:
From eq.1
P1 = F1-60.6
Put in eq.2
F1 = 251.2 lb/hr
P1 = 251.2 – 60.6
P1 = 190.63 lb/hr
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
PROBLEM 14:
A planting plant has a waste stream containing zinc & nickel in quantities in excess of that
allowed to be discharged into the sewer. The proposed process to be used as a first step in
reducing the concentration of zinc & nickel is shown in fig. Each stream contains water.
The concentrations of several of the streams are listed in the table. What is the flow (in
L/hr) of the recycle stream R32 if the feed is 1L/hr?
R21 W(100%)
F=1L/hr
1
P1 P2 D
PO R32=?
SOLUTION:
BASIS: 1 L/hr
MATERIAL BALANCE ACROSS OVERALL BOUNDARY
Overall Balance:
F + W = PO + D
1 + W = PO + D (1)
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
COMPONENT BALANCE:
Zn Balance:
Ni Balance:
(10) 1= (17.02) PO + D
10 = (17.02) PO + D
D = 10 – 17.02 PO(3)
99 = 188.398 PO
PO = 0.525 L/hr
From eq.3
D = 10 – 17.02 (0.525)
D = 1.06 L/hr
From eq.1
1 + W = 0.5 + 1.06
W = 0.585 L/hr
Overall Balance:
P2 + W = D + R32
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
COMPONENT BALANCE:
Zn Balance:
PROBLEM 15:
Calculate the composition & percent of each component in stream E.
D1 D2 D3=10 lb
A=0.50 A=0.17
B=0.23 B=0.10 C=0
C=0.27 C=0.73
Feed,
F=100 lb
1
P3
P1 P2 C=0 A=0.70
A=0.50
B = 0.20 B=0.30
C = 0.30
A=?
E=? B=?
C=?
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
GIVEN DATA:
SOLUTION:
BASIS: 100 lb
P3 = 3 D3 = 3×10 = 30 lb
P2 = D2
100 = D1 + D2 + 10 + 30
D1 + D2= 60
D1= 60 - D2(1)
COMPONENT BALANCE:
A Balance:
C Balance:
30 = (60-D2)0.27 + 0.73 D2
13.8 = 0.46 D2
D2 = 30 lb
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
D1= 60 – 30
D1 = 30 lb
-1 = -0.33×30 + 10 AD3
AD3 = 0.89
Overall Balance:
P2 + E = D3 + P3
30 + E = 10 + 30
E = 10 lb
COMPONENT BALANCE:
A Balance:
30 AP2 + 10 AE = 8.9 + 21
As AP2/BP2 = 4
AP2= 4 BP2
Eq.4 becomes
4 BP2+ BP2= 1
5 BP2= 1
BP2 = 0.2
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
(30)(0.8)+ 10 AE = 29.9
AE = 0.59
BE= 1 – 0.59
BE = 0.41
PROBLEM 16:
To save energy, stack gas from a furnace is used to dry rice the flow sheet & known data is
given. What is the amount of recyclegas in lbmolesper 100 lbof P if the concentration of
water in the gas stream entering the dryer is 5.2%?
SOLUTION:
Recycle, R
Wet Gas (lbmol),W
Dry Gas=0.95
Water=0.0473 Water=0.0931
b F1 Dry Gas=0.90
Stack Gas, S
(lbmol)
Water=0.052 DRYER
Rice Feed,F Rice Feed,P (lb)
(lb) Rice=0.75 Rice=0.95
Water=0.25 Water=0.05
COMPONENT BALANCE:
Rice Balance:
0.75 F = 0.95 P
0.75 F = 0.95×100
F = 126.67 lb
Water Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Gas Balance:
0.95 S = 0.90 W
W = (0.95/0.90) S (2)
1.323 = 0.0454 S
S = 29.14 lbmol
Overall Balance:
S + R = F1
F1 = 29.14 + R (3)
COMPONENT BALANCE:
Water Balance:
0.0411 R = 0.142
R = 3.4 lbmol
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
PROBLEM 17:
M1=100 Kg M2=75 Kg
z
a b
Feed,F=100 Kg R1=?
P=43.1 Kg
EXTRACTOR 1 EXTRACTOR 2
A=0.5 R1M=? A=0.053
W=0.5 R1A=? M=0.016
R1W=? W=0.931
E1 A=0.09 E2
c M=0.88
A=0.275 W=0.03
M=?
W=?
E
EM=?
EA=? V=?
EW=? A=0.97
M=0.02
d W=0.01
STILL
B=?
BA=?
BM=?
BW=?
SOLUTION:
BASIS:
MATERIAL BALANCE ACROSS BOUNDARY (z)
Overall Balance:
F + M1 + M2 = E1 + E2 + P
COMPONENT BALANCE:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Acetone Balance:
From eq.1
E1 = 231.9 - E2
Put in eq.2
16.0525 = 0.185 E2
E2 = 86.77 Kg
E1 = 145.13 Kg
M Balance:
ME1 = 0.674
WE1 = 0.051
Overall Balance:
F + M1 = E1 + R1
R1 = 54.87 Kg
COMPONENT BALANCE:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Acetone Balance:
R1A = 0.184
M Balance:
M1 = (E1)(ME1) + (R1)(R1M)
R1M = 0.0397
R1W = 0.7763
Overall Balance:
E1 + E2= E
E = 145.13 + 86.77
E = 231.9 Kg
COMPONENT BALANCE:
Acetone Balance:
EA = 0.206
M Balance:
EM= 0.751
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
EW = 1 – 0.751 – 0.206
EW= 0.043
Overall Balance:
E=V+B
V + B = 231.9
Assumption:
So, BA = 0
COMPONENT BALANCE:
Acetone Balance:
(231.9)(0.206) = 0.97 V + 0
Put V = 231.9 – B
47.77 = 0.97(231.9 – B)
B = 182.61 Kg
V = 231.9 – 182.61
V = 49.29 Kg
M Balance:
182.61 BM = 173.17
BM = 0.95
BW = 1 – 0.95
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
BW = 0.05
PROBLEM 18:
Ultrafiltrationis a method for cleaning up input & output streams from a number of
industrial processes. The lure of the technology is its simplicity, merely putting a
membrane across a stream to sieve out physically undesirable oil, dirt, metal particles,
polymers, & the like. The trick, of course, is coming up with the right membrane. The
screening material has to meet a formidable set of conditions. It has to be very thin (less
than 1 micron), highly porous, yet strong enough to hold up month after month under
severe stressesof liquid flow, pH, particle abrasion, temperature & other plant operating
characteristics.
Calculate the recycle rate in gallons per day (g.p.d) for the set up shown in fig & calculate
the concentration of oil plus dirt in the stream that enters the filtration module.
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
W
Makeup Water, 1 3
M=90 gpd 2
K N To
H2O=0.98855 process
H2O=1.00 H2O=0.9927
OD=0.01145 P1=2910 gpd
Oil,Dirt OD=0.00 OD=0.0073
Overall Balance:
M + P= K
90 + 2910 = K
K = 3000 g.p.d
COMPONENT BALANCE:
Water Balance:
In P stream
XH2O= 0.9925
XOD = 1-0.9925=0.0075
Overall Balance:
K+R=N
3000 + R = N (1)
COMPONENT BALANCE:
Water Balance:
STOICHIOMETRY-II
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
From eq.1
12.45 = 0.21755 R
R = 57.23 g.p.d
Overall Balance:
N = P1 + D
3057.23 = 2910 + D
D = 147.23 g.p.d
COMPONENT BALANCE:
Water Balance:
In P1 stream
XH2O = 0.9996
XOD = 0.0004
PROBLEM 19:
Dry coke composed of 4% inert solids (ash), 90% carbon & 6% hydrogen is burned in a
furnace with dry air.The solid refuse left after combustion contains 10% carbon & 90%
inert (ash) & no hydrogen in it.The inert ash content does not enter into the reaction.
STOICHIOMETRY-II
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
The orsatanalysis of flue gas gives 13.9% CO2, 0.8% CO, 4.3% O2& 81%
N2.Calculate the % excess air based on complete combustion of the coke?
SOLUTION:
Water, W=100%
CO2=0.139
CO=0.008
Product, P
O2=0.043
Feed, F N2=0.81
FURNACE
Inert=0.04
Carbon, C=0.9
Hydrogen,
H=0.06 Air, A Solid Refuse, R
40% Carbon=0.1
Inert=0.9
O2=0.21
N2=0.79
ATOMIC BALANCE:
Carbon Balance:
Oxygen Balance:
0.42 A = 0.372×100 + W
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Nitrogen Balance:
(2)(0.79)(A) = (2)(P)(0.81)
1.58 A = (2)(100)(0.81)
1.58 A = 162
A = 102.53 kgmol
Hydrogen Balance:
(1)(0.06/1)(F) = (2)(1)(W)
0.06 F = 2 W (3)
From eq.2
0.42×102.53 = 37.2 + W
W = 5.8626 kgmol
0.06 F = 2×5.8626
0.06 F = 11.7252
F = 195.42 kgmol
% Excess Air:
C + O2 CO2 (1)
2C + O2 2CO (2)
H2 + O2 2H2O (3)
Reaction 1:
CO2 : O2
1 : 1
13.9 : 13.9
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Reaction 2:
CO : O2
2 : 1
0.8 : 0.4
Reaction 3:
H2 O : O2
2 : 1
1 : ½
5.8 : 2.9
O2 Supplied = 0.21 A
= 0.21×102.53
STOICHIOMETRY-II
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
= × 100
PROBLEM 20:
A distillation column separates 10,000 kg/hrof a mixture containing equal mass of benzene
and toluene. The product D recovered from the condenser at the top of the column contains
95% benzene, and the bottom W from the column contains 96% toluene. The vapor V
entering the condenser from the top of the column is 8000 kg/hr. A portion of the product
from the condenser is returnedto the column as reflux R, and the rest is withdrawn as the
final product D. Assume that V, R, and D are identical in composition since V is condensed
completely. Find the ratio of the amount refluxedR to the product withdrawn D.
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Overall Balance:
F=D+W
10,000 = D + W (1)
COMPONENT BALANCE:
Benzene Balance:
D = 5050 kg/hr
From eq.1
10,000 = 5050 + W
W = 4950 kg/hr
Overall Balance:
V=R+D
8000 = R + 5050
R = 2950 kg/hr
Now,
PROBLEM 21:
A waste stream from a plant is being disposed of by burning in a flare with air. The waste
gas has the composition CH4:30%, CO2: 10%, CO: 8%, H2: 10%, O2: 2%, H2S: 2%, H2O:
2%, N2: 36%.
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
The Orsat analysis of the exit gas shows 0.3% SO2 along with CO2, O2 and N2. Calculate
the percent excess air and complete Orsat analysis.
SOLUTION:
Water, W=100%
Feed, F Product, P
COMBUSTION
PROCESS
CH4=0.3 SO2=0.003
CO2=0.1 CO2=?
CO=0.08 O2=?
H2=0.1 N2=?
Air, A
O2=0.02
H2S=0.02
H2O=0.02 O2=0.21
N2=0.36 N2=0.79
ATOMIC BALANCE
Oxygen Balance:
SulphurBalance:
(1)(0.02)(100) = (1)(0.003)P
P = 666.7 kgmol
Nitrogen Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Put in (1)
Carbon Balance:
= 0.071996
= 48
Hydrogen Balance:
W = 74 kg mole
Oxygen Balance:
A= 602.67 kg mole
Now,
Now,
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
CO + ½ O2 CO2 (2)
H2 + ½ O2 2H2O (3)
PROBLEM 22:
A furnace fuel gas of the following composition: 70% methane (CH4), 20% Hydrogen (H2)
and 10% Ethane (C2H6) with excess air. An oxygen probe placed at the exit of the furnace
reads 2% oxygen in the exit gases. The gases are then passed through a long duct to a heat
exchanger. At the entrance to the heat exchanger the Orsat analysis of the gas reads 6%. Is
the discrepancy due to the fact that the first analysis is on wet basis and the second analysis
is on dry basis (no water condenses in the duct), or due to the air leak in the duct? If the
former, give the Orsat analysis of the exit gas from the furnace. If the latter, calculate the
amount of the air that leaks into the duct per 100 moleof fuel gas burned.
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Water, W (kgmol)=100%
1 2
Feed, F (kgmol) E (kgmol) P-87
FURNACE DUCT
CH4=0.7 CO2=? nCO2
O2=0.02 CO2=?
H2=0.2 nO2
H2O=? O2=0.06
C2H6=0.1 nH2O
Air, A N2=? Air, B N2=?
1.00 nN2
(kgmol) (kgmol)
E
O2=0.21 O2=0.21
N2=0.79 N2=0.79
ATOMIC BALANCE
Hydrogen Balance:
( ) = 190 (1)
Oxygen Balance:
NitogenBalance:
(2)(0.79)(A) = ( )(E)(2)
1.58A= 2E( )
1.58A= 2E(1-0.02- - )
Carbon Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
90 = ( )(E) (4)
By (3)
A = 1.96E-560/1.58 (5)
Put in (2)
E= 1079.160 kg moles
From (5)
A=984.274kg moles
Boundry2 (Duct)
Hydrogen Balance:
(190)(2) = (1)(W)(2)
W= 190kg moles
Carbon balance
Hydrogen Balance
F = 100kg moles
Put in (6)
90 = (P)( ) (7)
Nitrogen Balance
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Oxygen Balance
B = 211 kg moles
From (8)
P = 1100 kg moles
PROBLEM 23:
A synthesis gas analyzing 6.4% CO2, 0.2% O2, 40% CO, and 50.8% H2, (the balance is N2),
is burnt with 40% dry excess air. What is the composition of the flue gas?
SOLUTION:
O2=0.21
N2=0.79
CO + ½ O2 CO2
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
CO : CO2
1 : 1
0.4 : 0.4
H2 + 1/2 O2 H2O
H2 : H2O
1 : 1
0.508 : 0.508
CO + ½ O2 CO2 (1)
H2 + ½ O2 H2O (2)
Reaction 1:
O2 : CO
0.5 : 1
0.5×0.4 : 0.4
Reaction 2:
H2 O : O2
1 : 0.5
0.508 : 0.5×0.508
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
PROBLEM 24:
A low-grade pyrite containing 32% S is mixed with 10 lb of pure sulfur per 100 lb of
pyrites so the mixture will burn readily, forming a burner gas that analyzes (Orsat) 13.4%
SO2 , 2.7% O2 and 83.9% N2. No sulfur is left in the cinder. Calculate the percentage of the
sulfur fired that burned to SO3. (The SO3 is not detected by the Orsat ananlysis.)
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
SO3, Y (lbmole)
Cinder, C
Feed, F = 100 lb (lb)
S=0.32 Product, Z
(lbmole)
SO2=0.134
Air, X O2=0.027
(lbmol) N2=0.839
O2=0.21
N2=0.79
Solution:
Material Balance:
SulphurBalance:
42/32 = Y=0.134Z
O2 Balance:
Nitrogen Balance:
(o.79)(X) = (0.839)(Z)
X = o.839Z/0.79 (3)
X = 9.46
Y = 0.1074
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Z = 8.9
PROBLEM 25:
One of the products of sewage treatment is sludge. After microorganismsgrow in the
activated sludge process to remove nutrients and organic material, a substantial amount of
wet sludge is produced. This sludge must be dewatered; one of the most expensive parts of
most treatment plant operations. How to dispose of the dewatered sludge is a major
problem. Some organizations sell dried sludge for fertilizer, some spread the sludge on
farmland, and in some places it is burned. To burn a desired sludge, fuel oil is mixed with
it, and the mixture is in a furnace with air. If you collect the following analysis for the
sludge and for the stack gas
(a) Determine the weight percent of carbon and hydrogen in the fuel oil.
(b) Determine the ratio of pounds of dry sludge to pounds of fuel oil in the mixture fed to
the furnace.
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Water, W (lbmol)=100%
C=?
H=?
Fuel oil, F (lb) Flue Gas, G
BURNER
SO2=0.0152
Sludge, S (lb) CO2=0.1014
O2=0.0465
Air, A CO=0.0202
S=0.32
(lbmol) N2=0.8167
C=0.4
H2=0.04
O2=0.24 O2=0.21
N2=0.79
ATOMIC BALANCE:
Nitrogen Balance:
(0.79)(2)A = (2)(100)(0.8167)
1.58 A = 163.34
A = 103 lbmol
SulphurBalance:
(0.32/32)(1)S = (1)(100)(0.0152)
0.01 S = 1.52
S = 152 lbmol
Oxygen Balance:
45.69918 = W + 34.64
W = 11.06 lbmol
Hydrogen Balance:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
Carbon Balance:
From (1 )
F-F = 16.04
Put in (2)
F = 101.150 lb
= 0.842
= 0.158
S/F = 1.5
PROBLEM 26:
Refined sugar (sucrose) can be converted to glucose and fructose by the inversion process
C12H22O11 + H2O C6H12O6 (G) + C6H12O6 (F)
The combined glucose and fructose is called inversionsugar. If 90% conversion of sucrose
occurs on one pass through the reactor, what would be the recycle stream flow per 100
lbmfresh feed of sucrose? What is the concentration of inversionsugar in the recycle and
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
product streams? Assume that the two streams are identical in composition. Refer to the
flowchart for the other details of the process.
SOLUTION:
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UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, KSK CAMPUS
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