Assignment title: Information
MODULE TITLE : MASS AND ENERGY BALANCE
TOPIC TITLE : EXAMPLES OF MASS BALANCE
LESSON 2 : CHEMICAL CHANGES
MAEB - 2 - 2
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1
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INTRODUCTION
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In this lesson we will look at solving further examples of mass balances
involving both non-reacting (physical changes) and reacting (chemical
changes) systems. The principles used to solve the problems can be applied to
most processes as we will be simply using blocks to indicate processes, thus
the method is independent of the process actually occurring.
The problems will all be steady state problems where the conditions within the
process do not change and will involve both overall, component and atomic (of
individual elements) mass balances.
Some of the work will reinforce the work already done in earlier lessons.
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YOUR AIMS
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At the end of this lesson, you should be able to solve mass balance problems
for physical and chemical changes involving:
• stoichiometric and non-stoichiometric (limiting and excess reactants)
reactions
• reactions which have less than 100% conversion
• recycle streams
• purge streams.
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STUDY ADVICE
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The examples will involve both simple mathematics and the use of
simultaneous equations and you should be familiar with solving these
equations.
You will need a copy of the Periodic Table and a calculator to solve the
problems in this lesson.
Note that in this lesson the phase (solid(s)
solution (aq)
others it has. When balancing chemical equations and in some calculations the
phase is unimportant. However, when describing a process or in certain
calculations, the phase may be important and as you progress through this
module this importance will become clearer.
, liquid(l)
, gas(g) or in water
) of a material has not been included in some equations whilst in
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SIMPLE BALANCE INVOLVING STOICHIOMETRIC QUANTITIES
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Example 1
15 m3 h–1 of an acidic effluent contains 12 grams per litre (g l–1) of
hydrochloric acid (HCl) and 5.5 g l–1 of sulphuric acid (H2SO4). It is to be
neutralised using 1 molar sodium hydroxide (NaOH) solution.
The reaction equations (not balanced) are
HCl NaOH NaCl H O
+ ⎯→⎯ +
2
H SO NaOH Na SO H O
+ ⎯→⎯ +
24 24 2
Calculate the rate of addition of sodium hydroxide solution.
Solution
The first thing to notice here is that the units used for concentrations are
different; two values are in g l–1 and the other in moles. The first thing we
need to do is to standardise the units. We will do this both ways to show that,
whichever unit is used, it will produce the same result.
What is the formula for changing g l–1 to molarity or vice versa?
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molarity g l of A
= ( )
molecular mass of A
1
–
Mr
or gl molarity –1 = × Mr
For HCl, from the Periodic Table, H = 1, Cl = 35.5 so the Mr of HCl
= 1 + 35.5 = 36.5
molarity of HCl solution in the effluent 12 = .0
36.5
= 0 329 .
For H2SO4, from the Periodic Table, H = 1, S = 32, O = 16, so the Mr of
H2SO4 = 2 × 1 + 32 + 4 × 16 = 98
molarity of H SO solution in the effluent 2 4 = 5.5
98
= 0 056 .
For NaOH, from the Periodic Table, H = 1, Na = 23, O = 16, so the Mr of
NaOH = 23 + 16 + 1 = 40
Amount of NaOH molarity g l 1 = × =× = Mr 1 40 40 –
We now need to check that the equations for neutralisation are balanced.
HCl NaOH NaCl H O2 + ⎯→⎯ +
This equation is balanced as it has 1 Na, 1 O, 2 H and 1 Cl atoms on both
sides.
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H SO NaOH Na SO H O 24 24 2 + ⎯→⎯ +
This equation is not balanced as it has only 1 Na on the left and two on the
right and unequal numbers of Hs. We balance the Na by putting a 2 in front of
the NaOH and then balance the Hs and Os by putting a 2 in front of the H2O to
H SO NaOH Na SO H O 24 24 2 + ⎯→ 2 2 ⎯ +
Doing the calculation based on molarity: for the HCl
HCl + NaOH NaCl H O
1 mole 1 mole
0.329 mols re
But as the NaOH is 1 mole per litre, we require 1 × 0.329 litre = 0.329 litres
per litre of effluent for neutralisation of the HCl.
But we have15 m3 h–1 (15 000 litres h–1) of effluent, so
⎯ →⎯ + 2
quires 0.329 mols
volume of NaOH required 0.329 15 000
= ×
= 4935 itres h
l
– = 4 935 .
1
1
–
m h
3 1
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Similarly for
H SO + 2NaOH Na SO 2H O
24 24 2 ⎯ →⎯ +
1 mole 2 moles
6 mols requires 2 0.056 0.112 mols × =
But the NaOH is 1 mole per litre so we only require 0.112 litres per litre of
effluent for neutralisation of the H2SO4.
But we have 15 m3 h–1 not just 1 litre, so
volume of NaOH required 0.112 15 000
= ×
1
1
–
= 1680 tres h
li
– = 1 68 .
m h
3 1
Therefore total amount of 1 molar NaOH required
= += 4 935 1 68 6 615 . .. m h 3 1–
Repeating this calculation but this time using mass rather than moles:
HCl + NaOH NaCl H O
1 mole 1 mole
⎯ →⎯ + 2
36.5g require
s 40 g
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There are 12 grams HCl present in 1 litre of effluent and we have 15 m3 h–1
= 15 000 litres. So mass of HCl = 12 × 15 000 g = 180 000 g
mass of NaOH required 40 180 000
∴ =
×
36.5
197 260 g
=
The NaOH solution is 40 g l–1, so
number of litres 197 260
m3 = == 4932 4 932 .
40
Similarly for
H SO + 2NaOH Na SO 2H O
24 24 2 ⎯ →⎯ +
1 mole 2 moles
98 g
There are 5.5 g l–1 of H2SO4 and we have 15 m3 h–1 (15 000 l h–1), so
requires 2 4 8 g × = 0 0
mass of
∴
mass of H SO g
NaOH required 80 82 500
5 5 15 000 82 500 .
2 4 =× =
× = 67 347
98 = g
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The NaOH solution is 40 gl–1, so
number of litres 67 347
3 = ==
t
otal amount of 1 molar NaOH required 4.932 = +
1684 1 684 .
40
1.684
6.616 m h 3 –1
=
This figure is basically the same as calculated on a mole basis, the slight
differences being due to rounding errors.
This is a simple example where the reactants are present in stoichiometric
quantities. Let us now move on to look at another question where non-
stoichiometric quantities are present.
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NON-STOICHIOMETRIC BALANCE
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Example 2
2600 kg of a 20% w/w sulphuric acid (H2SO4) solution in water is mixed with
7200 kg of a 10% w/w solution of potassium hydroxide (KOH) in water. The
balanced reaction equation is
H SO KOH K SO H O 2 4 ( ) ( ) ( ) () aq aq aq l 2 4 2 + ⎯→ 2 ⎯ +
(a) Determine the limiting reactant and the excess reactant (show your
calculations).
(b) Calculate the % excess of the excess reactant.
(c) Assuming 100% conversion of the limiting reactant, determine the
composition of the final mixture as percentages by weight (i.e. % w/w).
Solution
(a) The limiting reactant is the one that determines the amount of product
formed and the excess reactant is present in greater amounts than that
required by stoichiometry to react with the limiting reactant.
The first step, therefore, is to determine the amounts of each material that
react together. To do this we need to calculate the relative molecular
masses (Mr) of each reactant. As the question later asks for the
composition of the final product, we will also determine the relative
molecular masses of the products.
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For this we need the Periodic Table to find the values of relative atomic
mass (Ar
Look up the relative atomic mass of each element present from your Periodic Table
and use these values to calculate the relative molecular mass of each reactant and
product. The answers are in the text that follows.
) of each element present.
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The elements present are hydrogen (H), sulphur (S), oxygen (O), and
potassium (K). The Ar values are H = 1, S = 32, O = 16 and K = 39.
of H SO
M
r
M
r
of K SO
M
r
M
2 4 = ×+ + × =
of KOH
2 4
of H O
2 ×+ = 1 16 18
r
2 1 32 4 16 98
39 16 1 =
= ++
2 39 32 4 16 174
=× + +× =
2
=
56
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From the balanced equation, and substituting the Mr values to find the
reacting masses by stoichiometry:
H SO 2KOH K SO H O
2 4 aq aq ( ) ( ) ( ) () 2 4 aq 2 + ⎯→⎯ + 2 l
1 mol
e 2 moles 1 mole 2 moles
98 2 56 112 2 18 36 × = ×= 174
We now need to calculate the actual masses of reactants present and see
how they compare with the stoichiometric quantities.
2600 kg of 20% w/w H2SO4 will contain
0 2 2600 520 . × = kg H SO 2 4
7200 kg of 10% w/w KOH will contain
0 1 7200 720 . × = kg KOH
From the equation,
98 kg H2SO4 requires 112 kg KOH
1 kg H2SO4 requires kg KOH
112
98
112 520
594 3 × = .
98
520 kg H2SO4 requires kg KOH
As we have 720 kg of KOH, H2SO4 is the limiting reactant and KOH is
present in excess.
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(b) We found in part (a) that the KOH was the excess reactant, only 594.3 kg
being required but 720 kg was available. The excess is therefore
720 – 594.3 = 125.7 kg KOH
The % excess is given by
amount of excess 100
stoichiometric amount r
equired = × = 127 5 100
×
21 15 .
. . %
594 3
(c) To find the composition of the final mixture we need to look at what
(materials and amounts) is produced in the reaction and what is
unchanged.
We know that:
• all the sulphuric acid is used up as the reaction has 100% conversion
of the limiting reactant so it will not be present in the product
solution
• some KOH is un-reacted and will be present in the product solution
• the water present in the reactants will be unchanged in the reaction
• the products of the reaction (K2SO4 and water) will be present in the
product solution.
Let us draw a diagram to summarise this information.
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2600 kg
20% w/w H2
80% w/w H2
SO4
O
Reaction
7200 kg
10% w/w KOH
90% w/w H2
T kg
x% w/w K2
y% w/w H2
z% w/w KOH
O
If we let the amount of product solution formed = T
Doing an overall mass balance over the reactor
T =+= 2600 7200 9800 kg
Thus we know the total amount of solution.
We can convert the masses out into % w/w composition using the formula
% w/w composition of A amount of material A = × 100
total amount of material
In part (b), we calculated that the amount of un-reacted KOH
= 125.7 kg so the % KOH in the product solution is = 1.28%.
We now need to calculate the amounts of product formed in the reaction.
From the balanced equation earlier:
125 7 100
. ×
9800
H SO 2KOH K SO 2H O
2 4 ( ) ( ) ( ) () aq aq aq l 2 4 2 + ⎯→⎯ +
98 174
2 18 36 × =
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We have 520 kg of H2SO4 (20% w/w of 2600 kg) present so
amount of K SO formed 174 520
2 4
= × = 923 3.
98 kg
%K
2
4 = × = 923 3 100
SO
9 42 . . %
9800
amount of water for
med 36 520
98 = kg × = 191 0.
and we also have the water in the reactant solutions, which remains
unchanged, so the amount entering = amount leaving.
Water entering = 80% w/w of 2600 kg (in acid solution) + 90% w/w of
7200 kg (in the KOH solution) = 0.8 × 2600 + 0.9 × 7200 = 8560 kg.
This will leave unchanged and will need to be added to the water formed
in the reaction to give the total
8560 191 0 8751
+ =
% water 8751 100
kg
.
= × =
89 30
%
.
9800
Putting these values into a table to summarise the process gives us:
Material
KOH
H2
SO4
K2
SO4
H2
O
Total
IN (kg) Out
720
520
0
8560
9800
(kg) % composition
125.7
0
923.3
8751
9800
1.28
0
9.42
89.30
100
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Let us look at a slightly more complex example.
Example 3
Butane (C4H10) is burned with 15% excess air based on the complete
combustion of butane to carbon dioxide and water. However, the combustion
is incomplete and only 90% of the carbon is converted to carbon dioxide, the
other 10% forming carbon monoxide. The unbalanced reaction equations are:
C H O CO H O
+ ⎯→⎯ +
4 10 2 2 2
C H O CO H O
+ ⎯→⎯ +
4 10 2 2
(a) Balance the reaction equations.
(b) Determine the molar composition of the product gases.
Assume the water formed is a vapour and that air contains 79 mol% nitrogen
and 21 mol% oxygen.
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Solution
(a) First we need to balance the equations, i.e. the number of atoms of each
element present must be the same either side of the arrow.
Looking at the first equation, balance the carbons first. There are 4 on the
left and only 1 on the right so we need to put a 4 in front of the carbon
dioxide. The hydrogens are next with 10 on the left and 2 on right, so we
need to put a 5 in front of the water giving:
C H O CO H O 4 10 2 2 2 + ⎯→⎯ + 4 5
That only leaves the oxygen, 2 on left and 13 on the right! We could put
6.5 in front of the O2 on the left to balance the oxygens but chemists do
not normally like decimals or fractions in reaction equations so we double
everything to give:
2C H O CO H O 4 10 2 2 2 + ⎯→ 13 8 10 ⎯ +
Try to balance the second equation given below.
C H + O CO + H O 4 10 2 2 ⎯ →⎯
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Looking at the carbons first, there are 4 on the left and only 1 on the right
so we need to put a 4 in front of the carbon monoxide. The hydrogens are
next with 10 on the left and 2 on right, so we need to put a 5 in front of
the water giving:
C H O CO H O 4 10 2 2 + ⎯→⎯ + 4 5
That only leaves the oxygen, 2 on left and 9 on the right. We could put
4.5 in front of the O2 on the left to balance but, as before, we double
everything to give:
2C H O CO H O 4 10 2 2 + ⎯→ 9 8 10 ⎯ +
(b) Having balanced the equations we now need to look at the quantities
involved in the complete combustion reaction in order to calculate the
amount of excess air (15%) used.
The answers required are in moles (% molar composition) and the
composition of air is in moles so we will use moles in our answer. We
will also assume that 100 moles of butane are available (to help with
percentages).
By stoichiometry
2C H O CO H O 4 10 2 2 2
2 moles 13 moles 8 mo
+ ⎯→ 13 8 10 ⎯ +
les 10 moles
We have available 100 moles butane and for this we need
13 100
650 × = moles O2
2
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The question states that 15% excess air (oxygen) is used = 15% of 650
= 97.5 moles giving a total oxygen of 650 + 97.5 = 747.5 moles.
As oxygen is 21 mol % of the air, the nitrogen present (79%) with the
oxygen
= × = 747 5 79
2812 . moles.
21
All this nitrogen takes no part in the reactions and so will pass through
unchanged (it could be ignored from a mass balance point of view but
will be required for the composition of product gases). It is an example of
a tie substance.
To see what we have let us put this information onto a diagram.
Air
747.5 moles O2
2812 moles N2 Products
Butane (C4
100 moles
H10)
Combustion
Oxygen
2812 moles nitrogen
Carbon dioxide
Carbon monoxide
Water
As we are told in the question that 90% of the carbon forms carbon
dioxide and 10% forms carbon monoxide we can say that 90% of butane
forms carbon dioxide and 10% forms carbon monoxide (butane is the
only source of carbon).
Looking at the equation for producing carbon dioxide
By stoichiometry
2C H O CO H O 4 10 2 2 2
2 moles 13 moles 8 mo
+ ⎯→ 13 8 10 ⎯ +
les 10 moles
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If the reaction were complete, 100 moles of butane would produce
8 100
10 100
and require 13 100
× =
2
× =
2
650 mo × = les of oxygen
2
400 moles of carbon dioxide
500 moles of water
However, only 90% conversion is achieved so:
carbon dioxide actually produced 90% of 400 = =
360 moles
water produced 90% of 500 4 0 mo 5 les
oxygen used 90% of 650 5 moles = = 58
= =
Similarly for the carbon monoxide reaction
2C H O CO H O 4 10 2 2
2 moles 9 moles 8 moles
+ ⎯→ 9 8 10 ⎯ +
10 moles
If we have 100 moles of butane we would get 400 moles of carbon
monoxide , 500 moles of water and use 450 moles of oxygen.
However, as only 10% is converted:
carbon monoxide produced 10% of 400 40 mole = = s
water produced 10% of 500 0 moles
oxygen
= = 5
used 10% of 450 moles = = 45
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From the above two sets of calculations the total oxygen used = 585 + 45
= 630 moles.
Since we started with 747.5 moles oxygen, the amount un-reacted
= 747.5 – 630 = 117.5 moles
Adding these figures to the diagram and also putting the results into a
table gives.
Air
747.5 moles O2
2812 moles N2 Products
Butane (C4
100 moles
H10)
Combustion
117.5 moles oxygen
2812 moles nitrogen
360 moles carbon dioxide
40 moles carbon monoxide
500 moles water
Material
Butane
Oxygen
Nitrogen
Carbon dioxide
Carbon monoxide
Water
Total
IN OUT
100
747.5
2812
0
0
0
3659.5
Moles
0
117.5
2812
360
40
500
3829.5
Mol%
We can complete the table by calculating the mol % composition using
mol % composition of A moles of A 100
total = ×
number of moles
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Do this for yourself and check your answers with ours.
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Material
Butane
Oxygen
Nitrogen
Carbon dioxide
Carbon monoxide
Water
Total
IN OUT
100
747.5
2812
0
0
0
3659.5
Moles Mol%
0
117.5
2812
360
40
500
3829.5
0
3.07
73.43
9.40
1.04
13.06
100
Looking at this, there appears to be an 'imbalance'. The number of moles IN
does not equal the number of moles OUT. This is because we have done the
balance using individual materials and the stoichiometry of the reactions
means there is a change in the number of moles during the reactions, e.g. for
the formation of CO:
2C H 9O 8CO 10H O
4 10 2 2 + ⎯→⎯ +
2 moles 9 moles 8 moles
10 moles
11 moles IN produce 18 moles OUT
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We could have solved this problem in other ways. We could have done a
balance on the moles of each element present (C, H, O, and N) or we could
have converted moles to mass and done a mass balance. In both these cases
the numbers in and out would be the same.
We will do both the element and mass balance to check our answer is correct.
We can ignore the nitrogen as it passes through the process unchanged.
As the formula represents the number of atoms of each element present,
e.g. C4H10 has 4 carbons and 10 hydrogens and O2 has 2 atoms of oxygen, on
an element basis and using the figures in the previous table:
Material IN (moles)
OUT (moles)
Butane (C4
Oxygen (O2
Carbon dioxide (CO2
Carbon monoxide (CO)
Water (H2
Total
Material C H O Material C H O
H10)
)
)
O)
100
400
0
747.5
0
0
0
0
0
0
847.5
400
1000
0
1495
0
0
0
1000
1495
0
0
117.5
360
0
40
0
500
0
1017.5
0
0
360
40
0
1000
400
1000
0
0
0
0
To do the mass balance we need the relative molecular mass Mr of each of the
materials. (The Ar values from the Periodic Table are C = 12, O = 16,
H = 1).
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M
r
of butane C H
of oxyg
M
r
2 16 32
of carbon dioxide CO
Mr 12 2 16 44
( ) 4 10 = × + ×= 4 12 10 1 58
en O
( ) =× =
2
( ) =
2
M ( ) = + r of carbon monoxide CO =
+× =
12 16
28
M 2 1 16 18 r of water H O2
( ) = ×+ =
We can now calculate the mass of each material entering and leaving and put
these into the new table.
Material IN
Butane (Mr
Oxygen (Mr
Carbon dioxide (Mr
Carbon monoxide (Mr
Water (Mr
= 58)
= 32)
= 44)
= 28)
= 18)
Total
Moles Mass g
100
747.5
0
0
0
847.5
(Moles × Mr
5800
23920
0
0
0
29720
Moles Mass g
)
0
117.5
360
40
500
1017.5
OUT
(Moles × Mr
0
3760
15840
1120
9000
29720
You can see from these two checks that we do have a balance.
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CHEMICAL CHANGE WITH RECYCLE
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Example 4
Propene (C3H6) is to be chlorinated to produce dichloro-propane (C3H6Cl2) in
a reaction which has the balanced equation:
C H Cl C H Cl 3 6 2 ( ) ( ) () gg l 3 6 2 + ⎯→⎯
An excess of propene is required in the reactor as this suppresses any side
reactions and allows 96% conversion of the chlorine to dichloro-propane. The
reaction mixture is then separated into product, un-reacted propene and
chlorine, the latter two being recycled back to the reactor for further
processing.
(a) Draw a block diagram of the process.
(b) If 500 kg h–1 of dichloro-propane is required as product, determine the
mass of propene and chlorine required in the fresh feed at the steady state.
(c) If the mole ratio of propene to chlorine at the reactor inlet is 5:1
determine:
(i) the composition of the feed to the separator
(ii) the flow rate and composition of the recycle stream.
Data: The relative atomic masses (Ar
) of C = 12, H = 1 and Cl = 35.5.
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Solution
As in a previous example, there appears to be a conflict of units. In part (b) we
are given 500 kg h–1 of dichloro-propane but in part (c) we are given mole
ratio. As we saw, the actual choice of unit is unimportant as they should give
the same result. At this stage we will stick with mass as it demonstrates the
alternative not covered earlier when we used moles.
(a) The first thing the question asks for is the block diagram.
Try to do this first for yourself before looking at our answer.
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Reactor Product
Propene
Chlorine
Dichloro-propane
Feed
Propene
chlorine
Reactor
Separator
Product
Dichloro-propane
Recycle
Propene
Chlorine
We have not put any figures on it as yet as we are not given much in the
question!
(b) Where do we start? Well looking at our block diagram we see that, for
mass, the feed in = product out as there are no other streams entering or
leaving. So at the steady state the fresh feed in of propene and chlorine is
all converted to dichloro-propane.
We are given five bits of information in the question, the balanced
reaction equation, excess propene present, 98% conversion of chlorine,
500 kg h–1 of dichloro-propane are required and the relative atomic
masses of the elements present.
Let us look at the equation
C H Cl C H Cl
3 6 2 ( ) ( ) () gg l 3 6 2 + ⎯→⎯
1 mole 1 mole 1 mo
le
From this equation we see that for every mole of product we need 1 mole
of propene and 1 mole of chlorine. We require 500 kg h–1 of dichloro-
propane, how is this changed into moles or how do we change moles into
mass in kg? Well we use the equation we met earlier.
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moles of A mass of A in grams
relative molec = ular mass of A
( )
M
of propene C H3 6
M
r
( ) = × 3 12 6 1 42 + ×=
M ( ) = r of dichloro-propane C H Cl 36 2 3 12 6 1 2 35 5 113
× + ×+ × =
.
Mr of chlorine Cl2 35 5 71 . =
( ) = ×
2
So we can now use grams instead of moles in our reaction
C H Cl C H Cl
3 6 2 ( ) ( ) () gg l 3 6 2 + ⎯→⎯
42 g 71 g 113 g
So 42 g of propene + 71 g of chlorine will produce 113 g dichloro-
propane. But the question says we need 500 kg h–1 of dichloro-propane.
By ratio, to produce 500 kg h–1 of dichloro-propane we need
and 71
42 500
185 84 × = . – kg h of propene
113
× = 500
113 kg h of chlorine 1 314 16 . –
1
These figures answer part (b) and we can now put them onto the block
diagram.
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Feed
185.84 kg h–1
Propene
314.16 kg h–1
Chlorine
Reactor Product
Propene
Chlorine
Dichloro-propane
Reactor
Separator
Product
500 kg h–1
Dichloro-propane
Recycle
Propene
Chlorine
(c) For part (c) we are told that the feed to the reactor is in the mole ratio of 5
moles propene to 1 mole chlorine (i.e. for every 6 moles of feed entering
there are 5 moles propene and 1 mole chlorine) and that 96% of the
chlorine is converted.
From the reaction equation, we saw 1 mole of propene requires 1 mole of
chlorine to produce 1 mole of dichloro-propane.
Therefore for every 6 moles of feed, only 0.96 moles of chlorine along
with 0.96 moles of the propene are converted to 0.96 moles of dichloro-
propane. Thus the reactor product composition on this basis is
0.96 mole dichloro-propane formed
( ) = mole chlorine unreacted
1 0 96 0 04 –. .
( ) 5 0.96 = 4 04 . mole propene unreacted
–
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Convert this to mol %
⎟ moles of A 100
⎜ ⎞
⎛ ×
=
⎝
total moles
⎠
Dichloro-propane is
0 96 100
... . % ×
19 05 .
+ + =
0 96 0 04 4 04
Chlorine is
0 04 100
... . % ×
0 79 .
+ + =
0 96 0 04 4 04
Propene is
4 04 100
... . % ×
80 16 .
+ + =
0 96 0 04 4 04
The dichloro-propane is removed by the separator and the composition of the
recycle stream in terms of relative number of moles will be 4.04 moles propene
to 0.04 moles of chlorine, which as a mole ratio of propene: chlorine
= 4.04: 0.04 = 101:1.
This recycle stream is mixed with 500 kg h–1 of fresh feed having a mole ratio
of 1:1 to give a reactor feed which has a molar ratio of propene:chlorine = 5:1.
Let us look at that part of the process more closely and put the figures on the
diagram.
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Fresh feed 500 kg h–1
185.84 kg h–1
314.16 kg h–1 chlorine
propene
Unknown amount of reactor feed
Propene: chlorine = 5:1
Unknown amount of recycle
Propene: chlorine = 101:1
To solve this problem we need to convert all concentrations to the same units.
We could use moles or mass. We used moles in the last example, let us now
look at mass.
mass number of moles = × Mr
But the figures given are as ratios so we will need to work out relative masses
and hence % w/w in the mixtures.
Reactor feed
relative mass of propene number of moles = × Mr of propene
relative mass of chlor
% w/w propene mass of propene
=× =
ine number of moles of chlorine
= ×
=× =
total mas = × 100
= × = 210 100
5 42 210
Mr
17 7
1 1
s of reactor feed
74 73 . %
281
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% w/w chlorine mass of chlorine
total ma = × 100
= × = 71 100
ss of reactor feed
25 27 . %
281
Recycle stream
relative mass of propene number of moles = × Mr of propene
relative mass of ch
Mr
1
w/w propene mass of propene
100 %
= ×= 01 242
1 42 4
lorine number of moles of chlorine
= ×
1 7
= ×
= ×
1 =
7
total
mass of recycle
= × = 4242 100
98 35 . %
4313
% hlorine mass of chlorine
w/w c
total mass of = × 100
= × = 71 100
recycle
1 65 . %
4313
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Redoing the diagram, and letting the amount of recycle be x kg h–1
Fresh feed 500 kg h–1
185.84 kg h–1
314.16 kg h–1 chlorine
propene
(500 + x) kg h–1reactor feed
74.73% w/w propene
25.27% w/w chlorine
x kg h–1recycle
98.35% w/w propene
1.65% w/w chlorine
Doing an overall mass balance gives:
fresh feed recycle reactor feed
+ =
500 500
+= + x x
No help!!
Doing a balance over propene gives:
185 84 98 35 74 73 500
. .% .%
185 84
185 84 0 9835 373 6
0 9835 0 7473 373 65 185 84
. –. . – .
+= + ( )
0 9835 0 7473 500
+
.
. .
+ = 5 0 7473
.. .
x x
0 23
.
of of x x
= ×+ ( )
x x
x
+
.
x
=
62 187 81
x
=
795 1
x
=
.
. – kg h 1
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The amount of recycle is 795.1 kg h–1 and it composition is 98.35% w/w
propene and 1.65% w/w chlorine (mole ratio 101:1).
To check do a mass balance on the chlorine
314 16 795 1 0 0165 1295 1 0 2527
+× = ×
. .. ..
327 27 327
= .27
.
A balance!
Try doing the last part using moles instead of mass. Remember this is just a mixing
stage and what goes in as propene stays propene and the same with chlorine.
Fresh feed 500 kg h–1
185.84 kg h–1
314.16 kg h–1 chlorine
propene
Unknown amount of reactor feed
Propene: chlorine = 5:1
Unknown amount of recycle
Propene: chlorine = 101:1
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Fresh feed
Convert mass rate into moles, remember the mass quoted in the feed is in
kilograms and needs to be converted to grams:
moles propene mass of propene in g
moles chlorine mass
of pro = Mr pene
185 84 1000
= × =
4425 .
42
of chlorine in g
=
M of chlorine r
= 314 16 100 . × 0
71 = 4425
The feed is correct in mole terms (need to be equal), i.e moles propene = moles
chlorine.
Total feed = 8850 moles.
Recycle
This is 101 moles propene to 1 of chlorine by ratio which means for every
102 moles, 101 are propene and 1 is chlorine.
mole fraction propene moles propene
total mo = les
101
= =
102
0 99 .
mole fraction chlorine mole
s chlorine
total moles = = 1
=
0 01 .
102
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But we do not how many moles in total there are per hour. Let this be T.
The amount of propene in moles in the recyle
=× = T T mole fraction of propene 0.99
The amount of chlorine in moles in the recyle
=× = T T mole fraction of chlorine 0.01
Reactor Feed
The reactor feed is a molar ratio of propene:chlorine of 5:1 so for every
6 moles entering 5 are propene and 1 is chlorine.
mole fraction of propene moles propene
mole fraction of chlorine m
total = moles
oles chlorine
total moles = = 1
=
5
0 833 .
= =
6
0 167 .
6
The amount of reactor feed in moles
= + =+ fresh feed recycle 8850 T
Therefore, the amount of propene in moles will be
( ) 8850 8850 + T × = mole fraction of propene 0.833( ) + T
and the amount of chlorine in moles will be
( ) 8850 885 + T × = mole fraction of chlorine 0.167 0 ( ) + T
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We now have consistent units so we can do balances.
Over propene
4425 0 99 0 833 8850
4425 0 99 7372 0
+ =× + ( )
T T
. .
T 833T
+ =+
. .
Rearranging
( ) =
0 99 0 833 7372 4425
. –. –
T
0 157 2947
T
.
T moles
=
18 771
=
Check using balance over chlorine:
×+ = × + ( )
0 5 8850 0 01 0 167 8850
. ..
4425 0 01 147
T T
T 8 0 167
+ =
.
2947 0 157
T moles
+
T
=
.
18 771
=
T
.
(We may have used less significant figures in our calculation and ended up
with not quite as good a balance.)
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We could do a further check by seeing if this matches the mass balance.
If T = 18 771 moles, of which mole fraction chlorine is 0.01 and propene is
moles chlorine
∴ ==
mass moles
moles propene =× =
mass 18 583 42
0 01 187 71 . . T
= ×
13327 4
. g 13.33 kg
= =
0.99 18 771 18 583
= ×
780 486 g 780.49 kg
= =
Mr = ×
187 71 71
.
total mass in recycle 780.49 13.33 79 = += 3.82 kg
This compares to the 795.1 kg we calculated earlier, a difference of 0.15%.
This difference would be due to rounding errors. The summary table is:
Fresh feed
Propene
Chlorine
Dichloropropane
Total
Moles
Mass
4425
185.84
4425
314.16
0
0
8850
500
Final
products
Moles
Mass
0
0
0
0
4425
500
4425
500
Reactor
feed
Moles
23010
4603
0
27613
Mass
Moles
966.34
18585
327.49
185.7
0
4425
1293.8
23196
Reactor
products
Mass
780.5
13.33
500
1293.8
Recycle
stream
Moles
18585
185.7
0
18771
At this stage we should point out then when you solve problems, whilst we
would expect enough explanation to enable us to see what you were trying to
do, it would not need a lot of the additional explanation we have given here.
You would also only need to do one check not all the ones we have done in this
example!!
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________________________________________________________________________________________
BALANCE INVOLVING A PURGE STREAM
________________________________________________________________________________________
Example 5
In the production of ethanol from ethene and steam, equimolar quantities of
ethene and steam are passed into a high temperature catalytic converter and the
resultant reaction conversion is 70%.
Equation:
C H H O C H OH
24 2 25 + ⎯→⎯
ethene steam ethanol
1 mole 1
mole 1 mole
The ethanol is separated by absorption and the un-reacted ethene and steam are
recycled to the reactor. The ethene contains 0.01% impurity, which if allowed
to build up to above 1% in the feed to the catalytic converter causes the
catalyst to be poisoned.
Estimate the minimum % of the recycle stream leaving the separator that must
be removed (purged) to prevent the catalyst being poisoned.
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Solution
50 moles ethene
(0.01% impurity)
50 moles steam
Reactor
Separator
Recycle
Purge
All we are given is that the feed is equimolar (equal quantities), with 0.01%
impurity in the ethene, and that in the reactor, the conversion rate = 70%.
To solve this we need to make some assumptions. Let the feed be 100 moles,
50 moles ethene and 50 moles steam. (We could have chosen any number of
moles but 100 is used as it is a good number to work out values on!) There is
0.01% impurity in the feed ethene = 0.01% of 50 = 0.05 moles.
We will assume that the ethene impurity is additional and not part of the ethene
moles. If not then the problem is made much more difficult. The amount of
ethene will only be 49.95 moles compared to 50 moles steam, and hence steam
will build up in the recycle to produce a recycle stream richer in steam than
ethene. This will be reflected in changes in the purge concentration and the
amount of steam compared to ethene removed in the purge and entering the
reactor. Let us remove that complication by making it extra – the question
does say estimate not calculate!
Let the amount of recycle entering the reactor = x moles and the amount of
purge = y moles
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total moles entering reactor 100
= + x
maximum m
oles of impurity that can enter 1% of tota = l feed
= + 0.01 100 moles ( ) x
This stays with the un-reacted ethene and steam
moles ethene and steam leaving system 30% o = f feed
[70% is converted to ethanol which is removed]
∴
mole fraction of impurity in recycle
= + ( )
0 3 100 . x
0.01 = ( ) +
( ) + = 100
. x .
0 3 100
x
0 033
The purge rate must be such as to remove all the fresh impurity added
= 0.05 moles.
Thus (mole fraction of impur 0 05 0 033 . . = y ity purge rate in moles)
×
y = 1 515 .
moles
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Doing a balance around the purge point:
moles ethene and steam leaving reactor 0.3 1 = ( ) 00 + x
This must equal the amount recycled + the amount of purge
( ) + = +
0.3 100
But 1 515
30 0 3
+ =
28 485 0 7
. .
= +
x y
x xy
y
=
.
. +
x x
1 515
.
% of reactor recyle leaving as purge 1.515 100
=
x moles
40 69
=
.
40.69 1. = ×
x
+ 515 = 3 59 . %
This lesson has covered quite a lot of calculations and has offered several
different alternative ways of solving the problems. If done correctly, the result
should be independent of the method chosen and in practice engineers often
use alternatives to check their answers. When dealing with the large quantities
of potentially hazardous chemicals being processed, getting it wrong can have
serious consequences, on both economic and safety issues.
That completes our look at mass balances. Test your understanding of this
lesson by attempting the Self-Assessment Question which follows.
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________________________________________________________________________________________
SELF-ASSESSMENT QUESTION
________________________________________________________________________________________
Note: This is a complex problem incorporating all the types of calculation met
during the lesson. It is not a real process. It is important to break it up into
stages, calculating each answer in turn.
150 kg h–1 of tri-iron tetroxide (Fe3O4) is reacted with 10% excess hydrogen
gas containing 0.01% v/v chlorine gas at 500°C to produce iron and water with
60% conversion of the tri-iron tetroxide. The resultant iron/tri-iron tetroxide
/hydrogen/water mix is separated into hydrogen gas, liquid water and solid
iron metal/tri-iron tetroxide mix. The solid is recycled back to the reactor and,
at the steady state, the solid removed from the process is 95% w/w iron metal,
5% w/w Fe3O4. The hydrogen gas is also recycled back to the reactor taking
with it the chlorine impurity. The level of chlorine leaving the reactor must not
exceed 1% by volume to prevent side reactions.
The unbalanced reaction is:
Fe O H Fe H O 34 2 2 + ⎯→⎯ +
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(a) Balance the reaction equation.
(b) Calculate:
(i) the amount in kg h–1 of pure iron removed
(ii) the composition of the solid feed to the reactor
(iii) the amount in kg h–1 of recycled iron/ tri-iron tetroxide mix
(iv) the amount in kg h–1 of hydrogen required in the reactor
(v) the amount in kg h–1 of recycled hydrogen purged
(vi) the amount in kg h–1 of hydrogen recycled to the reactor
(vii) the amount in kg h–1 of fresh hydrogen feed.
Data:
Relative atomic mass of
hydrogen (H) = 1, oxygen (O) = 16, iron (Fe) = 56, chlorine (Cl) = 35.5.
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________________________________________________________________________________________
ANSWER TO SELF-ASSESSMENT QUESTION
________________________________________________________________________________________
(a) The unbalanced equation is
Fe O H Fe H O 34 2 2 + ⎯→⎯ +
First balance the Fe, 3 on the left 1 on the right, so multiply Fe on right by
3
Fe O H Fe H O 34 2 2 + ⎯→⎯ + 3
Now balance the O, 4 on left 1 on right, multiply H2O by 4
Fe O H Fe H O 34 2 2 + ⎯→⎯ + 3 4
Finally balance the H, 2 on left 8 on right, multiply H2 by 4
Fe O H Fe H O 34 2 2 + ⎯→ 4 34 ⎯ +
(Your answer may be a single line: Multiply Fe by 3, H2O by 4 and H2
by 4.)
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(b) Diagram:
150 kg h–1 Fe3
99.99% v/v hydrogen
0.01% v/v chlorine
O4
Reactor
Separator
95% w/w Fe
5% w/w Fe3
Water
Hydrogen purge Hydrogen recycle containing max
1% v/v chlorine
First we need to determine the relative molecular masses (Mr
within the equation:
) of materials
a
( ) =
M
r
of Fe O
M
r
M
ctually of Fe
A
r
of H O2
M
r
=× + × =
3 4
of H
= ×=
2
r
= ×+ =
3 56 4 16 232
212
56
2 1 16 18
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(i) To calculate the amount of iron produced we need to look at the
balance over iron. 150 kg h–1 of Fe3O4 enter and the product is 95%
w/w iron (Fe) and 5% w/w of Fe3O4
Do an element balance over Fe
In Fe3O4, the amount of iron is 168 (3 × 56) out of a total of 232
= × = 168 100
72 41 . %
232
150 kg h–1 of Fe3O4 feed contains 72.41% of Fe, so the amount of
Fe = 0.7241 × 150 = 108.62 kg h–1
The product out is 95% w/w pure iron and 5% w/w of Fe3O4.
Let the total mass of solid out = x
amount of Fe out 95% of of 72.41% of = + x x 5%
0 95 0 0362
= +
. .
x x
108 62 0 9862
∴ =
. .
110 14
x
=
x
kg h
–1
.
95% of this is pure iron 0.95 110.14
5 0 05 110 14 5 51 kg % . ..
is Fe O
=× =
=× =
3 4
104.63 kg h
1
–
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(ii) To produce the reactor product, the Fe3O4 feed is 60% converted into
pure iron. If we assume 100 kg of solid reactor product, 95 kg
(95% w/w) are iron and 5 kg (5% w/w) are un-reacted Fe3O4. The
5 kg of un-reacted Fe3O4 in the product must therefore be 40% of the
feed Fe3O4. (60% of the feed of Fe3O4 is converted.)
Therefore, the total feed into the reactor of Fe3O4 to give 100 kg of
final product is = 12.5 kg, 7.5 kg being converted to iron.
The iron is produced according to the following reaction equation
which, when Mr values and stoichiometry are considered, gives:
Fe O 4H 3Fe 4H O
34 2 2 + ⎯→⎯ +
232 g 4 2 8 g 3 56 168 g
5 100
×
40
×= × =
7.5 kg of Fe3O4 will produce
168 7 5
232
5 43 × = . . kg of Fe.
But 95 kg of iron are in the reactor product so 95 – 5.43 = 89.57 kg
must be present in reactor feed.
Thus to give a reactor product composition of 95 kg iron, 5 kg Fe3O4,
the feed must be 89.57 kg iron and 12.5 kg Fe3O4, giving a total
reactor feed of 102.07 kg.
The % composition of the reactor feed is
Fe 89.57 100
= × =
Fe O 1 100
= ×
3 4
. . %
102 07
2 5
.
.07 = 12 25 . %
102
87 75
This is on the basis of 100 kg of product but as we have converted to
% these will be valid for any amount of product formed.
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(iii) At the junction of solid feed and recycle and let the recycle be
y kg h–1.
y kg h–1
95% w/w Fe
5% w/w Fe3
O4
150 kg h–1
100% w/w Fe3
O4
150 + y kg h–1
87.75% w/w Fe
12.25% w/w Fe3
Doing a balance over pure Fe, as it is only present in the recycle
stream in and reactor feed out and no chemical change occurs:
95 87 75
% .%
.
y
0 0725 131 625
. .
of of 150 y y
= + ( )
0 95 0 8775 150
= ×+ 0 8775
y
. .
y
=
= kg h 1
1815 5
y
. –
(iv) The reactor feed is 12.25% Fe3O4 and the total amount of feed is
150 150 1815 5
+= +
y .
1965 5
. – kg h
=
1
amount of
∴
Fe O in reactor feed of 1965.5 kg h 3 4 = 12 25 . % –1
1 = 240 77 . kg h–
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According to the reaction equation
Fe O 4H 3Fe H O
34 2 2 + ⎯→⎯ +
232 4
So for 240.77 kg Fe3O4, the amount of hydrogen required by
stoichiometry = .
A 10% excess is required so the actual amount of hydrogen required
in the reactor = 8.30 + 10% of 8.30 = 9.13 kg h–1.
4
g 2 8 g
× =
8 240 77
8 30 × = . . – kg h 1
232
(v) However, only 60% of the Fe3O4 is converted so only 60% of the
stoichiometric quantity of hydrogen is used.
Hydrogen used in reactor = 60% of 8.30 = 4.98 kg h–1 so hydrogen
leaving the reactor = (9.13 – 4.98) = 4.15 kg h–1.
Amount of fresh hydrogen entering reactor
If the amount of purge kg h of hydrogen
total hydrogen
entering will be 4.98 kg h
chlorine
in this feed 0.01% by volume
= + amount used purged hydrogen
1 = z –
1 = + ( )z –
=
= 0 0001 . er mole of hydrogen
=
mole p
1 mole hydrogen 2 g
1 mole chlorine 71 g =
and
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so moles hydrogen in feed = ( ) + ×
=
4 98 1000
. z
2
24
90 500 +
so moles chlorine in feed 2490 500 ( )z ×
= +
z
0 0001
.
mass of chlorine in feed
= +
= + ( ) ×
= + ( )
=
0 249 0 05
. . z
0 249 0 05 71
. .
17.68 3.55 g h 1
. . – ( ) + z kg h 1
7 68 3 55
1
1000
z
–
z
This must be removed in the purge.
The reactor outlet has 4.15 kg h–1 of hydrogen and the chlorine is 1%
by volume.
So moles of hydrogen in reactor outlet mass = of hydrogen in kg 1000
= ×
=
and moles of chlori
ne in reactor outlet (1%) = ⎡ ×
=
molecular mass of h
4 15 1000
.
2
2075
⎢ ⎤
4 15 1000
. ×
⎣
2
20 75
.
ydrogen
0 01
⎥
⎦
.
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mass of chlorine in recycle 20.75 71 1473.2 = ×= 5 g
=
1.473 kg
% = f chlorine in recycle 100
w/w chlorine in recycle mass o
= ×
total mass of rec
+ = 1 473
. . . %
26 2 .
4 15 1 473
ycle
100
% en in recycle = = 100 26 2 73 8 – . .%
w/w hydrog
If the purge is z kg h–1 of hydrogen, then chlorine removed will be
26 2
× =
z
.
. .
73 8
0 355
z
z hence
( ) +
17 68 3 55
. .
1000
17 68 3 55 355
+ =
. .
351 45 17 68
=
0 355
z
.
z z
z
=
. .
z 0 05 . – kg h 1
=
(We could have estimated this answer as we started with 0.01%
chlorine and end up with 1%, a factor of hundred. Since the recycled
amount is similar to the feed amount, the amount removed will be
about 1/100 of the feed entering. This will not be the case if the
recycle was much greater or less than the feed).
(vi) Thus the amount of hydrogen recycled to the reactor = 4.15 – 0.05
= 4.10 kg h–1.
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(vii)
Fresh hydrogen 4.98 0.05 5.03 kg h 1 =+= – .
As a check if 0.05 kg h–1 of hydrogen is removed in the purge then
chlorine removed kg h = × = 0 05 26 2
0 0178 . .
73 8
g h
17 8. –
=
. . –1
1
hydrogen entering 5.03 kg h
1
–
=
5 03
. ×
=
2
2515
moles
=
1000
so chlorine entering 0.01% of 2515
=
0.0001 moles
= ×
0.2515 moles
=
2515
mount of chlorine in grams moles = ×= Mr 0 2515 . ×
a
71
17 8. g
=
A balance!
So even with very little information, by breaking down the process
into individual steps the problem can be solved. It is very important
to look at the question and the diagram of the system to get clues as
to where to start on a problem.
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53
________________________________________________________________________________________
SUMMARY
________________________________________________________________________________________
In this lesson we have looked at how we can do balances when the materials
taking part in the process are changed by chemical reaction. It is important
that the right concentration units are used throughout, whether moles or mass,
and you know the way to change from one to the other, covered in the early
lessons. The actual answer will not change, whichever method is used, but
experience often tells us which the simplest method is. There are no hard and
fast rules as to where you start and how you get the answer, once again this
tends to come from experience. For those only involved in the educational
side, a good clue as to the method to use is given by the order in which you are
asked to calculate values in a problem but out in industry you haven't got the
luxury of someone else doing the calculation first!!
The most important things in doing mass balances are:
• always break down the process into simple stages
• use a block diagram to visualise what is going on
• put all existing information onto the diagram and update when new
values are obtained
• keep the units consistent
• use all types of mass balance if necessary;
– overall
– over items of equipment
– over junctions/splits of flows
– over all materials
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54
– over individual materials
– over elements
– over reactions
• check your answers by making sure that all materials balance not just
the one you are asked to calculate.
We will be doing some more materials balances later in this module when we
look at energy balances, so be certain you understand how to do them.
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(Engineering)
© Teesside University 2011