Neraca Massa ATK

MATERIAL BALANCES WITH CHEMICAL REACTIONS

Introduction:

Chemical reactions play a vital role in manufacturing process. For design of chemical

process equipment, the operating conditions such as pressure, temperature,

composition and flow of the streams should be known. The material balance and energy

balance calculations come to the rescue of the designer and allows him to calculate the

various flow rates and temperature of the streams. Assuming that the kinetic data of the

reaction is available, the overall material balance of the steady state condition will be

discussed here.

Material balances:

The general mathematical statement can be written as

Total mass entering the unit = Total mass of products leaving the unit

It should be noted that in chemical reactions, the total mass of the input remains

constant, but the total moles may or may not remain constant.

Example: Consider the shift reaction

CO +H2O CO2 +H2

In this, it can be observed that two moles of reactants react with each other and

produce also two moles, thus the number of moles of the reactants entering the reaction

equals the number of the products leaving the reaction. The ammonia synthesis

reaction can be written as

N2 +3H2 2NH3

It can be observed that four moles of reactant only produce two moles of ammonia. The

total mass of the reactant entering and the product leaving the reaction are equal. For

the reaction of first type

1 mole CO ≡ 1 mole H2O

≡ 1 mol H2

≡ 1 mole CO2

Similarly for second type reaction

1 mole of N2 ≡ 3 mole of H2

≡ 2 mole of NH3

The above equalities decide the stoichiometric requirements of the components. In the

processes involving chemical reaction, the total mass of various compounds entering a

reaction is equal to the total mass of various components leaving the reaction, but

entering moles of components need not be equal to the moles of components leaving.

While doing material balance calculations in such case, it is very convenient to use

basis of calculations in molar units. Generally calculations should be based on limiting

reactant and quantity of new products formed should be calculated with the help

of chemical reactions and amount of limiting reactants reacted.

For any reactant material the balance of material can be written as

Material entering= material reacted +material un reacted

For products we can write

Material leaving = material produced by the reaction.

If the material is produced by more than one reaction, then the material leaving is sum

of the materials produced by all he reactions.

Definitions:

Stoichiometry:

It is a theory of the proportions in which chemical species combine with one another.

Stoichiometric equations:

Stoichiometric equation of a chemical reaction is a statement indicating relative moles of

reactant and products that take part in the reaction. Any balanced reaction equation is a

stoichiometric equation.

For example consider the stoichiometric equation

CO + 2H2 CH3OH

According to the equation one molecule (mol or kmol) of CO reacts with two molecules

(mol or kmol) of hydrogen to produce one molecule (mol or kmol) of methanol.

Stoichiometric co-efficient:

It is the number that precedes the formula of each component involved in a chemical

reaction.

Thus in the above example, the Stoichiometric co-efficient of CO is one. The

stoichiometric co-efficient of H2 is 2 and stoichiometric co-efficient of methanol is one.

The stoichiometric requirements of components are given as

1kmol of CO ≡ 1kmol of CH3OH ≡ 2kmol of H2

Stoichiometric ratio:

Two reactants A and B are said to be present in Stoichiometric proportions if the ratio of

moles of A present to the moles of B present is equal to the Stoichiometric ratio

obtained from the balanced equation.

Consider a reaction

CO + 2H2 CH3OH

For the reactants in the above equation to be present in Stoichiometric proportion there

must be 2 moles of H2 for every mole of CO (then nH2/ n CO = 2/1 = 2) present in the

feed to the reactor.

When the reactants are fed to a reactor in Stoichiometric proportion and the reaction

goes to completion all of the reactants will be consumed. In case of above cited

reaction, for example, if 200 kmol of H2 and 100kmol of CO are initially present, the H2

and CO would disappear at the same instant. It follows that if we start with 100ml of CO

and less than 200 mol of H2 (i.e., of H2 is present in less than its Stoichiometric

proportion) H2 disappears before the CO. On the other hand, if there are more than 200

moles of H2 initially present, the CO disappears first.

Limiting reactant / component:

Limiting reactant is the reactant which is present in such proportions that its

consumption will limit the extent to which the reaction can proceed. This reactant will not

be present in excess of that required to combine with any of the other reacting

materials.

It is the reactant which is present in such proportion that it’s compete consumption by

the reaction will limit the extent to which the reaction can proceed.

It is the reactant that would disappear first if a reaction goes to completion.

Excess Reactant:

Excess reactant is the reactant which is taken more than the requirement to combine

with other reactant as per the Stoichiometry.

It is the reactant which will be present in the product even if the reaction goes to

completion. It is the one which is in excess of theoretical or stoichiometric requirement.

E.g.: Consider the reaction

C2H4 + ½O2 C2H4O

In industrial practice of producing ethylene oxide by oxidation of ethylene, oxygen/ air

fed to the reactor is always in excess of theoretically required, thus ethylene is a limiting

reactant and oxygen/air is an excess reactant.

When all reactants are present in Stoichiometric proportion then none of the reactants

involved is limiting.

Percentage excess:

Percentage excess reactant is defined as % excess quantity taken based on theoretical

requirement.

It is the amount in excess of Stoichiometric (theoretical) requirements expressed as the

percentage of Stoichiometric / theoretical requirement.

Consider a reaction

A + B C

Where B is the excess reactant, then

Percent excess of B = mole of B supplied or fed – moles of B theoretically required x100

Moles of B theoretically required

Moles of B theoretically required are the moles of B that would correspond to the

stoichiometric proportions.

The above formula can be rearranged to calculate quantity of excess reactant actually

supplied.

Moles of B actually fed = stoichiometric requirement x (1 + fraction of excess of B)

The quantity of excess reactant must be found based on quantity of limiting reactant fed

to the reactor.

Consider for example, the reaction

SO2 + 1/2O2 SO3

And suppose that 100 moles SO2 /hr and 75 moles O2 /h are fed to the reactor. SO2 is

clearly the limiting reactant and to be in stoichiometric proportions, moles of O2 would

have to be 50 kmol / hr. The percent excess of O2 is therefore

= [(75 – 50) / 50] x 100 = 50

CONVERSION:

Conversion or fractional conversion or degree of completion is defined with respect to

limiting reactant and it gives idea regarding degree of completion of reaction. The

unreacted quantities of raw materials are easily obtained knowing the charged

quantities with the help of conversion then in turn it gives idea in case of unit process

whether recycling is to be done or not for process to be economically feasible.

Consider a chemical reaction

A + B C

Where A = a limiting reactant

B = the excess reactant.

Then the conversion or fractional conversion of A is the ratio of amount of A reacted to

the amount of A charged or fed to a reactor. The percentage conversion of A is the

amount of A reacted expressed as the percentage of amount of A charged or fed to a

reactor.

The amount of A can be expressed in moles or weight of the amount of A is expressed

in moles then

% conversion of A = Moles of A reacted x 100

Moles of A charged or fed

In case of recycle operations, the term namely per pass conversion is commonly used.

The per pass conversion is defined as the quantity of limiting reactant reacted /

consumed expressed as a percentage of limiting reactant in the mixed feed.

If 100 moles of a reactant are fed and 75 moles of the reactant reacts, the fractional

conversion is 0.75 (the % conversion is 75%) and the fraction unreacted is 0.25. If 50

moles of a reactant are fed and the percentage conversion is 80% then (50) x0.8 = 40

moles have reacted and (50) (0.20) = 10 moles remain reacted.

YIELD and SELECTIVITY:

In most of the chemical processes, though the objective is to produce a desired product

by reacting raw materials, the raw material may also undergo a series of parallel

reactions (side reaction) resulting into production of undesired material which has

reverse effect on economics of the process. In such cases, in industrial practice, the

steps are taken to depress the side reaction by use of selective catalyst or by other

means like maintaining proper concentration of reactants and by varying the

temperature of reaction. In addition the inhibitors are also used.

The terms yield and selectivity are used in case of multiple reactions to give information

regarding the degree to which a desired reaction predominates over side reaction or

reactions involved.

Consider the multiple reactions, namely a series parallel reaction, a series reaction and

a parallel reaction

Series parallel reaction A + B C

C + B D

Series reaction A C D

Parallel reaction A C

A D

Where C is a desired product, D is an undesired product, A is a limiting reactant.

Then yield of C is given as

Yield of C = moles of A reacted to produce C x 100

Total moles of A reacted

Yield of desired product is the quantity of limiting reactant reacted to produce the

desired product. It is expressed as a percentage of quantity of limiting reactant reacted

totally to produce the desired product .

Consider parallel reaction

A C and A D

Where C is a desired component, D is an undesired component

In such cases, the selectivity is given as

Selectivity of C relative to D = moles of C (desired product) formed

Moles of D (undesired product) formed

UNIT PROCESSES:

A Unit Process is a step in manufacturing in which chemical reaction takes place like

oxidation of toluene to benzaldehyde, hydrogenation of oils to vanaspathi.

Following are some of the examples of unit processes

Oxidation, hydrogenation, nitration, hydrolysis, esterification, Neutralization, alkylation,

amination etc.

OXIDATION:

Oxidation is defined in many ways and one simple definition of oxidation is the

interaction between oxygen molecules and all the different substances they may contact

CH3CHO + 1/2O2 CH3COOH

Acetaldehyde oxygen Acetic acid

C6H6 + 1/2O2 C4H2O3 + 2CO2 + 2H2O

Benzene maleic anhydride

4HCl + O2 � 2Cl2 + 2 H2O

HYDROGENATION:

This unit process specifically refers to the chemical reaction of a substance with

molecular hydrogen in the presence of a catalyst.

CO2 + 3H2 CH3OH + H2O

R CH = CH R' + H2 RCH2 – CH2 R'

NITRATION:

Those reactions where one or more nitro group (- NO2)n are introduced into the reacting

molecule.

CH3

C6H5 CH3 + HNO3 C6H4 + H2O

NO2

Toluene nitric acid mono nitro toluene

HYDROLYSIS:

It is applied to the reaction where in water effect a double decomposition with another

compound, hydrogen going to one component hydronyl to other.

C5H10Cl2 + H2O HCl + C5H11OH

Mono chloro pentane hydro chloride n – pentanol

KCN + H2O HCN + KOH

Pot.cyanide hydrogen cyanide pot.hydroxide

NEUTRALISATION:

The reaction taking place between an acid and alkali resulting in the formation of salt

and water is called neutralization.

CH3COOH + NaOH CH3COONa + H2O

Acetic acid sodium hydroxide sodium acetate

ALKYLATION:

It may be defined as introduction of alkyl radical by substitution or addition into an

organic compound.

CH3Br + 2Na + C2H5Br CH3C2H5 + 2NaBr

Methyl bromide sodium ethyl bromide propane sodium bromide

AMINATION:

The amination reaction is one which results in the formation of amines. This may be by

reduction or by ammonolysis

Amination by reduction:

It refers to those reactions which involve the synthesis of amines by reductive

methods.

Amines may be defined as derivatives of ammonia where one or more of the hydrogen

is replaced by alkyl, aryl, hydronyl, aralkyl or heterocyclic groups.

NO2 NH2

Zn

Acid

Nitro benzene aniline

Amination by ammonolysis:

It is the process of forming amines by the action of ammonia. It also includes the

use of the primary and secondary amines as aminating agents.

H2

RCHO + NH3 R CN (H2O + H2) RCH2 NH2

SULFONATION:

It may be defined as any chemical processes by which the sulfonic acid group (-

SO2OH) or the corresponding salt or sulfonyl halide group (- SO2Cl) is introduced into

an organic compound.

R CH2COOH + SO3 R CH (SO2OH) COOH

NH2COOH + H2S2O7 NH2SO3H + CO2

EMULSIFICATION:

All chemical reactions resulting in the formation of esters are all under this category.

CH3COOH + CH3CH2OH CH3COOC2H5 + H2O

Acetic acid n-propanol ethyl acetate water

HALOGENATION (Chlorination):

It can be defined as the process where by one or more halogen atom is introduced into

an organic compound.

C6H6 + Cl2 C6H5Cl + HCl

Benzene chlorine mono chloro benzene

C6H5Cl + Cl2 C6H4Cl2 + HCl

Mono chloro benzene chlorine dichlorobenzene

CH2 = CH2 + Br2 CH2Br - CH2Br

Ethylene bromide ethylene bromide

Oxidation reactions:

Note:

Air is used as source of oxygen

Air is a mixture of oxygen and nitrogen

Air contains 79% nitrogen and 21% oxygen by mole

Solution Procedure:

Assume a suitable basis

Write the possible reactions

Calculate the oxygen required assuming complete oxidation

Calculate oxygen supplied from the excess

Calculate nitrogen entering along with oxygen

Identify the products of oxidation for the problem

Calculate the quantities of each of the products

Calculate the other required quantities as per problem requirement

1. Ethylene oxide is produced by oxidation of ethylene. 100 kmol of ethylene are fed to

a reactor and the product is found to contain 80 kmol ethylene oxide and 10 kmol CO2.

Calculate

a) the % conversion of ethylene and b) the percent yield to ethylene oxide.

Basis: 100 kmol ethylene fed to the reactor.

Reactions:

C2H4+1/2 O2�C2H4O ( 1 )

C2H4+3O2�2CO2+2H2O (2)

80 kmol ethylene oxide is produced and this is possible only by reaction 1. As per

Stoichiometry 1kmol ethylene oxide will be produced per kmol of ehylene. Therefore

ethylene reacted for reaction 1 is 80 kmoles.

10 kmol of CO2 is produced and this possible through reaction 2. As per Stoichiometry 2

moles CO2 will be produced per mol ethylene. Therefore kmol ethylene reacted by

reaction 2 is 10/2=5 kmol.

Total ethylene reacted towards equation1and 2 = 80 + 5 = 85 kmol.

Total kmol of ethylene taken = 100 kmol

Therefore % conversion = (ethylene reacted / ethylene taken) x100

= (85/100) x100= 85%

% yield of ethylene oxide= (moles ethylene reacted to ethylene oxide /Total moles of

ethylene reacted) x100

= (80/85)100=94.12%

2. In the production of chlorine gas by oxidation of hydrochloric acid gas, air is used

30% in excess of that theoretically required. Based on 4 kmol HCl, Calculate a) the

weight ratio of air to hydrochloric aid gas in feed. b) if the oxidation is 80% complete,

find the composition of the product stream on mole basis.

Basis: 4 kmol HCl

4HCl+ O2 � 2Cl2 + 2 H2O

O2 required = 1 kmol

O2 supplied= 1x1.30 = 1.3 kmol because 30% excess air is used

N2 supplied along with O2= 1.3 x 79/21=4.89

Air supplied= O2 supplied + N2 supplied= 1.3+4.89=6.19 kmol

Weight of air supplied= 6.19x29=179.51 kg

HCl taken= 4 kmol= 4x36.5= 146 kg

Weight ratio of air to hydrochloric acid in feed = 179.51/ 146=1.23

If oxidation is 80% complete composition of the gases in mole%

Gases leaving:

Cl2, H2O, HCl,O2, N2

HCl reacted= 4x0.8=3.2 kmol

HCl leaving unreacted= 4 - 3.2=0.8 kmol

O2 leaving = O2 supplied – O2 consumed= 1.3 – 3.2x1/4

=1.3-0.8= 0.5kmol

N2 leaving = N2 entering= 4.89 kmol

Cl2 leaving = Cl2 produced= 3.2x2/4= 1.6 kmol

H2O leaving = H2O produced= 3.2x2/4= 1.6 kmol

Composition of gases leaving:

Components Kmoles Mole%

Cl2 1.60 17.04

H2O 1.60 17.04

HCl 0.80 8.52

O2 0.50 5.32

N2 4.89 52.08

Total 9.39 100.00

3. Calculate the composition of gases obtained by the burning of pure FeS2 with 60%

excess air. The oxidation proceeds according to the reaction

4FeS2 +11O2 � 2Fe2O3 + 8 SO2 .

Assume complete conversion of the reaction

Basis: 4 kmol of FeS2

O2 required= 11 kmol

O2 supplied =11x1.6= 17.6 kmol

N2 supplied= 17.6x79/21=66.21 kmol

Gases leaving: SO2,O2,N2

SO2 leaving = SO2 produced= 8 kmol

O2 leaving= O2 supplied- O2 utilized= 17.6- 11 =6.6 moles (O2 required= O2 utilized

because of complete conversion of reaction)

N2 leaving = N2 supplied=66.21kmol

Composition of gases leaving:

Component moles Mole%

SO2 8.00 9.90

O2 6.60 8.17

N2 66.21 81.93

Total 80.81 100.00

4. Iron pyrites FeS2 is burnt with air 100 % in excess of that required to oxidize all iron

to Fe2O3 and all sulphur to sulphur dioxide. Calculate the composition of the exit gases

in mole% and weight %, if 80% of sulphur is oxidized to sulphur dioxide and the rest to

sulphur trioxide. All iron is oxidized to Fe2O3.

Basis: 4 moles of FeS2

4FeS2 +11O2 � 2Fe2O3 + 8 SO2

SO2 +1/2O2� SO3

O2 required= 11 moles

O2 supplied =11x2= 22 moles

N2 supplied= 22x79/21=82.76 moles

Gases leaving: SO2, SO3,O2,N2

SO2 leaving= SO2 produced- SO2 converted to SO3

= 8 – 8 x(1-0.8) = 8 -1.6 = 6.4 moles

SO3 leaving = SO3 produced= 1.6 moles

O2 leaving = O2 supplied – O2 consumed

= 22 - 11- (1.6 x1/2 ) = 22-11- 0.8 =10.2

N2 leaving = N2 supplied= 82.76 moles

Compositions of gases leaving

5. Orthoxylene on oxidation gives Phthalic anhydride as per the reaction

C8H10 +3O2 � C8H4O3+3H2O.

20% excess air is used. Conversion is 50% and yield of phthalic anhydride is

90%.Calculate the requirement of orthoxylene and air for 100 kmole of phthalic anhydrid

Basis: 1 mol orthoxylene

C8H10 +3O2 � C8H4O3+3H2O

C8H10+21/2 O2 � 8CO2+5H2O

O2 required= 3 kmol

O2 supplied=3x1.2= 3.6 kmol

N2 supplied= 3.6x79/21=13.54 kmol

From Stoichiometry kmol phthalic anhydride produced assuming 100% yield= 0.5 kmol

Since yield is 90%, phthalic anhydride produced=0.5x0.9=0.45 kmol

To produce 0.45 kmol phthalic anhydride , kmol of orthoxyene used = 1 kmol

Therefore to produce 100 kmol phthalic anhydride, orthoxylene required=

(1/0.45)x100=222.22 kmol

Weight of orthoxylene required= 222.22x106=23555.55 kg.

Component Moles Mole% Mol.wt weight Weight%

SO2 6.40 6.34 64 409.46 12.87

SO3 1.60 1.58 80 128.00 4.02

O2 10.20 10.10 32 326.40 10.26

N2 82.76 81.97 28 2317.28 72.84

Total 100.96 99.99 3181.14 99.99

6. In the oxidation of SO2 to SO3, the conversion is 75% by using 70% excess air.

Calculate a) composition of gases leaving the reactor in mole basis b) kg mole air fed

per kg mole SO2.

Basis: 1mol SO2

SO2+1/2O2� SO3

Mol O2 required= 0.5 mol

O2 supplied= 0.5x1.7= 0.85

N2 supplied=0.85x79/21=3.20

Air supplied= O2 supplied +N2 supplied= 0.85+3.2=4.05

Gases leaving the reactor

SO2,SO3,O2,N2

Moles SO2 reacted=1x0.75= 0.75 kmol

Moes SO2 leaving = SO2 unreacted= 1x(1-0.75)=0.25

Moles SO3 leaving = mol SO3 produced= 0.75x1/1=0.75

Mol O2 leaving= mol O2 supplied-O2 consumed = 0.85-0.75x1/2= 0.85-0.375=0.475

Mol N2 leaving= mol N2 supplied= 3.20

Composition of gases leaving

Component Mol Mol%

SO2 0.250 5.35

SO3 0.750 16.04

O2 0.475 10.16

N2 3.200 68.45

Total 4.675 100.00

7. In production of sulphur trioxide (SO3) 100 k mol of SO2 and 200 k mol of O2 are fed

to the reactor. The product stream is found to contain 80 k mol of SO3. Find the percent

conversion of SO2.

BASIS: 100 k mol of SO2 entering the reactor

100 k mol of SO2 product

200 k mol O2 SO3, SO2 ,O2 REACTOR

Reaction SO2 + 1/2O2 SO3

1 k mol of SO2 ≡ 1 k mol of SO3

1 k mol SO3 requires 1 k mol of SO2 to be reacted by above reaction

1 k mol of SO3 ≡ 1 k mol of SO2

80 k mol of SO3 = ?

SO2 reacted = 80 x (1/1) = 80 k mol

% conversion of SO2 = k mol of SO2 reacted * 100 = 80 * 100 = 80%

K mol of SO2 fed 100

8.10 kg of PbS and 3 kgs of O2 react to yield 6 kgs of Pb and 1 kg of PbO2 according to

the reaction

PbS + O2 Pb+ SO2

PbS+2O2 PbO2+SO2

Calculate a. the amount of PbS that does not react b. %excess O2 based on the amount

of PbS that actually reacts c. amounts of SO2 formed d. % conversion of PbS to Pb.

Basis: 10 kg PbS and 3 kg O2 taken

PbS + O2 Pb+ SO2 ( 1)

PbS+2O2 PbO2+SO2 (2)

The atomic weight values are Pb-207.2, S-32, O-16. Therefore weights of PbS = 239.2,

PbO2= 239.2 SO2=64, O2=32 , 2O2=64

Pb leaving= 6 kg = Pb produced according to reaction 1.

PbS reacted for reaction 1 = (239.2/ 207.2)x6= 6.93 kg

PbS reacted for reaction 2= (239.2/239.2)x1= 1 kg

Total PbS reacted= 6.93+1=7.93 kg

Amount of PbS that does not react= 10-7.93= 2.07 kg

O2 required= (32/239.2)6.93 + (64/239.2)1=0.927+0.268=1.195

O2 taken =3 kg

% excess O2= [(O2 taken –O2 required)/ O2 required] x100

= [(3-1.195) / 1.195] x 100 = 151.05%

SO2 formed by reaction 1= (64/207.2) x 6.93 = 2.14

SO2 formed by reaction 2 = (64/207.2) x 1 = 0.31

Total SO2 formed= 2.14+0.31=2.45 kg

Conversion of PbS to Pb = (PbS reacted to Pb / PbS taken) x100 = (6.93/10) x100

= 69.3%.

9. Vapor phase oxidation of ethyl alcohol using Cu catalyst at 550°C results into formation

of acetaldehyde

C2H5OH + 1/2O2 CH3CHO + H2O

The conversion is 30% and excess air used is 10%. Calculate the composition of

product stream if 100 kg of C2H5OH per hour is fed to the reactor.

BASIS: 1 hr operation

100 kg/hr of C2H5OH is fed which is equal to = 100/46 = 2.1739 k mole

O2 required for 2.1739 k mole of C2H5OH = 2.1739x1/2 =1.087

O2 supplied = 1.087 x 1.1 = 1.1957 k mole

N2 accompanying = 1.1957 x (79/21) = 4.4981 k mole

Quantity of aldehyde formed = 2.1739 x 0.3 = 0.6522 (the conversion is 30%)

Alcohol (C2H5OH) remaining unconverted = supplied – consumed

= 2.1739 – (2.1739 x 0.3) = 1.5217 k mole

Water formed during the reaction = 0.6522 k mole

Oxygen leaving= oxygen supplied- oxygen utilized

=1.1957-2.1739x0.3x0.5=1.1957-0.3261=0.8696 kmol

Nitrogen leaving= 4.4981 kmol

Component Moles Mole%

Alcohol 1.5217 18.57

Formaldehyde 0.6522 7.96

Water 0.6522 7.96

Oxygen 0.8696 10.61

Nitrogen 4.4981 54.90

8.1938 100.00

10. Ethylene and oxygen are fed to a tubular reactor filled with silver catalyst at 250°C.

The product ethylene oxide is recovered from gaseous effluent by absorption in water in

presence of acid at 60°C. The reaction taking place is as follows

C2H4 + 1/2O2 C2H4O - (1)

C2H4O + H2O CH2OH - (2)

CH2OH

Conversion for reaction (1) is 90% which for reaction (2) it is 65%. If 100 kg of ethylene

glycol (CH2OH - CH2OH) is formed, Calculate the quantity of ethylene supplied to the

tubular reactor.

BASIS: 100 kg of CH2OH - CH2OH formed

Ethylene glycol formed= 100/62 = 1.6129 k mol

As per reaction (2) the conversion is 65%

Quantity of C2H4O required for 1.6129 k mole of (CH2OH) 2 will be

= 1.6129 x1/1x (100/65)= 2.4814 k mol

As per reaction (1) conversion is 90%

Quantity of C2H4 required for the formation of 2.4814 k mole of C2H4O will be

= 2.4814 x (100/90) = 2.757 kmol

Mass of C2H4 required = 2.757 x 28 = 77.196 kg

Mass of C2H4 required for 100 kg of ethylene glycol formed is 77.196 kg.

11. Ethylene oxide is produced by oxidation of ethylene .100 k mole of ethylene and 100

k mole of oxygen is charged to the reactor. The conversion of ethylene is 85% and

percent yield of C2H4O is 94.12.Calculate the composition of product leaving the

reactor.

BASIS: 100 k mol each of C2H4 and O2 charged

100 k mole of C2H4

100 k mole of O2 Product

C2H4, C2H4O, CO2, H2O

REACTOR

C2H4 + 1/2O2 C2H4O

C2H4 + 3O2 2CO2 + 2H2O

Percent conversion = 85

Moles of C2H4 totally reacted = 0.85 x 100 = 85 k mol

Percent yield of C2H4O = 94.12

Moles of C2H4 reacted to C2H4O by reaction (1)

= 0.9412 x85 = 80 kmol

Moles of C2H4 reacted by reaction (2) = 85 – 80 = 5 k mole

1 kmol C2H4 ≡ 2 kmol of CO2

Moles of CO2 produced = (2/1) x 5 = 10 k moles

From reaction 1, 1 kmol of C2H4 ≡ 0.5 k mol O2

O2 reacted by reaction 1 = (0.5/1) x 80 = 40 kmol

From reaction 2, 1 k mol C2H4 ≡ 3 kmol of O2

O2 reacted by reaction 2 = (3/1) x 5 = 15 k moles

O2 totally reacted = 40 + 15 = 55 k moles

O2 fed = O2 reacted + O2 unreacted

O2 unreacted = 100 – 55 = 45 kmoles

1 kmol of C2H4 ≡ 2 kmol of H2O

Moles of H2O produced = (2/1) x 5 = 10 kmoles

C2H4 fed = C2H4 reacted + C2H4 unreacted

C2H4 unreacted = 100 – 85 = 15 kmoles

Composition of product stream:

Component Quantity, kmol Mole%

C2H4 15 9.38

C2H4O 85 50.00

O2 45 28.12

CO2 10 6.25

H2O 10 6.25

Total 160 100.00

12. Oxidation of ethylene to produce ethylene oxide is given by reaction

C2H4 + 1/2 O2 C2H4O

If air is used 20%b in excess of that theoretically required. Calculate the quantity of air

supplied based on 100 k mol of ethylene fed to the reactor.

BASIS: 100 kmol of ethylene fed to the reactor.

Reaction: C2H4 + 1/2 O2 C2H4O

From reaction we have

1 kmol of C2H4 ≡ 0.5 kmol of O2

Stoichiometric or theoretical requirement of oxygen for

1 kmol of C2H4 ≡ 0.5 kmol of O2

Theoretical requirement of oxygen for

100 kmol of C2H4 = (0.5/1) x 100 = 50 k moles

Oxygen supplied 20% in excess of theoretical requirements,

Oxygen in supplied air = theoretical requirement of oxygen

= 50 x 1.2= 60 kmol

Moles of air supplied = (100/21)x 60 = 285.71 kmoles

Amount of air supplied = 285.71 x 29 = 8285.59 kg

13.In manufacture of acetic acid by oxidation of acetaldehyde, acetaldehyde and

oxygen were fed to reactor and the product leaving the reactor contains 14.81%

acetaldehyde, 59.26% acetic acid and rest oxygen (mole basis). Find the % conversion

of acetaldehyde.

BASIS: 100 k mol of products leaving

Acetaldehyde Acetic acid- 59.26%

Oxygen acetaldehyde-14.81%

Oxygen-25.93

Reaction: CH3CHO + 1/2O2 CH3COOH

Acetic acid formed= 59.26 kmoles

Acetaldehyde reacted= 59.26x1/1= 59.26 kmol

Acetaldehyde unreacted= 14.81 kmol

Total acetaldehyde entering= reacted +unreacted= 59.26+14.81= 74.07 kmol

% conversion of acetaldehyde= (59.26/ 74.07) x100= 80%

14. In production of sulphur trioxide, 100 k mol of SO2 and 100 k mol of O2 are fed to the

reactor. If the % conversion of SO2 is 80, calculate the composition of the product

stream on mole basis

BASIS: 100 k mol of SO2 and 100 k mole of O2 charged to the reactor

100 k mol of SO2

100 Kmol of O2

Product SO3, SO2, O2

Reaction SO2 + 1/2O2 SO3

Percent conversion = 80%

Moles of SO2 reacted = 100x0.8= 80 kmol

SO2 unreacted = 100 – 80 = 20 k mol

From reaction 1 k mole of SO2 ≡ 1 k mole of SO3

SO3 produced = (1/1) x 80 = 80 k mol

From reaction

1 k mol of SO2 ≡ ½ k mol of O2

REACTOR

REACTOR REACTOR

O2 reacted = (0.5 / 1) 80 = 40 k molx

O2 unreacted = 100 – 40 = 60 k mol

Composition of product stream

Component k mol mole%

SO2 20 12.5

SO3 80 50.0

O2 60 37.5

Total 160 100.0

15. In manufacture of SO3,the feed to the reactor consists of 50 k moles of SO2 and 150

k moles of air. Calculate % excess of air used.

BASIS: 50 k moles of SO2 fed to the reactor.

Reaction: SO2 + 1/2 O2 SO3

Air used = 150 k mole

Oxygen in supplied air = 150 x 0.21 = 31.5 kmol

From reaction 1 kmol of SO2 ≡ 0.5 kmol of O2

Theoretical (stoichiometric) requirement of O2 for 50 kmol SO2

= (0.5/1) x 50 = 25 kmol

% excess O2 used = oxygen supplied – oxygen theoretically required

Oxygen theoretically required

= [(31.5 – 25) / 25] x100 = 26

16.In the manufacture of benzaldehyde a mixture containing dry air and toluene is fed to

the reactor (converter) at a temperature of 448°K (175°C) and 100 kpa of pressure. The

mixture passes through the catalyst bed in the converter. The product gas stream

leaves the converter at 468°K (195°C). The reaction takes place as follows

C6H5CH3 (g) + O2 (g) C6H5CHO (g) + H2O (g)

The dry air is supplied at 100% excess so as to maintain the high yield of

benzaldehyde. The side reaction taking place is

C6H5CH3 (g) + 9O2 (g) 7CO2 (g) + 4H2O (g)

At the above mentioned conditions the overall conversion is 13% based on toluene.

Approximately 0.5 of the toluene charged burns to CO2 and H2O. calculate the

composition of gas stream leaving the converter.

BASIS: 100 kmol of toluene charged

C6H5CH3 (g) + O2 (g) C6H5CHO (g) + H2O (g)

C6H5CH3 (g) + 9O2 (g) 7CO2 (g) + 4H2O (g)

Overall conversion = 13%

Toluene totally reacted = 0.13 x 100 =13 kmol

Toluene unreacted = 100 – 13 = 87 kmol

Toluene

Air C6H5CHO, C6H5CH3, CO2, H2,N2

Toluene reacted by reaction 2 = (0.5/100) x 100 = 0.5 kmol

Toluene reacted by reaction 1 = 13 – 0.5 = 12.5 kmol

From reaction 1 kmol of toluene ≡ 1 kmol of benzaldehyde

≡ 1 kmol of water

Benzaldehyde produced = (1/1) x 12.5 = 12.5 kmol

Water produced by reaction 1 = (1/1) x 12.5 = 12.5 kmol

Carbon dioxide produced by reaction 2 = (7/1) x 0.5 = 3.5 kmol

Water produced by reaction 2 = (4/1) x 0.5 = 2 kmol

Total water produced = 12.5 + 2 = 14.5 kmol

Theoretical oxygen required = (1/1)x 100 = 100 kmol

Oxygen in supplied air = 100 x2 = 200 kmol (100% excess )

Nitrogen in supplied air = 200 x 79 /21 = 752.4 kmol

Oxygen reacted = oxygen reacted by reaction 1 + oxygen reacted by reaction 2

= (1/1) x 12.5 + (9/1)x 0.5 = 17 kmol

Oxygen unreacted = 200 – 17 = 193 kmol

Component kmol mole%

C6H5CH3 87.0 8.18

C6H5CHO 72.5 1.18

N2 752.4 70.79

O2 193.0 18.16

CO2 3.5 0.33

H2O 14.5 1.36

1062.9 100.00

Hydrogenation

17.A mixture of pure carbon dioxide and hydrogen is passed over a nickel catalyst. The

temperature of the catalyst bed is 588K and the reactor pressure is 2.02 mPag. The gas

mixture leaving the reactor is analyzed to contain 57.1% CO2, 41.1% H2, 1.68% CH4

and 0.12% CO by volume on dry basis. The reactions taking place in the reactor are

CO2+4H2� CH4+2H2O

CO2+H2�CO+H2O

Find a) the conversion of CO2 per pass b) the yield of CH4 in terms of CO2 reacted and

c) the composition of feed.

Basis: 100 kmol of dry gases leaving the reactor

CO+4H2� CH4+2H2O (1)

CO2+H2�CO+H2O (2)

Methane leaving = 1.68 kmol

Therefore CO2 reacted by equation 1= 1/1x1.68= 1.68 kmol

Co leaving =1.2 kmol

Therefore CO2 reacted by equation 2= 1/1x0.12= 0.12 kmol

Total CO2 reacted = 1.68+0.12= 1.8 kmol

CO2 supplied= CO2 reacted + CO2 un reacted = 1.8 + 57.1= 58.9 kmol

% conversion of CO2 per pass= (1.8/58.9)x100= 3.056 %

% yield of CH4 = (kmol CO2 reaced to CH4/ kmol CO2 totally reacted) x100

= (1.68/1.8)x100=93.33

Hydrogen reacted by equation 1= 4/1x1.68= 6.72 kmol

H2 reacted by reaction 2= 1/1x0.12= 0.12

Total hydrogen reacted= 6.72 + 0.12 = 6.84

Hydrogen supplied= hydrogen reacted + hydrogen un reacted = 6.84+41.1= 47.94 kmol

Analysis of feed

Component Mol Mol%

H 2 47.94 44.87

CO2 58.90 55.13

Total 106.84 100.00

18.Gaseous benzene C6H6 reacts with hydrogen in presence of Ni catalyst as per the

reaction

C6H6 + 3H2 � C6H12

Benzene cyclohexane

30% excess hydrogen is used above that required by the above reaction. Conversion is

50% and yield is 90%. Calculate the requirement of benzene and hydrogen gas for 100

moles of cyclohexane.

Basis: 1 mol benzene

H2 required= 3 mol

Hydrogen supplied= 3x1.3= 3.9 mol

Cyclohexane produced for a 50% conversion and 100 % yield= (1/1) x 1 x 0.5 = 0.5 mol

Since yield is 90 %, cyclohexane produced = 0.5 x 0.9 = 0.45 mol

0.45 mol cyclohexane is produce by 1 mol hexane

Therefore mole hexane required t produce 100 mole cyclohexane= (1/0.45) x100

= 222.22 mol

Hydrogen supplied to produce 100 mol cyclohexane = (3.9/1)222.22 = 866.658 mol

19.An organic ester of formula C19H36O2 is to be hydrogenated at a rate of 100 kg / hr to

yield C19H38O2. The hydrogen required for the plant, which run continuously, is available

as 50 lit cylinders in which the gas is contained at 70 bar and 300 K. How many cylinder

should the company order per week?

Basis: 100 kg / hr of C19H36O2

C19H36O2 + H2 � C19 H38O2

Mol weight of C19H36O2 = 228 + 36 + 32 = 296

Mol weight of C19H38O2 = 228 + 38 + 32 = 298

Mol C19H36O2 charged = 100/ 296 kmol

Hydrogen required per hour= (1/1)100/296= 100/296 kmol

Volume of hydrogen required= (100/296) 22.414 x 1000 x 300 /273 x1.013 / 70

= 120.42 lit/hr

Volume hydrogen required per week= 120.42 x 24 x 7= 20230.56 lit

Volume of 1 cylinder= 50 lit

Therefore number of cylinders required= 20230.56 / 50= 404.61

20. Methanol is produced by the reaction of carbon monoxide with hydrogen.

CO+2H2�CH3OH (1)

The side reaction is

CO+3H2�CH4+H2O .(2)

The conversion per pass is 12.5%.Of this 87.5% is reached via equation (1) and 12.5%

reach via equation (2). The feed mixture entering the reactor contains 32 mol% CO and

68 mol% hydrogen. The stream leaving the reactor passes through a condenser and a

separator. The carbon monoxide and hydrogen leaving this unit are recycled. The

methane leaves as a gas and the liquid mixture of methanol and water passes to a

distillation column for the concentration of methanol.Compute a) the analysis of hot

gases leaving the reactor by mol% and weight % b)the methanol content weight% of

the liquid leaving the condenser.

Basis: 100 mol feed to the reactor

Conversion per pass= 12.5%

Gase leaving: CO,

Therefore CO consumed = 32 x 0.125 = 4 kmol

CO consumed by reaction (1) is = 4 x 0.875 = 3.5 kmol

CO consumed by reaction (2) is = 4 x 0.125 = 0.5 kmol

Total CO reacted = 3.5 + 0.5 = 4 k mol

CO leaving= CO unreacted = CO entering – CO reacted= 32- 4 = 28 kmol

H2 consumed in reaction (1) = (2/1) 3.5 = 7 kmol

H2 consumed in reaction (2) = (3/1)0.5 = 1.5 kmol

Total H2 consumed = 7+1.5 = 8.5 kmol

H2 leaving = H2 unreacted = H2 supplied – H2 consumed = 68 - 8.5 = 59.5 kmol

Methanol leaving the reactor= methanol formed = (1/1) x 3.5 = 3.5 kmol

Water leaving the reactor = Water formed = (1/1) x 0.5 = 0.5 kmol

Methane leaving= methane formed = (1/1) x 0.5 = 0.5 kmol

Analysis of gases leaving:

Component mol Mol% Mol

weight

weight Weight%

CO 28 30.43 28 784 75.97

H2 59.5 64.68 2 119 11.53

CH3OH 3.5 3.81 32 112 10.85

H2O 0.5 0.54 18 9 0.87

CH4 0.5 0.54 16 8 0.78

Total 92 100.00 1032 100.00

Methanol content of the liquid = x100 = 92.56% by weight

21. The carbon monoxide is reacted with hydrogen to produce methanol. Calculate from

the reaction, the stoichiometric ratio of H2 to CO. k mol of CH3OH produced per k mol

CO reacted. The weight ratios of CO to H2 if both are fed to the reactor in stoichiometric

proportions. The quantity of CO required to produce 1000 kg of CH3OH.

reaction CO + 2H2 CH3OH

Stoichiometric co-efficient of CO = a =1

Stoichiometric co-efficient of H2 = b = 2

Stoichiometric ratio of H2 to CO = b/a = 2/1 = 2

From the reaction 1 k mol of CO ≡ 1 k mol of CH3OH

CO reacted to produce 1 k mol of CH3OH = 1/1 * 1 = 1 k mol.

CO = feed to the reactor = 100 kg

CO = feed to the reactor = 1 * 28 = 28 kg

H2 fed to the reactor in stoichiometric proportion with CO = 2 k mol.

H2 fed to the reactor = 2 * 2 = 4

Weight ratio of CO to H2 fed = 28/4 = 70

For reaction 28 kg CO ≡ 32 kg CH3OH

Amount of CO required to produce 1000 kg CH3OH = 28/32 * 100 = 875 kg.

22. Ammonia is produced by following reaction

N2 + H2 2NH3

Calculate the molal flow rate of hydrogen corresponding to nitrogen feed rate of 25

k mol/hr if they are at the stochiometric proportions. The kg of ammonia produced per

hour if percent conversion is 25 and nitrogen feed rate is 25 k mol/hr.

BASIS: 25 kmol / hr of N2 fed to the reactor.

Reaction: N2 + H2 2NH3

Stoichiometric proportion of N2 to H2 from the reaction is 1:3

1 k mol of N2 ≡ 3 k mol of H2

Molal flow rate of hydrogen in stoichiometric proportion to nitrogen flow rate

= 3/1x25 = 75 k mol/hr

No of moles of N2 reacted = 25x0.25 = 6.25 k mol/hr

From reaction we have 1 k mol of N2 ≡ 2 k mol of NH3

2 k mol of NH3 is produced per 1 k mol of N2 reacted

NH3 produced = (2/1) x 6.25 = 12.50 k mol/hr

Amount of NH3 produced = 12.5x molecular weight of NH3 = 12.5x 17 = 212.5 kg/hr.

23.Gaseous benzene (C6H6) reacts with hydrogen in presence of Ni catalyst as per the

reaction

C6H6 (g) + 3H2 (g) C6H12 (g)

30% excess hydrogen is used above that required by the above reaction. Conversion is

50% and yield is 90%. Calculate the requirement of benzene and hydrogen gas for 100

kmol of cyclo hexane.

BASIS: 100 kmol of cyclo hexane

Reaction: C6H6 (g) + 3H2 (g) C6H12 (g)

Cyclo hexane produced = 100 kmol

1 kmol of cyclo hexane needs 1 kmol of benzene

Benzene reacted to cyclo hexane = 1/ 1x100 =100kmol

% yield of cyclo hexane = kmol of benzene reacted to produce cyclo hexane x100

kmol of benzene totally reacted

90 = 100 x 100

Kmol of benzene totally reacted

Kmoles of benzene totally reacted = 100 x 100 = 111.11 kmol

90

% conversion of benzene = 50

% conversion of benzene = kmol of benzene totally reacted x 100

Kmol of benzene totally fed

50 = 111.11 x 100

Kmol of benzene charged

Kmol of benzene charged = 100 x 111.11 = 222.22 kmol

50

From reaction 1 kmol of benzene ≡ 3 kmol of hydrogen

Theoretical requirement of hydrogen gas = 3/1x 222.22 = 666.66 kmol

Requirement of benzene = 222.22 kmol

Requirement of hydrogen = 666.66 kmol

Nitration

24.Benzene reacts with nitric acid to produce nitrobenzene and water

C6H6+HNO3� C6H5NO2 +H2O

Nitrobenzene undergo further nitration to form dinitrobenzene

C6H5NO2 +HNO3�C6H4 (NO2)2+H2O

The percent conversion of benzene is 90 and acid used is 65% excess over theoretical

required by reaction 1. If the mole ratio of benzene to dinitrobenzene in product stream

is 17:1, calculate the quantities of benzene and nitric acid required for production of

2000 kg/h of nitrobenzene.

Basis: 1kmol / h benzene fed to the reactor

C6H6+HNO3� C6H5NO2 +H2O

C6H5NO2 +HNO3�C6H4 (NO2)2+H2O

Benzene converted= 0.9 kmol because conversion is 90%

Kmol nitrobenzene produced = 0.9 x 17/18= 0.85

Kmol dinitrobenzene produced= 0.9 x 1/18= 0.05

(Ratio of nitrobenzene to dinitrobenzene in product is 17:1)

Nitric acid required for the reaction= 1kmol

Nitric acid fed = 1 x 1.65 = 1.65 kmol

Weight of Nitric acid fed = 1.65 x 63 = 103.95 kg/h

Weight of benzene fed to the reactor = 1x78= 78 kg/h

Nitrobenzene produce /h = 0.85 x 123 = 104.55 kg/h

Benzene required for the production of 2000 kg/h of nitrobenzene

= (78/104.55) x 2000 = 1492.11kg/h

Nitric acid required for the production of 2000 kg/h of nitrobenzene

= (103.95/104.55) x 2000 = 1988.52 kg/h

25. Benzene reacts with nitric acid to produce nitrobenzene and water. Some of the

nitrobenzene may further undergo nitration to form dinitrobenzene. If the product

streams contain 10 mol nitrobenzene,2 mol dinitrobenzene and 4mol benzene, calculate

the percent conversion of benzene, yield of nitrobenzene and selectivity to

nitrobenzene.

Basis: 10 mol nitrobenzene, 2 mol dinitrobenzene and 4 mol benzene in the

product stream

C6H6+HNO3� C6H5NO2 +H2O (1)

C6H5NO2 +HNO3�C6H4 (NO2)2+H2O (2)

Benzene reacted to give nitrobenzene = (1/1) x10= 10 mol

Benzene reacted to give dinitrobenzene = (1/1) x 2 = 2 mol

Total benzene reacted= 10+2 =12 mol

Benzene unreacted = 4 mol

Benzene charged = benzene reacted + benzene un reacted = 12+4=16 mol

Conversion benzene = (benzene reacted/ benzene charged) x100 = (12/16) x100=75%

Yield of nitrobenzene= (benzene reacted to nitrobenzene/ total benzene reacted) x 100

= (10/12) x 100= 83.33%

Selectivity to nitrobenzene = (mol nitrobenzene formed /mol dinitrobenzene formed)

=5/2

26. Picric acid is obtained by nitrating phenol according to the reaction

C6H5OH+ 3HNO3� C6H2OH (NO2)3+3H2O.

In a particular operation, 100 kg of phenol was reacted with 250 kg of HNO3 to produce

200 kg of picric acid. Which is the limiting reactant? What is the conversion of the

reaction? Find the composition of the product stream by weight% and mol %.

Basis: 100 Phenol

C6H5OH+ 3HNO3� C6H2OH (NO2)3+3H2O

94 189 229 54

Mol weight of phenol= 94 , mol weight of nitric acid=63 ,molecular weight of picric

acid=229, molecular weight of water=18.

Weight of HNO3 required to react with 100 kg phenol= (189 /94)100= 201.06 kg

HNO3 used = 250 kg. Since HNO3 used is more than the required, it is an excess

reactant. Therefore phenol is the limiting reactant.

200 kg of picric acid is produced.

Therefore weight of phenol reacted to give 200 kg picric acid as per stoichometry

= (94/229) x 200 = 82.1 kg.

Conversion of the reaction= (limiting reactant reacted / limiting reactant taken) x100

= (82.1/100)100= 82.1 %

Components leaving the reactor: Picric acid, phenol, nitric acid, water

Picric acid leaving = picric acid produced = 200 kg

Water leaving = water produced = (54/229)x200 = 47.16 kg

Phenol leaving = phenol unreacted = phenol taken –phenol reacted= 100- 82.1= 17.9 kg

Nitric acid leaving = nitric acid unreacted = nitric acid taken - nitric acid reacted

= 250 - (189/94) x 82.1= 250-165.07= 84.93 kg

Composition of the product stream:

Component Weight Weight % Mol weight mol Mol%

Picric acid 200.00 57.14 229 0.873 17.70

Phenol 17.90 5.11 94 0.190 3.85

Water 47.16 13.47 18 2.620 53.13

Nitric acid 84.93 24.27 63 1.248 25.31

Total 349.99 99.99 4.931 99.99

27. Toluene reacts with nitric acid to form ortho nitro toluene and water. After carrying

out the nitration operation which can be represented by the equation

C6H5CH3 +HNO3� C6H4CH3NO2+H2O

The product mixture was analyzed and found to contain 75% ortho nitro toluene 10%

water, 15% nitric acid by weight. Comment about the conversion of the reaction. What is

the % excess of the reactant used? What is the quantity of toluene used for the

reaction?

Basis: 100 kg of products

The product does not contain toluene and therefore the conversion is 100% and also

toluene is the limiting reactant.

Amount of ortho nitro toluene produced= 75 kg

HNO3 reacted to produce 75 kg o nitro toluene= (63/137) x 75 = 34.49 kg

HNO3 unreacted= 15 kg

HNO3 fed= HNO3 reacted+HNO3 unreacted= 34.49+15= 49.49 kg

% excess HNO3= (15/ 34.49) x100= 43.49%

Toluene fed for the reaction= (92/ 137) x75= 50.36 kg.

28.Benzoic acid is nitrated to obtain nitro benzoic acid according to the reaction

C6H5COOH+HNO3� C6H4 (COOH) NO2 + H2O.

122 63 167 18

Calculate the weight of 90% HNO3 required for reacting with 250 kg benzoic acid. If

50% excess acid is used and if the conversion is 85% find the composition of the

products leaving the reactor.

Basis: 250 kg benzoic acid

90% HNO3 required to react with 250 kg benzoic acid = (63/122) x250 x 100/90 =143.44

Acid used = 143.44 x 1.5 = 215.16 kg

100% acid = 215.16 x 0.9 = 193.64 kg

Water entering with acid = 215.16 x 0.1= 21.52kg

Conversion is 85%

Benzoic acid reacted = 250 x 0.85 = 212.5 kg

Benzoic acid unreacted = 250 x 0.15 = 37.5 kg

HNO3 reacted = (63/122) x 212.5 = 109.73 kg

HNO3 unreacted = 193.64-109.73 = 83.91 kg

Water leaving = 21.52 kg

Nitrobenzoic acid produced= (267/122) x 212.5= 465.06 kg

Water produced= (18/122) x 212.5 = 31.35 kg

Water leaving= water produced +water in acid = 31.35+21.52 = 52.87 kg

Composition of the products leaving the reactor:

Component Weight Weight%

Benzoic acid 37.50 5.87

Nitric acid 83.91 13.12

Nitro benzoic acid 465.06 72.74

Water 52.87 8.27

Total 639.34 100.00

29. Chlorobenzene is nitrated using a mixture of nitric acid and sulphuric acid. During

the pilot plant studies, a charge of 100 kg chlorobenzene 106.5 kg 65.5% (by weight)

nitric acid and 108.0 kg 93.6% (by weight) sulphuric acid. After two hours of operation,

the final mixture was analysed.It was found that the final product contained 2%

unreacted chlorobenzene. Also the product distribution was found to be 66% p – nitro

chloro benzene and 34% O – nitro chloro benzene. Calculate a) The analysis of charge

b)The % conversion of chloro benzene and c)The composition of product mixture.

BASIS: 100 kg Chloro benzene

The charge consists of chloro benzene and mixed acid

HNO3 charged = 106.5 x 0.655 = 69.758 kg

H2SO4 charged = 108.8x 0.936 = 101.088 kg

Water in the charge = 106.5 x 0.345 + 108.0 x 0.064 = 43.655 kg

The analysis of the reactant can be tabulated as shown

Component mol weight Qty Charged Weight.%

Chloro benzene 112.5 100.00 31.80

Nitric acid 63.0 69.76 22.18

Sulphuric acid 98.0 101.09 32.14

Water 18.0 43.66 13.88

314.51 100 .00

The reaction taking place are

Cl

Cl NO2

+ H2O

+ HNO3

Chloro benzene nitric acid O – nitro chloro benzene water

Cl Cl

+ H2O

+ HNO3

NO2

Chloro benzene nitric acid P – nitro chloro benzene water

As given in the problem, the yield of P – NCB is 66%. Since the total charge (weight)

remains constant.

Unreacted C.B in the product = 314.5 x 0.02 = 6.29 kg.

Amount of C.B that has reacted = 100 – 6.29 = 93.71 kg.

Conversion of C.B = 93.71 x 100 = 93.71%

100

Sulphuric acid remains unreacted

From reaction I it is clear that 1 kmol of C.B ≡ 1 kg mol HNO3

≡1 kg mol NCB

≡1 kg mol H2O

112.5 kg of CB ≡ 63 kg of HNO3

Thus 63 kg of HNO3 will be consumed for converting 112.5 kg C.B into NCB

Total HNO3 consumed = (63/112.5) x 93.71

=52.478 kg

Unreacted HNO3 = 69.758-52.478 =17.28 kg

Mol wt of NCB=157.5

Total NCB produced = (157.5/112.5) x 93.71=131.194 kg

p-NCB = 0.66x131.194=86.588 KG

o-NCB = 0.34x131.194 =44.606 kg

Water produced = (15/112.5)x 93.71 =14.994 kg

Total water produced =43.665+14.994=58.649 kg

Final product analysis:


CB 6.27 2.00

P NCB 86.59 27.33

O NCB 44.61 14.18

HNO3 17.28 5.49

H2SO4 101.09 32.15

H2O 58.65 18.65

TOTAL 314.48 100.00

30. Acetylene is produced according to the following reaction:

CaC2+2H2O�C2H2+ Ca(OH)2

If 100 lit of gas is burnt per hour at 298 K (25°C) and 98.68 kpa pressure, calculate the

amount of CaC2 in kg which must be used in the acetylene lamp, at the above

temperature and pressure, to get 15 hours service of the lamp. CaC2 reacts to give

acetylene gas to burn in the lam.

Basis: 15 h of operation.

Gas burning rate of acetylene= 100 lit/ hr

Acetylene gas burned= 100x15= 1500 lit

C2H2 generated=C2H2 burned=1500 lit

31.Ethyl alcohol is produced industrially by fermentation of molasses. The molasses

sample contains 45% by weight fermentable sugar in the form of sucrose. The reactions

taking lace in the fermenter are.

Sucrose (d-Glucose) (d-Fructose)

(Monosaccharide) Alcohol

Calculate the theoretical production of rectified spirit of density (0.785 kg/l) in liters per

ton of molasses.

Basis: 1000 kg of molasses.

Sucrose (d-Glucose) (d-Fructose)

Sucrose in molasses = 0.45 X 1000 = 450 kg

From the reactions, we have

32.100 kmol of ethyl acetate is charged and hydrolyzed to produce ethyl alcohol and

acetic acid. The product stream is fond to contain 45 kmol ethyl alcohol. What is the

conversion? What will be the amount of acetic acid produced?

Basis: 100 kmol ethyl acetate

CH3COOC2H5 + H2O� C2H5OH +CH3COOH

88 18 46 60

Kmol ethyl alcohol produced= 45 kmol

As per Stoichiometry 1kmol ethyl acetate gives 1 kmol ethyl alcohol

Therefore kmol ethyl acetate reacted = (1/1) x 45 = 45 kmol

Conversion of ethyl acetate= (ethyl acetate reacted / ethyl acetate taken) x100

= (45/100) x 100=45%

Acetic acid produced / 45 kmol ethyl acetate = (1/1) x 45 = 45 kmol

33. Acetyl chloride is hydrolyzed for the production of acetic acid and hydrochloric acid.

The product stream is found to contain 38.46 mol% acetic acid, 38.46% hydrochloric

acid and the remaining water. Find the ratio of acetyl chloride to water used for reaction.

Basis: 100 kmol products

CH3COCl + H2O� CH3COOH + HCl

kmol acetic acid produced = 38.46 kmol

As epr Stoichiometry 1 kmol acetic acid will be produced from 1 kmol acetyl chloride

Therefore actyl chloride reacted = (1/1) x 38.46 = 38.46 kmol

Water reacted= (1/1) x 38.46 = 38.46 kmol

Water present in product = 100- 38.46-38.46 = 23.08 kmol = water unreacted

Water fed = water reacted + water unreacted = 38.46+23.08 = 61.54 kmol

Ratio of acetyl chloride to water in feed = 38.46 / 61.54 = 0.625

Neutralization

34.100 kg hydrochloric acid is neutralized by 150 kg of sodium hydroxide. What is the

quantity of sodium chloride that will be produced? Find the composition of neutralized

mixture by eight and mol.

Basis: 100 kg HCl and 150 kg NaOH

HCl+ NaOH� NaCl + H2O

36.5 40 58.5 18

NaOH required to react with 100 kg HCl= (40/36.5) x 100 = 109.59 kg

NaOH taken= 150 kg

Since NaOH is taken more than the required, it is excess reactant

As per Stoichiometry 58.5 kg sodium chloride is produced for 36.5 kg HCl

Therefore sodium chloride produced= (58.5/36.5) x 100 =160.27 kg

Components leaving

NaCl,H2O,NaOH

Water leaving=Water produced = (18/36.5) x 100 = 49.59 kg

NaOH leaving = NaOH entering- NaOH reacted = 150-109.59 = 40.41 kg

Sodium chloride leaving = sodium chloride produce = 160.27 kg

Composition of neutralized mixture

Component weight Weight % Mol wt mol Mol%

NaCl 160.27 64.11 58.5 2.74 42.22

NaOH 40.41 16.16 40 1.01 15.56

H2O 49.32 19.73 18 2.74 42.22

Total 250 100.00 6.49 100.00

35. Ammonium hydroxide is used for neutralizing acetic acid according to the reaction.

CH3COOH +NH4OH �CH3COONH4 +H2O

If 120 kg of acetic acid is to be neutralized what will be the quantity of ammonium

hydroxide required? If 25% excess NH4OH is used and if the reaction is 90% complete,

what will be the composition of the product streams?

Basis: 120 kg acetic acid fed

CH3COOH + NH4OH � CH3COONH4 +H2O

60 35 77 18

As per Stoichiometry 35 kg NH4OH s required for 60 kg acetic acid

Therefore NH4OH required to react with 120 kg acetic acid = (35/60) x 120 =70 kg

NH4OH supplied = 70 x 1.25 = 87.5 kg

Reaction is 90% complete

Acetic acid reacted = 120 x 0.9 = 108 kg

Acetic acid unreacted = acetic acid taken- acetic acid reacted = 120-108 = 12 kg

Ammonium hydroxide reacted = (35/60) x 108 = 63 kg

Ammonium hydroxide unreacted = ammonium hydroxide supplied- reacted

= 87.5-63 = 24.5 kg

Ammonium acetate produced = (77/60) x 108 = 32.4 kg

Composition of products

Component Weight Weight %

Ammonium acetate 138.6 66.80

Water 32.4 15.61

Ammonium hydroxide 24.5 11.81

Acetic Acid 12.0 5.78

Total 207.5 100.00

Esterifcation

36.Ethyl acetate ester is produced by the reaction between acetic acid and ethanol

using catalyst. The reaction goes to completion. The acetic acid used contained 10%

water. Calculate the quantity of ethyl acetate produced for 200 kg ethanol and complete

analysis of products.

Basis: 200 kg ethanol

CH3COOH+ C2H5OH � CH3COOC2H5 +H2O

60 46 88 18

As per Stoichiometry 88 kg ethyl acetate produced for 46 kg ethanol

Therefore ethyl acetate produced for 200 kg ethanol= (88/46) x 200 = 382.61 kg

46 kg ethanol requires 60 kg acetic acid

Therefore 200 kg ethanol requires = (60/46) x 200 = 260.87 kg

Acetic acid used is 90% pure

90% acetic acid used = 260.87x100/90 = 289.86 kg

Water in acetic acid = 189.86 - 260.87 = 28.99 kg

Water produced= (18/46) x 200 = 78.26 kg

Water leaving= water produced+ water with acid = 78.26 + 28.99 = 107.25 kg

Complete analysis of products


Ethyl acetate 382.61 78.11

Water 107.25 21.89

Total 489.86 100.00

37. In the manufacture of ethyl benzoate, benzoic acid is reacted with ethanol in

presence of catalyst. The feed mixture is taken in a batch reactor along with catalyst

and at the end of reaction the products were analyzed and found to contain 35% ethyl

benzoate, 35% water, 10% ether and 20% benzoic acid. What is the % conversion of

the reaction? Which is the excess reactant and what is the % excess?

Basis: 100 kmol reaction products

C6H5 COOH+ C2H5OH � C6H5COOC2H5+ H2O

Benzoic acid ethanol ethyl benzoate

Ethyl benzoate in products = 35 kmol

As per Stoichiometry 1 kmol ethyl benzoate requires I kmol benzoic acid and 1 kmol

ethanol

Therefore kmol benzoic acid reacted = (1/1) x 35 = 35 kmol

Kmol ethanol reacted= (1/1) x 35 = 35 kmol

Kmol benzoic acid fed = benzoic acid reacted + benzoic acid unreacted

= 35+20= 55 kmol

Kmol ethanol fed = kmol ethanol reacted + kmol ethanol unreacted = 35+10= 45 kmol

Since benzoic acid fed is more than the required to react with ethanol, benzoic acid is

excess reactant and ethanol is the limiting reactant

Conversion= (ethanol reacted/ ethanol fed) x100 = (35/45) x 100 = 77.78%

% excess benzoic acid

= [(benzoic acid fed- benzoic acid required) / benzoic acid required] x 100

= [(55-45)/45] x 100 = (10/45) x 100 = 22.22%

Neutralization

38.100 kg hydrochloric acid is neutralized by 150 kg of sodium hydroxide. What is the

quantity of sodium chloride that will be produced? Find the composition of neutralized

mixture by eight and mol.

Basis: 100 kg HCl and 150 kg NaOH

HCl+ NaOH� NaCl + H2O

36.5 40 58.5 18

NaOH required to react with 100 kg HCl= (40/36.5) x 100 = 109.59 kg

NaOH taken= 150 kg

Since NaOH is taken more than the required, it is excess reactant

As per Stoichiometry 58.5 kg sodium chloride is produced for 36.5 kg HCl

Therefore sodium chloride produced= (58.5/36.5) x 100 =160.27 kg

Components leaving

NaCl,H2O,NaOH

Water leaving=Water produced = (18/36.5) x 100 = 49.59 kg

NaOH leaving = NaOH entering- NaOH reacted = 150-109.59 = 40.41 kg

Sodium chloride leaving = sodium chloride produce = 160.27 kg

Composition of neutralized mixture

Component weight Weight % Mol wt mol Mol%

NaCl 160.27 64.11 58.5 2.74 42.22

NaOH 40.41 16.16 40 1.01 15.56

H2O 49.32 19.73 18 2.74 42.22

Total 250 100.00 6.49 100.00

39. Ammonium hydroxide is used for neutralizing acetic acid according to the reaction.

CH3COOH +NH4OH �CH3COONH4 +H2O

If 120 kg of acetic acid is to be neutralized what will be the quantity of ammonium

hydroxide required? If 25% excess NH4OH is used and if the reaction is 90% complete,

what will be the composition of the product streams?

Basis: 120 kg acetic acid fed

CH3COOH + NH4OH � CH3COONH4 +H2O

60 35 77 18

As per Stoichiometry 35 kg NH4OH s required for 60 kg acetic acid

Therefore NH4OH required to react with 120 kg acetic acid = (35/60) x 120 =70 kg

NH4OH supplied = 70 x 1.25 = 87.5 kg


Acetic acid reacted = 120 x 0.9 = 108 kg

Acetic acid unreacted = acetic acid taken- acetic acid reacted = 120-108 = 12 kg

Ammonium hydroxide reacted = (35/60) x 108 = 63 kg

Ammonium hydroxide unreacted = ammonium hydroxide supplied- reacted

= 87.5-63 = 24.5 kg

Ammonium acetate produced = (77/60) x 108 = 32.4 kg



Ammonium acetate 138.6 66.80

Water 32.4 15.61

Ammonium hydroxide 24.5 11.81

Acetic Acid 12.0 5.78

Total 207.5 100.00

Esterifcation

40.Ethyl acetate ester is produced by the reaction between acetic acid and ethanol

using catalyst. The reaction goes to completion. The acetic acid used contained 10%

water. Calculate the quantity of ethyl acetate produced for 200 kg ethanol and complete

analysis of products.

Basis: 200 kg ethanol

CH3COOH+ C2H5OH � CH3COOC2H5 +H2O

60 46 88 18

As per Stoichiometry 88 kg ethyl acetate produced for 46 kg ethanol

Therefore ethyl acetate produced for 200 kg ethanol= (88/46) x 200 = 382.61 kg

46 kg ethanol requires 60 kg acetic acid

Therefore 200 kg ethanol requires = (60/46) x 200 = 260.87 kg

Acetic acid used is 90% pure

90% acetic acid used = 260.87x100/90 = 289.86 kg

Water in acetic acid = 189.86 - 260.87 = 28.99 kg

Water produced= (18/46) x 200 = 78.26 kg

Water leaving= water produced+ water with acid = 78.26 + 28.99 = 107.25 kg

Complete analysis of products


Ethyl acetate 382.61 78.11

Water 107.25 21.89

Total 489.86 100.00

41. In the manufacture of ethyl benzoate, benzoic acid is reacted with ethanol in

presence of catalyst. The feed mixture is taken in a batch reactor along with catalyst

and at the end of reaction the products were analyzed and found to contain 35% ethyl

benzoate, 35% water, 10% ether and 20% benzoic acid. What is the % conversion of

the reaction? Which is the excess reactant and what is the % excess?

Basis: 100 kmol reaction products

C6H5 COOH+ C2H5OH�C6H5COOC2H5+ H2O

Benzoic acid ethanol ethyl benzoate

Ethyl benzoate in products = 35 kmol

As per Stoichiometry 1 kmol ethyl benzoate requires I kmol benzoic acid and 1 kmol

ethanol

Therefore kmol benzoic acid reacted = (1/1) x 35 = 35 kmol

Kmol ethanol reacted= (1/1) x 35 = 35 kmol

Kmol benzoic acid fed = benzoic acid reacted + benzoic acid unreacted

= 35+20= 55 kmol

Kmol ethanol fed = kmol ethanol reacted + kmol ethanol unreacted = 35+10= 45 kmol

Since benzoic acid fed is more than the required to react with ethanol, benzoic acid is

excess reactant and ethanol is the limiting reactant

Conversion= (ethanol reacted/ ethanol fed) x100 = (35/45) x 100 = 77.78%

% excess benzoic acid

= [(benzoic acid fed- benzoic acid required) / benzoic acid required] x 100

= [(55-45)/45] x 100 = (10/45) x 100 = 22.22%

Alkylation

42. Ethyl benzene is manufactured by the action of benzene with ethyl chloride using

anhydrous aluminium chloride as catalyst.The reaction is

C6H6 + ClC2H5 � C6H5C2H5 + HCl

Benzene ethyl chloride ethyl benzene

Ethyl chloride is used 15% in excess of theoretically required. The reaction is 95%

complete. During alkylation the liberated HCl is scrubbed with water in order to obtain

20%HCl solution by weight. Calculate the raw materials required for 3000 kg of ethyl

benzene produced and the amount of 20% HCl produced.

Basis: 1 kmol benzene

C6H6 + C2H5Cl � C6H5C2H5 + HCl

78 64.5 106 36.5

As per stoichometry I kmol benzene requires 1 kmol ethyl chloride

Ethyl chloride required= (1/1) x 1 = 1 kmol

Ethyl chloride fed= 1x1.15 = 1.15 kmol

Weight of ethyl chloride fed= 1.15 x 64.5= 74.175 kg


Therefore ethyl benzene produced = (1/1) x 1 x 0.95 = 0.95 kmol

Weight of ethyl benzene produced = 0.95 x 106 = 100.7 kg

HCl produced = (1/1) x 1 x 0.95 = 0.95 kmol

Weight of HCl produced = 0.95 x 36.5 = 34.675 kg

Benzene unreacted = 1-0.95 = 0.05 kmol = 0.05 x 78 = 3.9 kg

To produce 20% solution of HCl, water required = (80/20) x 34.675 = 138.7 kg

Total weight of HCl produced = 138.7+ 34.675 = 173.375 kg

Benzene required for 3000 kg ethyl benzene = (78/100.7) x 3000 = 2323.73 kg

Ethyl chloride fed for 3000 kg ethyl benzene = (74.175/100.7) x 3000 = 2209.78 kg

43. One method of producing toluene is by alkylation of benzene by chloromethane

using aluminium chloride catalyst. In a particular operation 200 kg of benzene and 150

kg of chloro methane are used to produce 200 kg of toluene. The HCl produced is

absorbed using 200 kg water to obtain HCl solution. Calculate a) the % excess

chloromethane used b)the composition of raw materials entering c)the degree of

completion of reaction and d) concentration of HCl solution produced by completely

absorbing the HCl gas produced.

Basis: 200 kg benzene and 150 kg chloromethane fed

C6H6+ CH3Cl� C6H5CH3 +HCl

78 50.5 92 36.5

Chloromethane required to react with 200 kg benzene = (50.5/78) x 200 = 129.49 kg

% excess chloromethane used = [(150-129.49) / 129.49] x 100 = 15.84%

Composition of raw material entering


Benzene 200 57.14

Chloromethane 150 42.86

Total 350 100.00

Toluene produced = 200 kg

Benzene reacted to produce 200 kg toluene = (78/92) x 200 = 169.57 kg

Conversion of benzene= (benzene reacted/benzene fed) x 100

= (169.57/200) x 100 = 84.785%

HCl produced = (36.5/92) x 200 = 79.35 kg

Water used for absorption= 200 g

Total HCl solution produced = 200 + 79.35 = 279.35 kg

Concentration of HCl solution = (79.35/279.35) x 100 = 28.41%

AMINATION

44. The reaction between acetone and ammonia to form amino propane is carried out in

a batch reactor. The final products stream is analyzed and found to contain 35 mol%

amino propane, 35 mol% water 20 mol% ammonia and the remaining acetone. The

feed to the reactor contains only acetone and ammonia. Calculate the fractional

conversion of the limiting reactant and the % by which the other reactant is in excess.

Basis: 100 kmol of products formed

CH3COCH3+ NH3� CH3-CH-NH2-CH3+H2O

Acetone Amino propane

The products contain 20 mol% NH3 and 10% acetone

As per Stoichiometry the requirement of acetone and ammonia are same, but the

product stream contains 20% unreacted NH3 and 10% unreacted acetone

Therefore NH3 is the excess reactant and acetone is limiting reactant

Mole amino propane in product = 35 kmol

Therefore acetone reacted = (1/1) x 35 = 35 kmol

Acetone fed= acetone reacted+ acetone unreacted = 35 +10 = 45 kmol

% conversion= (35/45) x 100 = 77.78%

NH3 reacted= (1/1) x 35 = 35 kmol

NH3 fed = NH3 reacted + NH3 unreacted = 35 + 20 = 55 kmol

Ammonia required to react with 45 kmol acetone = (1/1) x 45 = 45 kmol

% excess NH3 fed = [(55-45)/45] x 100 = (10/45) x 100 = 22.22%

45. Calculate the composition of products obtained by amination of benzyl chloride with

60% excess NH3.Assume the reaction proceeds according to the reaction

C6H5COCl+NH3� C6H5CONH2+HCl

Benzyl chloride Benzmide

The conversion is 90%.

Basis: 1 kmol benzyl chloride

C6H5COCl+NH3� C6H5CONH2+HCl

NH3 required= (1/1) x 1 = 1 kmol

NH3 fed= 1x1.6= 1.6 kmol

Conversion is 90%

Kmol benzyl chloride leaving = benzyl chloride unreacted = 1 x (1-0.9) = 0.1 kmol

Ammonia leaving = ammonia unreacted = ammonia fed- ammonia reacted

=1.6- (1/1) x 0.9 = 0.7 kmol

Benzamide leaving = benzamide produced = (1/1) x 0.9 = 0.9 kmol

HCl leaving = HCl produced = (1/1) x 0.9 = 0.9 kmol


Component Mol mol %

Benzamide 0.9 34.62

Hydrochloric acid 0.9 34.62

Benzyl chloride 0.1 3.85

Ammonia 0.7 26.92

Total 2.6 100.01

Neraca Massa ATK

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