material balance

1. M AT E R I A L B A L A N C E W I T H
C H E M I C A L R E A C T I O N .
U N D E R T H E G U I D A N C E O F
P R O F. K A R N E M A D A M
• AKSHAY GUNJAL (66)
• PAYAL KHANDAGALE (67)
• ARJUN PHAD (68)
• JAY JAWALGE (69)
• SHASHIKANT GAIKWAD (70)

2. INTRODUCTION
• Material balances are important first step when designing a
new process or analyzing an existing one.
• Material balances are nothing more than the application of
the law of conservation of mass, which states that mass can
neither be created nor destroyed.
• They are used in industry to calculate mass flow rates of
different streams entering or leaving chemical or physical
processes

3. THE GENERAL BALANCE EQUATION
• Suppose propane is a component of both the input and output streams of a
continuous process unit shown below, these flow rates of the input and output are
measured and found to be different.
Process unit
qin (kg propane/h) qout (kg
propane/h)
Input + generation − output − consumption =
accumulation

4. PROCEDURE FOR MATERIAL BALANCE
CALCULATIONS.
• Draw and label the process flow chart (block diagram). When labeling, write
the values of known streams and assign symbols to unknown stream variables.
Use the minimum number possible of symbols.
• Select a basis of calculation. This is usually the given stream amounts or flow
rates, if no given then assume an amount of a stream with known composition.
• Write material balance equations. Note in here the maximum number of
independent equations you can write for each system is equal the number of
species in the input and output streams of this system. Also note to first write
balances that involve the fewest unknown variables.

5. BALANCES ON REACTIVE SYSTEMS
• Theory of proportions in which chemical compounds react is called stoichiometry.
• A statement of the relative number of moles or molecules reacting to produce
products is given by a chemical equation known as stoichiometric equation.
• For example,
2 moles of SO2 and one mole of O2 react to produce 2 moles of SO3. Then
the stoichiometric equation will be 2SO2+O2→2SO3. Numbers that precede the
formulas are known as stoichiometric coefficients.
• In a stoichiometric equation, the number of atoms in both sides of the equation must
be balanced. Ratio of stoichiometric coefficients of two species is known as
stoichiometric ratio
• EXAMPLE: 2SO2+O2→ 2SO3. What is the stoichiometric ratio of SO2 to SO3?
• Solution stoichiometric ratio of SO2 to SO3 =2 mole of SO reacted/2 mole of SO
produced
2/2 = 1

6. •The chemical compound which is present less
than its stoichiometric amount, will disappear
first. This reactant will be the limiting reactant
and all the others will be excess reactants.

7. YIELD, SELECTIVITY&CONVERSION
• YIELD=
MOLES OF DESIRED PRODUCT FORMED
MOLES OF SPECIFIC REACTANT CONSUME
×STOICHIOMETRIC FACTOR
• SELECTIVITY=
𝐴𝑀𝑂𝑈𝑁𝑇 𝑂𝐹 𝐿𝐼𝑀𝐼𝑇𝐼𝑁𝐺 𝑅𝐸𝐴𝐶𝑇𝐴𝑁𝑇 𝑇𝐻𝐴𝑇 𝑅𝐸𝐴𝐶𝑇 𝑇𝑂 𝐺𝐼𝑉𝐸 𝐷𝐸𝑆𝐼𝑅𝐸 𝑃𝑅𝑂𝐷𝑈𝐶𝑇
𝐴𝑀𝑂𝑈𝑁𝑇 𝑂𝐹 𝑇𝐻𝐴𝑇 𝑅𝐸𝐴𝐶𝑇𝐴𝑁𝑇 𝑇𝑂 𝐺𝐼𝑉𝐸 𝑈𝑁𝐷𝐼𝑆𝐼𝑅𝐸 𝑃𝑅𝑂𝐷𝑈𝐶𝑇.
• CONVERSION=
𝑅𝐸𝐴𝐶𝑇𝐼𝑁𝐺 𝐴𝑀𝑂𝑈𝑁𝑇 𝑂𝐹 𝐶𝑂𝑀𝑃𝑂𝑁𝐸𝑁𝑇
𝐼𝑁𝐼𝑇𝐴𝐿 𝐴𝑀𝑂𝑈𝑁𝑇
×100

8. • EXAMPLE 1 : Monochloro acetic acid (MCA) is manf. In a semi batch reactor by action of acetic acid with
chlorine gas at 100°C in presence of PCl3 catalyst. (MCA) thus form will further react with chlorine to form
di-chloro acetic acid (DCA).To prevent formation of DCA excess acetic acid is use. A small scale unit which
reduce 5000 kg/day (MCA) required 4536 kg/per chlorine gas also 263 kg/day DCA is separated in
crystallizer to get almost pure MCA product find %conversion, %yield of MCA & selectivity.
• SOLUTION: Basis: one day operation.
REACTIONS: CH3COOH+CL2 → CH2CLCOOH + HCL
(MCA)
CH2CLCOOH+CL2 → CHCL2COOH+HCL
(DCA)
AA MCA
DCA
HCL
CL2 AA
HERE CL2 is limiting component.
Molar mass AA=60
CL2=71
MCA=94,5
DCA=129

9. CL2 in feed=4536Kg/day therefore
In moles =4536/71=63,88 kmol/day in feed.
MCA in product = 5000KG/day
In moles = 5000/94,5=52,91kmol/day in product.
1 Kmol of MCA = 1 Kmol of CL2
Therefore 52.91 Kmol of MCA = x Kmol of CL2
Therefore CL2 required to produce MCA = 52.91 Kmol
DCA in product = 263 kg/day
In moles= 263/129= 2.038 Kmol
1 Kmol of DCA = 2 Kmol of CL2
2.038 Kmol of DCA = x Kmol of CL2
Therefore CL2 required to produce DCA = 4.077 Kmol
Therefore total CL2 consumed = 56.987 Kmol

10. % Conversion = CL2 utilized×100
CL2 in feed
=
56.987 ×100
63.88
=89.2 %
Yield = Moles of desired product formed × Stiochiometric coeff
Moles of spc reactant consumed
=
52.91
56.987
× 1 = 92.84 %
Selectivity =
𝐴𝑀𝑂𝑈𝑁𝑇 𝑂𝐹 𝐿𝐼𝑀𝐼𝑇𝐼𝑁𝐺 𝑅𝐸𝐴𝐶𝑇𝐴𝑁𝑇 𝑇𝐻𝐴𝑇 𝑅𝐸𝐴𝐶𝑇 𝑇𝑂 𝐺𝐼𝑉𝐸 𝐷𝐸𝑆𝐼𝑅𝐸 𝑃𝑅𝑂𝐷𝑈𝐶𝑇
𝐴𝑀𝑂𝑈𝑁𝑇 𝑂𝐹 𝑇𝐻𝐴𝑇 𝑅𝐸𝐴𝐶𝑇𝐴𝑁𝑇 𝑇𝑂 𝐺𝐼𝑉𝐸 𝑈𝑁𝐷𝐼𝑆𝐼𝑅𝐸 𝑃𝑅𝑂𝐷𝑈𝐶𝑇.
= 52.91/4.078 = 12.9 %

11. APPLICATION:
• Material balances considering reactions are used to design
reactors.
• Material balance can also be used to identify the most
efficient operation of reactor.
• Material balance provide data for yield, conversion and
selectivity for particular reaction.

12. CONCLUSION
Material balance provide important information for
the design of the plant. Firstly, different assumptions could
be made for each equipment and streams based on the
material balance. Design implications is done based on the
calculation too. Material balance is required to calculate
energy balance for the entire process.

13. THANK YOU !