What is Liquid-liquid extraction

Report
Chapter 5 Liquid-Liquid Extraction
Subject: 1304 332 Unit Operation in Heat transfer
Instructor: Chakkrit Umpuch
Department of Chemical Engineering
Faculty of Engineering
Ubon Ratchathani University
Here is what you will learn in this chapter.
5.1 Introduction to Extraction Processes
5.2 Equilibrium Relations in Extraction
5.3 Single- Stage Equilibrium Extraction
5.4 Equipment for Liquid-Liquid Extraction
5.5 Continuous Multistage Countercurrent
Extraction
2
5.1 Introduction to Extraction Processes
“When separation by distillation is ineffective or very difficult e.g. closeboiling mixture, liquid extraction is one of the main alternative to
consider.”
What is Liquid-liquid extraction (or solvent extraction)?
Liquid-Liquid extraction is a mass transfer operation in which a liquid
solution (feed) is contacted with an immiscible or nearly immiscible liquid
(solvent) that exhibits preferential affinity or selectivity towards one or
more of the components in the feed. Two streams result from this contact:
a) Extract is the solvent rich solution containing the
desired extracted solute.
b) Raffinate is the residual feed solution containing little solute.
3
5.1 Introduction to Extraction Processes
Liquid-liquid extraction principle
When Liquid-liquid extraction is carried out in a test tube or flask the
two immiscible phases are shaken together to allow molecules to
partition (dissolve) into the preferred solvent phase.
4
5.1 Introduction to Extraction processes
An example of extraction:
Extract
Acetic acid in H2O
Organic layer contains most of acetic acid in
ethyl acetate with a small amount of water.
+
Raffinate
Ethyl acetate
Aqueous layer contains a weak acetic acid
solution with a small amount of ethyl
acetate.
The amount of water in the extract and ethyl acetate in the raffinate
depends upon their solubilites in one another.
5
5.2 Single-stage liquid-liquid extraction processes
Triangular coordinates and equilibrium data
Each of the three corners
represents a pure component A,
B, or C.
Point M represents a mixture of
A, B, and C.
The perpendicular distance from
the point M to the base AB
represents the mass fraction xC.
The distance to the base CB
represents xA, and the distance
to base AC represents xB.
xA + xB + xC = 0.4 + 0.2 + 0.4 = 1
Equilateral triangular diagram
xB = 1.0 - xA - xC
(A and B are partially miscible.)
yB = 1.0 - yA - yC
6
Liquid-Liquid phase diagram where components A and B are partially
miscible.
Liquid C dissolves completely in A or in B.
Liquid A is only slightly soluble in B and B slightly soluble in A.
The two-phase region is included inside below the curved envelope.
An original mixture of composition M will separate into two phases a and b which are on
the equilibrium tie line through point M.
The two phases are identical at point P, the Plait point.
7
8
Ex 5.1 Define the composition of point A, B, C, M, E, R, P and DEPRG in
the ternary-mixture.
Point A = 100% Water
Point B = 100% Ethylene Glycol
Point C = 100% Furfural
Point M = 30% glycol, 40% water, 30% furfural
Point E = 41.8% glycol, 10% water, 48.2% furfural
Point R = 11.5% glycol, 81.5% water, 7% furfural
The miscibility limits for the furfural-water binary
system are at point D and G.
Point P (Plait point), the two liquid phases have
identical compositions.
DEPRG is saturation curve; for example, if feed
50% solution of furfural and glycol, the second
phase occurs when mixture composition is 10%
water, 45% furfural, 45% glycol or on the
saturation curve.
Liquid-Liquid equilibrium, ethylene glycol-furfural-water, 25ºC,101 kPa. 9
Equilibrium data on rectangular coordinates
The system acetic acid (A) –
water (B) – isopropyl ether
solvent (C). The solvent pair B
and C are partially miscible.
xB = 1.0 - xA - xC
yB = 1.0 - yA - yC
Liquid-liquid phase diagram
10
EX 5.2
An original mixture weighing 100 kg and containing 30 kg of
isopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water (B) is
equilibrated and the equilibrium phases separated. What are the
compositions of the two equilibrium phases?
Solution:
Composition of original mixture is xc= 0.3, xA = 0.10, and xB = 0.60.
11
Liquid-liquid phase diagram
1. Composition of xC = 0.30, xA
= 0.10 is plotted as point h.
2. The tie line gi is drawn
through point h by trial and
error.
3. The composition of
extract (ether) layer at g
= 0.04, yC = 0.94, and
1.00 - 0.04 - 0.94 =
mass fraction.
the
is yA
yB =
0.02
4. The raffinate (water) layer
composition at i is xA = 0.12,
xC = 0.02, and xB = 1.00 –
0.12 – 0.02 = 0.86.
12
Phase diagram where the solvent pairs B-C and A-C are partially miscible.
“The solvent pairs B and C and also A and C are partially miscible.”
13
5.3 Single-Stage Equilibrium Extraction
Derivation of lever-arm rule for graphical addition
An overall mass balance:
A balance on A:
V LM
5.1
Vy A  LxA  MxAM
5.2
Where xAM is the mass fraction of A in the M stream.
A balance on C:
VyC  LxC  MxC M
5.3
14
Derivation of lever-arm rule for graphical addition
Sub 5.1 into 5.2
L y A  x AM

V x AM  x A
(5.4)
Sub 5.1 into 5.3
L yC  xC M

V xC M  xC
(5.5)
Equating 5.4 and
5.5 and rearranging
xC  xC M
x A  x AM

xC M  y C
x AM  y A
(5.6)
Eqn. 5.6 shows that points L, M, and V must lie on a straight line. By
using the properties of similar right triangles,
Lever arm’s rule
L(kg ) V M

V (kg ) L M
(5.7)
L(kg) V M

M (kg) L V
(5.8)
15
Ex 5.3 The compositions of the two equilibrium layers in Example 5.1 are for
the extract layer (V) yA = 0.04, yB = 0.02, and yC = 0.94, and for the raffinate
layer (L) xA = 0.12, xB = 0.86, and xC = 0.02. The original mixture contained
100 kg and xAM = 0.10. Determine the amounts of V and L.
Solution:
Substituting into eq. 5.1
V  L  M  100
Substituting into eq. 5.2, where M = 100 kg and xAM = 0.10,
V (0.04)  L(0.12)  100(0.10)
Solving the two equations simultaneously, L = 75.0 and V = 25.0. Alternatively, using
the lever-arm rule, the distance hg in Figure below is measured as 4.2 units and gi
as 5.8 units. Then by eq. 5.8,
L
L
h g 4.2



M 100 gi 5.8
Solving, L = 72.5 kg and V = 27.5 kg, which is a reasonably close check on the
material-balance method.
16
5.2 Single-stage liquid-liquid extraction processes
Single-state equilibrium extraction
We now study the separation of A from a mixture of A and B by a solvent C in a single
equilibrium stage.
An overall mass balance:
L0  V2  L1  V1  M
5.9
A balance on A:
L0 x A0  V2 y A2  L1 x A1  V1 y A1  Mx AM
5.10
A balance on C:
L0 xC 0  V2 yC 2  L1 xC1  V1 yC1  MxC M
5.11
x A  xB  xC  1.0
17
To solve the three equations, the equilibrium-phase-diagram is used.
1. L0 and V2 are known.
2. We calculate M, xAM, and xCM by
using equation 5.9-5.11.
3. Plot L0, V2, M in the Figure.
4. Using trial and error a tie line is
drawn through the point M, which
locates the compositions of L1 and V1.
5. The amounts of L1 and V1 can be
determined by substitution in
Equation 5.9-5.11 or by using leverarm rule.
18
Ex 5.4 A mixture weighing 1000 kg contains 23.5 wt% acetic acid (A) and
76.5 wt% water (B) and is to be extracted by 500 kg isopropyl ether (C) in a
single-stage extraction. Determine the amounts and compositions of the
extract and raffinate phases.
Solution Given:
L0  1000kg andV2  500kg
Given: L0  V2  1000 500  M  1500kg
xA0  0.235, xB0  0.765and y A2  1.0
L0 xA0  V2 y A2  MxAM
(1000)(0.235)  (500)(0)  (1500) xAM
x AM  0.157
Given: xc 0  1  xA0  xB0  1.0  0.235 0.765  0
L0 xC 0  V2 yC 2  MxC M
(1000)(0)  (500)(1)  (1500) xC M
xCM  0.33
19
V2 (0,1) = (yA2, yC2)
V1 (0.1,0.89) = (yA1, yC1)
M(0.157,0.33) = (xAM, xCM)
L1(0.2,0.03) = (xA1, xC1)
M
L0(0.235,0) = (xA0, xC0)
20
From the graph: xA1 = 0.2 and yA1 = 0.1;
L1 xA1  V1 y A1  MxAM
L1 (0.2) V1 (0.1)  (1500)(0.157)
L1  0.5V1  1,177.5
(1)
From the graph: xC1 = 0.03 and yC1 = 0.89;
L1 xC1  V1 yC1  MxC M
L1 (0.03) V1 (0.89)  (1500)(0.33)
L1  29.67V1  16,500
(2)
Solving eq(2) and eq(3) to get L1 and V1;
L1  914.86kg and V1  525.28kg
xA1  0.2, y A1  0.1, xC1  0.03 and yC1  0.89
Answer
21
5.3 Equipment for Liquid-Liquid Extraction
Introduction and Equipment Types
As in the separation processes of distillation, the two phases in liquidliquid extraction must be brought into intimate contact with a high
degree of turbulence in order to obtain high mass-transfer rates.
Distillation:
Rapid and easy because of the large difference in
density (Vapor-Liquid).
Liquid extraction:
Density difference between the two phases is not
large and separation is more difficult.
Liquid extraction equipment
Mixing by mechanical
agitation
Mixing by fluid flow
themselves
22
Mixer-Settles for Extraction
Separate mixer-settler
Combined mixer-settler
23
Plate and Agitated Tower Contactors for Extraction
Perforated plate tower
Agitated extraction tower
24
Packed and Spray Extraction Towers
Spray-type extraction tower
Packed extraction tower
25
5.4 Continuous multistage countercurrent extraction
Countercurrent process and overall balance
An overall mass balance:
A balance on C:
L0  VN 1  LN  V1  M
5.12
L0 xC 0  VN 1 yC N 1  LN xC N  V1 yC1  MxC M
Combining 5.12 and 5.13
Balance on component A gives
xCM 
x AM
L0 xC 0  VN 1 yCN 1
L0  VN 1

LN xCN  V1 yC1
5.13
5.14
LN  V1
L0 x A0  VN 1 y AN 1 LN x AN  V1 y A1


L0  VN 1
LN  V1
5.15
26
5.4 Continuous multistage countercurrent extraction
Countercurrent process and overall balance
1. Usually, L0 and VN+1 are known and
the desired exit composition xAN is set.
2. Plot points L0, VN+1, and M as in the
figure, a straight line must connect these
three points.
3. LN, M, and V1 must lie on one line.
Also, LN and V1 must also lie on the
phase envelope.
27
Ex 5.5 Pure solvent isopropyl ether at the rate of VN+1 = 600 kg/h is being
used to extract an aqueous solution of L0=200 kg/h containing 30 wt% acetic
acid (A) by countercurrent multistage extraction. The desired exit acetic acid
concentration in the aqueous phase is 4%. Calculate the compositions and
amounts of the ether extract V1 and the aqueous raffinate LN. Use equilibrium
data from the table.
Solution: The given values are VN+1 = 600kg/h, yAN+1 = 0, yCN+1 = 1.0, L0 = 200kg/h,
xA0 = 0.30, xB0 = 0.70, xC0 = 0, and xAN = 0.04.
In figure below, VN+1 and L0 are plotted. Also, since LN is on the phase
boundary, it can be plotted at xAN = 0.04. For the mixture point M,
substituting into eqs. below,
xCM 
x AM 
L0 xC 0  VN 1 yCN 1
L0  VN 1

200(0)  600(1.0)
 0.75
200 600
L0 x A0  VN 1 y AN 1 200(0.30)  600(0)

 0.075
L0  VN 1
200 600
28
Using these coordinates,
1) Point M is plotted in Figure below.
2) We locate V1 by drawing a line from LN through M and extending it until
it intersects the phase boundary. This gives yA1 = 0.08 and yC1 = 0.90.
3) For LN a value of xCN = 0.017 is obtained. By substituting into Eqs. 5.12
and 5.13 and solving, LN = 136 kg/h and V1 = 664 kg/h.
29
Stage-to-stage calculations for countercurrent extraction.
Total mass balance on stage 1
Total mass balance on stage n
From 5.16 obtain difference Δ in flows
Δ is constant and for all stages
L0  V2  L1  V1
Ln1  Vn1  Ln  Vn
L0  V1  L1  V2  
  L0  V1  Ln  Vn1  LN  VN 1  ....
x  L0 x0  V1 y1  Ln xn  Vn1 yn1  LN x N  VN 1 y N 1  ...
5.16
5.17
5.18
5.19
5.20
30
Stage-to-stage calculations for countercurrent extraction.
Δx is the x coordinate of point Δ
x 
L0 x0  V1 y1 Ln xn  Vn1 y n1 LN x N  VN 1 y N 1


L0  V1
Ln  Vn1
LN  VN 1
5.21
5.18 and 5.19 can be written as
L0    V1
Ln    Vn1
LN    VN 1
5.22
31
Stage-to-stage calculations for countercurrent extraction.
1. Δ is a point common to all streams passing each
other, such as L0 and V1, Ln and Vn+1, Ln and Vn+1,
LN and VN+1, and so on.
2. This coordinates to locate this Δ operating point
are given for x cΔ and x AΔ in eqn. 5.21. Since the
end points VN+1, LN or V1, and L0 are known, xΔ can
be calculated and point Δ located.
3. Alternatively, the Δ point is located graphically in
the figure as the intersection of lines L0 V1 and LN
VN+1.
4. In order to step off the number of stages using
eqn. 5.22 we start at L0 and draw the line L0Δ,
which locates V1 on the phase boundary.
5. Next a tie line through V1 locates L1, which is in
equilibrium with V1.
6. Then line L1Δ is drawn giving V2. The tie line
V2L2 is drawn. This stepwise procedure is
repeated until the desired raffinate composition LN
is reached. The number of stages N is obtained to
perform the extraction.
32
Ex 5.6 Pure isopropyl ether of 450 kg/h is being used to extract an aqueous
solution of 150 kg/h with 30 wt% acetic acid (A) by countercurrent multistage
extraction. The exit acid concentration in the aqueous phase is 10 wt%.
Calculate the number of stages required.
Solution:
The known values are VN+1 = 450, yAN+1 = 0, yCN+1 = 1.0, L0 = 150, xA0
= 0.30, xB0 = 0.70, xC0 = 0, and xAN = 0.10.
1. The points VN+1, L0, and LN are plotted in Fig. below. For the mixture point M,
substituting into eqs. 5.12 and 5.13, xCM = 0.75 and xAM = 0.075.
2. The point M is plotted and V1 is located at the intersection of line LNM with the
phase boundary to give yA1 = 0.072 and yC1 = 0.895. This construction is not shown.
3. The lines L0V1 and LNVN+1 are drawn and the intersection is the operating point Δ
as shown.
33
1. Alternatively, the coordinates of Δ can
be calculated from eq. 5.21 to locate
point Δ.
2. Starting at L0 we draw line L0 Δ, which
locates V1. Then a tie line through V1
locates L1 in equilibrium with V1. (The
tie-line data are obtained from an
enlarged plot.)
3. Line L1 Δ is next drawn locating V2. A tie
line through V2 gives L2.
4. A line L2 Δ is next drawn locating V2. A
tie line through V2 gives L2.
5. A line L2 Δ gives V3.
6. A final tie line gives L3, which has gone
beyond the desired LN. Hence, about
2.5 theoretical stages are needed.
34
5.4 Continuous multistage countercurrent extraction
Countercurrent-Stage Extraction with Immiscible Liquids
If the solvent stream VN+1 contains components A and C and the feed stream L0
contains A and B and components B and C are relatively immiscible in each other, the
stage calculations are made more easily. The solute A is relatively dilute and is being
transferred from L0 to VN+1.
 x 
 y

 x
L 0   V  N 1   L N
 1  x0 
 1  y N 1 
 1  xN

 y 
  V  1 
 1  y1 

 x 
 y

 x
L 0   V  n 1   L n
 1  x0 
 1  y n 1 
 1  xn

 y 
  V  1 
 1  y1 

5.23
5.24
Where L/ = kg inert B/h, V/ = kg inert C/h, y = mass fraction A in V stream, and x =
mass fraction A in L stream. (5.24) is an operating-line equation whose slope ≈ L//V/.
If y and x are quite dilute, the line will be straight when plotted on an xy diagram.
35
Ex 5.7 An inlet water solution of 100 kg/h containing 0.010 wt fraction
nicotine (A) in water is stripped with a kerosene stream of 200 kg/h
containing 0.0005 wt fraction nicotine in a countercurrent stage tower. The
water and kerosene are essentially immiscible in each other. It is desired to
reduce the concentration of the exit water to 0.0010 wt fraction nicotine.
Determine the theoretical number of stages needed. The equilibrium data are
as follows (C5), with x the weight fraction of nicotine in the water solution and
y in the kerosene.
X
y
x
y
0.001010
0.000806
0.00746
0.00682
0.00246
0.001959
0.00988
0.00904
0.00500
0.00454
0.0202
0.0185
36
Solution:
The given values are L0 = 100 kg/h, x0 = 0.010, VN+1 = 200 kg/h, yN+1 =
0.0005, xN = 0.0010. The inert streams are
L  L(1  x)  L0 (1  x0 )  100(1  0.010)  99.0kg water / hr
V /  V (1  y)  VN 1 (1  yN 1 )  200(1  0.0005)  199.9kgkerosene/ hr
Making an overall balance on A using eq. 5.23 and solving, y1 = 0.00497.
These end points on the operating line are plotted in Fig. below. Since the
solutions are quite dilute, the line is straight. The equilibrium line is also
shown. The number of stages are stepped off, giving N = 3.8 theoretical
stages.
37
38
Homework No.9
1. A single-stage extraction is performed in which 400 kg of a solution
containing 35 wt% acetic acid in water is contacted with 400 kg of pure
isopropyl ether. Calculate the amounts and compositions of the extract
and raffinate layers. Solve for the amounts both algebraically and by
the lever-arm rule. What percent of the acetic acid is removed?
39
Homework No.10
1. Pure isopropyl ether is to be used to extract acetic acid from 400 kg of
a feed solution containing 25 wt% acetic acid in water.
(a) If 400 kg of isopropyl is used, calculate the percent recovery in the
isopropyl solution in a one-stage process.
(b) If a multiple four-stage system is used and 100 kg fresh isopropyl is
used in each stage, calculate the overall percent recovery of the acid in
the total outlet isopropyl ether. (Hint: First, calculate the outlet extract
and raffinate streams for the first stage using 400 kg of feed solution
and 100 kg of isopropyl ether. For the second stage, 100 kg of
isopropyl ether contacts the outlet aqueous phase from the first stage.
For the third stage, 100 kg of isopropyl ether contacts the outlet
aqueous phase from the first stage. For the third stage, 100 kg of
isopropyl ether contacts the outlet aqueous phase from the second
stage, and so on.)
40
Homework No.9 (Solution)
1. A single-stage extraction is performed in which 400 kg of a solution
containing 35 wt% acetic acid in water is contacted with 400 kg of pure
isopropyl ether. Calculate the amounts and compositions of the extract
and raffinate layers. Solve for the amounts both algebraically and by
the lever-arm rule. What percent of the acetic acid is removed?
Solution
Given:
L0  V2  400 400  M  800kg
xA0  0.35, xB0  0.65 and y A2  1.0
L0 xA0  V2 y A2  MxAM
(400)(0.35)  (400)(0)  (800) xAM
x AM  0.175
Given:
xc0  1  xA0  xB0  1.0  0.35  0.65  0
L0 xC 0  V2 yC 2  MxC M
(400)(0)  (400)(1)  (800) xC M
xCM  0.5
41
V2 (0,1) = (yA2, yC2)
V1 (0.12,0.87) = (yA1, yC1)
M(0.175,0.5) = (xAM, xCM)
M
L1(0.22,0.03) = (xA1, xC1)
L0(0.35,0) = (xA0, xC0)
42
From the graph: xA1 = 0.22 and yA1 = 0.12;
L1 xA1  V1 y A1  MxAM
L1 (0.22) V1 (0.12)  (800)(0.175)
L1  0.54V1  636.36
(1)
From the graph: xC1 = 0.03 and yC1 = 0.87;
L1 xC1  V1 yC1  MxC M
L1 (0.03) V1 (0.87)  (800)(0.5)
L1  29V1  1,333.33
(2)
Solving eq(1) and eq(2) to get L1 and V1;
L1  623.12kg and V1  24.49kg
xA1  0.23, y A1  0.12, xC1  0.03 and yC1  0.88
Answer
43
V1 (0.12,0.88) = (yA1, yC1)
M(0.175,0.5) = (xAM, xCM)
M
L1(0.23,0.03) = (xA1, xC1)
Lever arm’s rule
L1(kg ) V M

M (kg ) L V
L1(kg) 0.38

 0.44
M (kg) 0.86
L(kg ) V M

V (kg ) L M
L(kg ) 0.38

 0.81 V (kg)  352/ 0.81  434.57kg
V (kg ) 0.47
L1(kg)  0.44x800  352kg
44
The amount and composition of extract are
L1  325kg *
xA1  0.23, xC1  0.03 and xB1  0.74
Answer
The amount and composition of raffinate are
V1  434.57kg *
y A1  0.12, yC1  0.88 and yB1  0
Answer
*The correction of amount of extract and raffinate depends on the
identification of points on the graph. The value obtained from Lever’s arm
rule is more reliable personally.
The percent of acetic acid removed is about
=(400x0.35-325x0.23)x100% = 46.6%
400x0.35
Answer
45
Homework No.10
1. Pure isopropyl ether is to be used to extract acetic acid from 400 kg of
a feed solution containing 25 wt% acetic acid in water.
(a) If 400 kg of isopropyl is used, calculate the percent recovery in the
isopropyl solution in a one-stage process.
(b) If a multiple four-stage system is used and 100 kg fresh isopropyl is
used in each stage, calculate the overall percent recovery of the acid in
the total outlet isopropyl ether. (Hint: First, calculate the outlet extract
and raffinate streams for the first stage using 400 kg of feed solution
and 100 kg of isopropyl ether. For the second stage, 100 kg of
isopropyl ether contacts the outlet aqueous phase from the first stage.
For the third stage, 100 kg of isopropyl ether contacts the outlet
aqueous phase from the first stage. For the third stage, 100 kg of
isopropyl ether contacts the outlet aqueous phase from the second
stage, and so on.)
46
Solution a):
Given:
L0  V2  400 400  M  800kg
xA0  0.25, xB0  0.75 and y A2  1.0
L0 xA0  V2 y A2  MxAM
(400)(0.25)  (400)(0)  (800) xAM
x AM  0.125
Given:
xc0  1  xA0  xB0  1.0  0.35  0.75  0
L0 xC 0  V2 yC 2  MxC M
(400)(0)  (400)(1)  (800) xC M
xCM  0.5
47
V2 (0,1) = (yA2, yC2)
V1 (0.09,0.9) = (yA1, yC1)
M(0.125,0.5) = (xAM, xCM)
M
L1(0.17,0.03) = (xA1, xC1)
L0(0.25,0) = (xA0, xC0)
48
From the graph: xA1 = 0.17 and yA1 = 0.09;
L1 xA1  V1 y A1  MxAM
L1 (0.17) V1 (0.12)  (800)(0.125)
L1  0.71V1  588.24
(1)
From the graph: xC1 = 0.03 and yC1 = 0.9;
L1 xC1  V1 yC1  MxC M
L1 (0.03) V1 (0.9)  (800)(0.5)
L1  30V1  1,333.33
(2)
Solving eq(1) and eq(2) to get L1 and V1;
L1  570.18kg and V1  25.43kg
xA1  0.17, y A1  0.09, xC1  0.03 and yC1  0.9
Answer
49
Solution b):
Given:
L0  V2  400 100  M  500kg
xA0  0.25, xB0  0.75 and y A2  1.0
First stage
L0 xA0  V2 y A2  MxAM
(400)(0.25)  (100)(0)  (500) xAM
x AM  0.2
Given:
xc0  1  xA0  xB0  1.0  0.35  0.75  0
L0 xC 0  V2 yC 2  MxC M
(400)(0)  (100)(1)  (500) xC M
xCM  0.2
50
V2 (0,1) = (yA2, yC2)
V1 (0.12,0.86) = (yA1, yC1)
M(0.2,0.2) = (xAM, xCM)
L1(0.17,0.03) = (xA1, xC1)
L0(0.25,0) = (xA0, xC0)
M
51
From the graph: xA1 = 0.17 and yA1 = 0.09;
L1 xA1  V1 y A1  MxAM
L1 (0.17) V1 (0.12)  (800)(0.125)
L1  0.71V1  588.24
(1)
From the graph: xC1 = 0.03 and yC1 = 0.9;
L1 xC1  V1 yC1  MxC M
L1 (0.03) V1 (0.9)  (800)(0.5)
L1  30V1  1,333.33
(2)
Solving eq(1) and eq(2) to get L1 and V1;
L1  570.18kg and V1  25.43kg
xA1  0.17, y A1  0.09, xC1  0.03 and yC1  0.9
% recovery in isopropyl solution is 96.93%
Answer
52

similar documents