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PROCESS CALCULATIONS

SECOND EDITION

V. VENKATARAMANI Formerly Professor

Department of Chemical Engineering National Institute of Technology Tiruchirappalli

N. ANANTHARAMAN Professor

Department of Chemical Engineering National Institute of Technology Tiruchirappalli

K.M. MEERA SHERIFFA BEGUM Associate Professor

Department of Chemical Engineering National Institute of Technology Tiruchirappalli New Delhi-110001 2011

PROCESS CALCULATIONS, Second Edition

V. Venkataramani, N. Anantharaman and K.M. Meera Sheriffa Begum

© 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher.

ISBN-978-81-203-4199-9

The export rights of this book are vested solely with the publisher.

Second Printing (Second Edition) … … February, 2011

Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Meenakshi Art Printers, Delhi-110006.

To My Parents

— V. Venkataramani

— K.M. Meera Sheriffa Begum

To My Mother

— N. Anantharaman

8 RECYCLE AND BYPASS 180–195 8.1 Recycle ... 180 8.2 Bypass ... 180 8.3 Purge... 180 Worked Examples ... 181 Exercises ... 195 CONTENTSix 9 ENERGY BALANCE 196–220 9.1 Definitions ... 196 9.1.1 Standard State ... 196 9.1.2 Heat of Formation ... 196 9.1.3 Heat of Combustion ... 197

9.1.4 The Heat of Reaction ... 197

9.1.5 Heat of Mixing ... 197

9.2 Hess’s Law ... 197

9.3 Kopp’s Rule ... 198

9.4 Adiabatic Reaction Temperature ... 198

9.5 Theoretical Flame Temperature ... 198

Exercises ... 217

10 PROBLEMS ON UNSTEADY STATE OPERATIONS 221–228 Worked Examples ... 221 Exercises ... 227

T ables ... 229–234 I Important Conversion Factors ...229

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III(a) Empirical Constants for Molal Heat Capacities of Gases at Constant Pressure ... 234

III(b)Molal Heat Capacities of Hydrocarbon Gases ... 234

Answers to Exercises... 235–245 Index... 247–248

Preface

The objective of this book is to enrich a budding chemical engineer the techniques involved in analyzing a process plant by introducing the concepts on units and conversions, mass and energy balances. This will enable him to achieve a proper design of process equipment. An attempt has been made to explain the principles involved through numerical examples. The problems are not only confined to SI system of units but also worked out in other systems like FPS, CGS and MKS systems. We feel that our attempt will be more rewarding if students come across data presented in FPS, CGS and MKS systems while designing equipment, since different reference books give standard values and data in various units.

The book covers various interesting topics such as units and dimensions, mass relations, properties of gases, vapour pressure, psychrometry, crystallization, mass balance including recycle and bypass, energy balance and unsteady state operations. The second edition is now enriched with additional worked examples and exercises to give additional exposure and practice to students.

The text is designed for a one semester programme as a four credit course and takes care of the syllabus on ‘Process Calculations’ of most of the universities in India.

V. Venkataramani N. Anantharaman K.M. Meera Sheriffa Begum

xi

Preface to the First Edition

Chemical engineers in process industries generally need to focus on design, operation, control and management of a process plant. It is, therefore, absolutely essential for them to be conversant with the mass and energy conservation techniques at every stage of the process to achieve economy in the design of process equipment in various units of the plant. This book aims at imparting knowledge of the basic chemical engineering principles and techniques used in analyzing a chemical process. By applying the relevant techniques, a chemical engineer is able to evaluate material and energy balances in different units and present the information in a proper form so that the data can be used by the

management in taking correct decisions. Keeping this in mind, an attempt has been made to give a brief theory on the principle involved and more emphasis on numerical examples.

Since data are generally obtained in different units, the worked examples are not confined to SI units, but to other systems as well, such as FPS, CGS and MKS systems of units. The examples

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The text is organized into ten chapters and appends three important tables. The organization is such that the topics are presented in order of easy comprehension rather than following a logical sequence, e.g. the chapter on unsteady state operations has been included as the last chapter so that students can absorb the problems easily. We strongly feel that once the student understands the topics presented in this book, he will find other advanced courses in chemical engineering simple and easy to follow. The topics covered in this book cater to the syllabi on ‘Process Calculations’ of most universities offering courses in chemical engineering and its allied branches at the undergraduate level.

V. Venkataramani N. Anantharaman xiii

Acknowledgements

At the outset, we wish to thank the almighty for his blessings.

V. Venkataramani wishes to acknowledge his wife, Prof. (Mrs.) Booma Venkataramani, sons Mr. V. Ravi Chandar, Mr. V. Hari Sundar and his daughters-in-law Mrs. Vandana Ravi Chandar and Mrs. Ramya Hari Sundar and granddaughters Miss Vaishnavi Ravi Chandar and Miss Sadhana Hari Sundar for their support, cooperation and patience shown during the preparation of this book.

N. Anantharaman wishes to thank his mother, wife, Dr. Usha Anantharaman, sons Master A. Srinivas and A. Varun for all their patience, cooperation and support shown during the preparation of this book. The encouragement received from his brothers and sisters and their family members is

gratefully acknowledged. He also wishes to place on record the support received from his brothers-in-law and sisters-brothers-in-law and their family members.

K.M. Meera Sheriffa Begum wishes to acknowledge her mother, husband Mr. S. Malik Raj and baby M. Rakshana Roshan for all their encouragement, support and cooperation while preparing this book. She also wishes to place on record the support received from her parents-in-law. The support

received from her brothers, sisters, in-laws and their families is gratefully acknowledged. We also thank Director, NIT, Tiruchirappalli for extending all the facilities and his words of appreciation.

We wish to acknowledge the support and encouragement received from the Head of Chemical Engineering Department and all our colleagues during the course of preparation of this book. We also wish to place on record the suggestions received from students, especially those at NIT, Tiruchirappalli, and also the faculty from other institutions.

xv xvi

ACKNOWLEDG EMENTS

We gratefully acknowledge all the well-wishers.

Finally, we wish to thank the publishers, PHI Learning, New Delhi for encouraging us to bring out the second edition of the book.

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Units and Dimensions 1

1.1 INTRODUCTION

Chemical engineers are concerned with the design and development of processes which involve changes in the bulk properties of matter. To make a quantitative estimation of these processes,

chemical equations showing the quantities of reactants and products are used. Though internationally we follow SI system of units, a chemical engineer is expected to be familiar and conversant with all the systems so far adopted for measuring and expressing various quantities. A review of literature and data over the years will be available in various units. These are used to express properties, process variables and design parameters in FPS, CGS, MKS and SI systems of Units. Hence, one has to be conversant with their use and applications. This chapter deals with the basic notations and conversion of a given quantity from one system of units to another.

The quantities used in our analysis are classified as fundamental quantities and derived quantities. The fundamental quantities comprise length, mass, time and temperature. The quantities such as force, density, pressure, mass flow rate derived from the fundamental quantities are called derived

quantities. While handling these quantities, we come across different systems of units as mentioned earlier. Now let us see in detail these systems of units and their conversion from one unit to another.

1.2 BASIC UNITS AND NOTATIONS

English Metric Engineering System Engineering,_{CGS MKS}International, FPS SI

Mass ( m) lb g kg kg Length (L)ft cmm m Time (t)ssss Temperature (T)°F °C °C K

1 Mass (m) 1 kg = 2.205 lb Length (L) 1 ft = 30.48 cm = 0.3048 m Time (t) 1 h = 3600 s Temperature (T ) °C=

°F 32_{(Celsius and Centigrade are same)}

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1.3 DERIVED UNITS

Area: = length ´ breadth (L2): 1 ft2 = 0.0929 m2

10.76 ft2 = 1 m2

Force: = mass ´ acceleration (mL/t2): 1 dyne = 1 g cm/s2

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(Force applied on a mass of 1 g, which gives an acceleration of 1 cm/s2) 1 Newton (N) = 1 kg m/s2

= (1000 g) (100 cm)/s2

1 N = 105 g cm / s2 = 105 dynes

Work/energy: = 1 kg m/s2

1 erg is the amount of work done on a mass of 1 g when it is displaced by 1 cm by applying a force of 1 dyne. 1 erg = [1 dyne] ´ [1 cm] = [1 g cm/s2] ´ [1 cm] = 1 g cm2/s2 1 Joule = (1 N) ´ (1 m) = 105 g cm/s2 ´ 100 cm = 107 g cm2/s2 1 Joule = 107 erg Heat Unit:

1 Btu = 0.252 kcal = 252 cal 1 cal = 4.18 J

1 J/s = 1 W

1.4 DEFINITIONS

System. This refers to a substance or group of substances under consideration, e.g. storage tank,

water in a tank, hydrogen stored in cylinder, etc.

Process. Changes taking place within the system is called process, e.g. burning of fuel, or reaction

between two substances like hydrogen and oxygen to form water.

Isolated system. Boundaries of the system are limited by a mass of material, and its energy content is

completely detached from all other matter and energy. In an isolated system, the mass of the system remains constant, regardless of the changes taking place within the system.

Extensive property. It is a state of system, which depends on the mass under consideration, e.g.

volume.

Intensive property. This state of a system is independent of mass. An example of this property is

temperature.

WORKED EXAMPLES

1.1 The superficial mass velocity is found to be 200 lb/h.ft2. Find its equivalent in kg/s.m2

G (Mass velocity) = (200) lb/h.ft2

=

(200)Êˆ¥¥ ¥11 1m2

Á˜Ë¯ (3600 s) (0.0929) = 0.2712 kg/s.m2

1.2 Convert the heat transfer coefficient of value 100 Btu/h.ft2.°F into W/m2 °C

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(0.0929 m2)

(1.8 °C) [1 degree variation in Farh. scale is equivalent to 1.8 times the variation in celsius scale] = 4.186 ¥ 10–2 kcal/s.m2 °C = 4.186 ¥ 10–2¥ 103¥ 4.18 W/m2 °C = 174.98 W/m2 °C

1.3 The rate of heat loss per unit area is given by (0.5) [(DT)1.25/(D)0.25] Btu/h ft2 for a process, where, DT is in °F and D is in ft. Convert this relation to estimate the heat flux in terms of kcal/h. m2 using DT in °C and D in m. We know that, 9_C 1 + 32 = F15 9_C 2 + 32 = F25 Therefore, 1.8 [DC] = (DF) qT()1.25 0.5()0.25 A q_{, Btu/h ft}2_{= 0.5}'(°F)1.25_{A (ft)}0.25 We know that, 1 Btu = 0.252 kcal 1 ft2 = 0.0929 m2 1 ft = 0.3048 m For DT °F = 1.8 DT °C ËÛ1.25 Btu/h ft 2 = 0.5 'ÌÜ (ft)0.25 (a) ÌÜ ÍÝ (1.8 ' T °C) 1.25 = (0.5)_(Dm/0.3048)0.25(b) = (0.7746) '(°C)1.25 0.25_(m) Btu/h ft2 = (0.252 kcal)/h (0.0929 m2) = 2.713 kcal/h m2 (c)

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From (b) and (c) we get the expression for heat flux in units of kcal/h m2 with temperature difference in Celsius and diameter in metre as:

Heat flux, kcal/h m2 =

(m)0.25 ËÛ_{T °C)}1.25 'ÌÜ_´ 2.713_ÌÜ ÍÝ (2.101)( ' T °C) 1.25 =_(m)0.25 (d)

Now let us check the conversion with the following data: D = 0.2 ft, i.e. D¢ = 0.06096 m DT = 18 °F, i.e. DT = 10 °C

From Eq. (a), heat flux is = 0.5 (18)1.25

0.25

= 27.72 Btu/h ft

2_{= 75.2 kcal/h m}2_(0.2)

Also, from Eq. (d), heat flux = 2.101 (10)1.25_(0.06096)0.25

= 75.2 kcal/h m2 Both the values agree.

1.4 If C_p of SO₂ is 10 cal/g mole K, what is the value in FPS units? The C_p value is the same in all units, i.e. 10 Btu/lb mole °R.

1.5 Iron metal weighs 500 lb and occupies a volume of 29.25 litres. Find the density in kg/m3.

Basis: 500 lb of Iron = 500/2.2 = 227.27 kg

29.25 lit = 29.25 ´ 10–3 m

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227.27_{´ 10}–3_{= 7770 kg/m}3_{Density =}

29.25

1.6 Etching operation follows the relation d = 16.2 – 16.2e–0.021t, where t is in s. and d is in microns. Convert this equation to evaluate d in mm with t in min.

d = 16.2 [1 – e–0.021t]

Let d¢ be in mm and t¢ be in min. (d = d¢ ´ 103 and t = t¢ ´ 60) Then, d = d¢ × 103 = 16.2 [1 – e–0.021t

× 60_]

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1.7 The density of fluid is given by r = 70.5 exp (8.27 × 10–7). Convert this equation to calculate the density in kg/m3 with pressure in N/m2.

1000 kg/m3 = 62.43 lb/ft3 14.7 psi = 1.0133 × 105 N/m2 1 kg/m3 = 62.43 × 10–3 lb/ft3 Let, r¢ be in kg/m3 and p¢ be in N/m2. Then, (r¢ × 62.43 × 10–3) = 70.5 × exp (8.27 × 10–7× p¢ × 14.7/1.0133 × 105) r¢ (kg/m3) = 1.129 × 103 × exp [119.97 × 10–12 × p¢ (N/m2)]

1.8 Vapour pressure of benzene in the range of 7.5 °C to 104 °C is given by log₁₀ (p) = 6.9057 – 1211/(T + 220.8), where T is in °C and p is in torr 1 torr = 133.3 N/m2. Convert it to SI units. Let p¢ be in N/m2and T¢ be in K. Then, log ÉÙ ÊÚ pÈØ_{= 6.9057 –}1211 (T273)_220.8 1211_{log p¢ – log 133.3 = 6.90305 –} T 52.2 log (p1211_{.¢) = 9.0305 –} T52.2 EXERCISES

1.1 Convert the following quantities:

(a) 42 ft2/h to cm2/s (b) 25 psig to psia (c) 100 Btu to hp-h (d) 30 N/m2 to lbf/ft2

(e) 100 Btu/h ft2 °F to cal/s cm2 °C (f) 1000 kcal/h m°C to W/m K

1.2 The heat transfer coefficient for a stream to another is given by h = 16.6 C_p G0.8/D0.2 where h = Heat transfer coefficient in Btu/(h)(ft)2(°F) D = Flow diameter, inches

G = Mass velocity, lb/(s)(ft)2 C_p = Specific heat, Btu/(lb) (°F)

Convert this equation to express the heat transfer coefficient in kcal/ (h)(m)2(°C)

With D = Flow diameter in m, G = Mass velocity in kg/(s)(m)2 and C_p = Specific heat, kcal/(kg) (°C)

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p/(T)0.5 where m is in lb/min, p is in psia and T is in °R, where T(°R) = T °F + 460. Obtain an expression for the mass flow rate in kg/s with p in atmospheres (atm) and T in K.

1.4 The flow past a triangular notch weir can be calculated by using the following empirical formula:

q = [0.31 h2.5/g0.5] tan F

where q = Volumetric flow rate, ft3/s

h = Weir head, ft

g = Local acceleration due to gravity, ft/s2

F = Angle of V-notch with horizontal plane

1.5 In the case of liquids, the local heat transfer coefficient, for long tubes and using bulk-temperature

properties, is expressed by the empirical equation,

h

= 0.023

G

0.8_k0.67_C0.5_/(D0.2_m0.47

p )

where G = Mass velocity of liquids, lb/ft2.s

k = Thermal conductivity, Btu/ft.h.°F C_p = Specific heat, Btu/lb °F

D = Diameter of tube, ft

m = Viscosity of liquid, lb/ft.s

Convert the empirical equation to SI units.

Mass Relations 2

2.1 MASS RELATIONS IN CHEMICAL REACTION

In stoichiometric calculations, the mass relations between reactants and products of a chemical

reaction are considered and are based on the atomic weight of each element involved in the reaction. For the following reactions the material balance is established as indicated below:

(i) CaCO₃ ® CaO + CO₂ (2.1) [40 + 12 + 3 ´ 16] ® [40 + 16] + [12 + 32] 100 ® 56 + 44

(ii) 3Fe + 4H₂O ® Fe₃O₄ + 4H₂ (2.2) (3 ´ 55.84) + 4(2 ´ 1 + 16) ® (55.84 ´ 3 + 4 ´ 16) + 4(2 ´ 1) 167.52 + 72 ® 231.52 + 8 239.52 ® 239.52

Based on the reactions given by Eqs. (2.1) and (2.2) we conclude that when 100 parts by weight of CaCO₃ reacts, 56 parts by weight of CaO and 44 parts by weight of CO₂are formed. Similarly, when 167.52 parts by weight of iron reacts with 72 parts by weight of steam (water), we get 231.52 parts by weight of magnetite and 8 parts by weight of hydrogen. Thus the total weight of reactants is always

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equal to the total weight of products.

Such computations will help one to estimate the quantity of reactants needed to obtain a specified amount of product.

gram atom (or g atom) = Mass in grams/Atomic weight katom (or kg atom) = Mass in kg/Atomic weight

gram mole (or g mole) = Mass in grams/Molecular weight kmole (or kg mole) = Mass in kg/Molecular weight

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The conclusions based on reactions (2.1) and (2.2) on material balance can be expressed in other forms too, as per the definitions given above:

1 kmole of CaCO₃ gives 1 kmole of CaO and 1 kmole of CO₂

Similarly, 3 kmoles of iron reacts with 4 kmoles of steam (water) to yield 1 kmole of oxide and 4 kmoles of hydrogen. When such balances (on molar basis) are made, the number of moles on the

reactants side need not be equal to the total numbers of moles on the product side.

One atom of oxygen weighs = 16 grams O One atom of hydrogen weighs = 1 gram H One molecule of oxygen weighs = 32 grams O One molecule of hydrogen weighs = 2 grams H In other words, 16 grams of oxygen 32 pounds of oxygen 2 g atoms of oxygen 1 gram of hydrogen 2 kg of hydrogen

2 kg atoms of hydrogen = 1 kmole of hydrogen

\ g atom or lb atom = Mass in grams or pounds/Atomic weight \ g mole or lb mole = Mass in grams or pounds/Molecular weight = 1 g atom of oxygen = 1 lb mole of oxygen = 1 g mole of oxygen = 1 g atom of hydrogen = 1 kmole of hydrogen

2.2 CONSERVATION OF MASS

The law of conservation of mass states that mass can neither be created nor be destroyed. It is the

basic principle adopted in solving the material balance problems in chemical process calculations, whether a chemical reaction is involved or not. However, while applying the law of conservation of mass, one should not apply it for the conservation of molecules. We frequently come across chemical reactions in which the total number of moles on the reactant side is not equal to the total number of moles on the product side. For example:

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Na₂CO₃ + Ca(OH)₂ ® CaCO₃ + 2NaOH

The total number of moles on the reactant side is 2 and on the product side 3. Here the mass balance is ensured but not the mole balance. Now consider the reaction:

2Cr₂O₃ + 3CS₂® 2Cr₂S₃ + 3CO₂

Here the total number of moles both on the reactant side and the product side is 5. Hence both the conservation of mass and the conservation of moles are observed.

2.3 AVOGADRO’S HYPOTHESIS

1 g mole of any gaseous substance at NTP occupies 22,414 cc or

22.414 litres and 1 lb mole of the same substance occupies 359 ft3at NTP. 1 kmole of any gaseous substance occupies 22.414 m3 at NTP.

Note: NTP = Normal temperature (273 K) and pressure (1 atmosphere) are also referred to at times

as standard conditions (SC).

2.4 LIMITING REACTANT AND EXCESS REACTANT

For most of the chemical reactions the reactants will not be used in stoichiometric proportion or quantities. One of the reactants will be present in excess and remain unreacted even when the other reactant has completely reacted. The reactant thus present in excess is termed excess reactant and the other reactant which is present in a lesser quantity and cannot react with whole of the other reactant (excess reactant) is called limiting reactant. All calculations involved in estimating the quantity of product and conversion are always based on the limiting reactant. The amount by which any reactant is present in excess to that required to combine with the limiting reactant is usually expressed as percentage excess. The percentage excess of any reactant is defined as the percentage ratio of the excess to that theoretically required by the stoichiometric equation for combining with the limiting reactant. A limiting reactant is the one, which will not be present in the product, whereas the excess reactant is the one, which will always be present in the product.

Let us consider that 18 kg of carbon is burnt with 32 kg of oxygen. As per stoichiometry C + O₂Æ CO₂ i.e. 12 kg of carbon will burn with 32 kg of oxygen to form 44 kg of CO₂.

Hence, for 18 kg of carbon to react fully we should have 48 kg of oxygen. Since 32 kg of oxygen alone is available, it is called the limiting reactant and carbon is called the excess reactant. For 32 kg of oxygen to react fully, it is sufficient to have 12 kg of carbon. However 6 kg of carbon is present in excess.

Hence % excess of carbon is =

6_{¥ 100 = 50%.} 12

2.5 CONVERSION AND YIELD These terms are used for a chemical reaction where the reactants give out new compounds or products.

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Conversion is the ratio of the amount of material actually converted to that present initially, whereas

the yield is the amount of desired product actually formed compared to that which can be formed theoretically.

Conversion for a reaction is based on the limiting reactant whereas the yield is based on the product formed.

2.6 COMPOSITION OF MIXTURES AND SOLUTIONS

Different methods are available for expressing composition of mixtures of gases, liquids and solids. Conventionally the composition of solids is either expressed on weight basis or mole basis. Let us consider a binary system comprising components A and B.

W = Total weight of the system.

W_A, W_B = Weight of components A and B respectively. M_A, M_B = Molecular weight of components A and B respectively,

if they are compounds.

A_A,A_B = Atomic weight of components A and B respectively, if they are elements.

V = Volume of the system.

V_A and V_B= Pure component volume of components A and B respectively.

2.6.1 Weight Percent

This is defined as the ratio of weight of a particular component to the total weight of the system in every 100 part, i.e.

Weight % of A =

WA_{¥ 100} W

This method of expressing composition is generally employed in solid and liquid systems and not used in gaseous system. One major advantage of weight percent is, its independence to changes in temperature and pressure.

The composition of a solid mixture is to be always taken as weight % when nothing is mentioned above its units.

2.6.2 Volume Percent

The ratio of the volume of each component and the total volume of the system, for each 100 part of the total volume is called volume percent Volume % of A =

VA_{´ 100} V

This method of expressing composition is employed always for gases, rarely for liquids and seldom for solids. The composition of a gas mixture is to be taken as volume % when nothing is mentioning about its units.

The volume % is also equal to mole % for ideal gases but not for liquids and solids. This is based on Avogadro’s law.

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These concepts are generally adopted in the case of a mixture containing molecules of different species.

W_AMAMole fraction of A = _{WA WB} M_A M_B Mole % of A = Mole fraction of A ´ 100

2.6.4 Atomic Fraction and Atomic Percent This is adopted when a mixture contains two or more

atoms. W_A Atomic fraction of

A

=

A_A W_AW_B A_A A_B

Atomic % of A = Atomic fraction of A ´ 100

2.6.5 Composition of Liquid Systems

In the case of liquids we come across more number of methods of expressing compositions of the liquid constituents.

(i) Weight ratio (ii) Mole ratio (iii) Molality (iv) Molarity

= Weight of solute/weight of solvent = g moles of solute/g moles of solvent = g moles of solute/1 kg of solvent

= Number of g moles of solute/1 litre of solution

(v) Normality (N) = Number of gram equivalents of solute/1 litre of solution

Hence, concentration in grams per litre = Normality ( N) ´ Equivalent weight of solute For very dilute aqueous solutions, molality = molarity

2.7 DENSITY AND SPECIFIC GRAVITY

Density is defined as mass per unit volume and it varies with temperature. Specific gravity is the ratio of density of a liquid to that of water. However, in the case of gases, it is defined as the ratio of its density to that of air at same conditions of temperature and pressure.

Over a narrow range of temperature, the variation in density of solids is not high. However, in the case of liquids and gases the variation in density is significant. Similarly, the densities vary

significantly with concentration also. This property of density and specific gravity varying with concentration is very widely used both in industries and markets as an index for finding the composition of a system comprising a specific solute and a specific solvent.

Several scales are in use in which specific gravities are expressed in terms of a degree, which are related to specific gravities and densities by arbitrary mathematical definitions.

2.7.1 Baume (°Be) Gravity Scale

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Degrees Baume’ =

140_{– 130}

G

is the specific gravity at 60_°F 15¦µ

60 15 §¶ ¨·

Thus, water will have a gravity of 10° Be’ and this degree decreases with increase in specific gravity.

For liquids heavier than water

145_{Degrees Baume’ = 145 –}

G

In this scale the degree increases with increase in specific gravity.

2.7.2 API Scale (American Petroleum Institute) This scale is used for expressing gravities of

petroleum products. This is similar to Baume’ scale for liquids lighter than water. Degrees API = 141.5– 131.5_G

2.7.3 Twaddell Scale

This scale is used for liquids heavier than water. Degrees Twaddell (Tw) = 200 (G – 1.0)

2.7.4 Brix Scale

This scale is used in sugar industry and 1 degree Brix is equal to 1% sugar in solution. Degree Brix = 400– 400_G

WORKED EXAMPLES 2.1 Convert 5000 ppm into weight %. 5000 ×100_{= 0.5%106}

2.2 The strength of H₃PO₄ was found to be 35% P₂O₅. Find the weight % of the acid. The acid can be split into

2H PO → P O + 3H O 34 25 2 (2 98) 142 (3 18) ×

196 units of the acid contains 142 units of the pentaoxide. The weight % of pentaoxide is (142/196) which is 72.5% for pure acid. When the strength of pentoxide is 35%, the weight % of acid is ₌

48.3%

2.3 What is the volume of 25 kg of chlorine at standard condition?

25 kg Cl₂ = 25kmoles of Cl_{22 ×35.46}

25 × 22.414_{= 7.9 m}3_{Volume =}

2 × 35.46

2.4 How many grams of liquid propane will be formed by the liquefaction of 500 litres. of the gas at

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1 g mole of any gas occupies 22.414 litres at NTP

500 litres of propane at NTP = 500 = 22.31 g moles_22.414

22.31 g moles of propane weighs = 22.31 ´ 44 = 981.52 g. 2.5 Find the volume of (a) 100 kg of hydrogen and (b) 100 lb of hydrogen at standard conditions?

(a) 100 kg of H₂ = 50 kmoles of hydrogen

volume occupied by 50 kmoles of hydrogen º 50 ´ 22.414 = 1120.7 m3 (b) 100 lb of H₂ º 50 lb moles of hydrogen

Volume occupied by 50 lb moles of hydrogen º 50 ´ 359 = 17950 ft3

2.6 A solution of naphthalene in benzene contains 25 mole % Naphthalene. Express the composition

in weight %. Basis: 100 g moles of solution

Component Molecular weight Weight, Actual g mole weight, g Composition in weight percent

Naphthalene C10H8 128 Benzene C6H6 78 25 25 ´ 128 = 3200 75 75 ´ 78 = 5850 3200 ´ 100/9050 = 35.35 5850 ´ 100/9050 = 64.65 Total 9050 100.00

2.7 What is the weight of one litre of methane CH₄ at standard conditions? 22.414 litres of any gas at NTP is equivalent to 1 g mole of that gas

1_{= 0.0446 g mole\ 1 litre of methane º 1 ´} 22.414

\ Weight of one litre methane = 0.0446 ´ 16 = 0.714 g

2.8 A compound whose molecular weight is 103 analyses C : 81.5, H : 4.9 and N : 13.6 by weight.

What is the formula?

Basis: 100 g of substance Element Atomic weight

Carbon 12 Hydrogen 1 Nitrogen 14

Weight, g Weight, Rounding of Weight of g atom atoms each element

81.5 81.5/12 = 6.8 7 84 4.9 4.9/1 = 4.9 5 5 13.6 13.6/14 = 0.9 1 14 Total 103

Hence the formula obtained after rounding is correct. So the molecular formula is C₇H₅N

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Na₂O : 7.8%, MgO : 7.0%, ZnO : 9.7%, Al₂O₃: 2.0%, B₂O₃ : 8.5% and rest SiO₂.

Basis: 100 g of glass sample

Component Weight, g Molecular weight g mole mole %

Na₂O 7.8 62.0 0.1258 7.665 MgO 7.0 40.3 0.1737 10.583 ZnO 9.7 81.4 0.1192 7.262 Al₂O₃ 2.0 102.0 0.0196 1.194 B₂O₃ 8.5 69.6 0.1221 7.439 SiO₂ 65.0 60.1 1.0815 65.857 Total 100.0 — 1.6419 100.0

2.10 A gaseous mixture analyzing CH₄: 10%, C₂H₆: 30% and rest H₂ at 15 °C and 1.5 atm is flowing through an equipment at the rate of 2.5 m3/min. Find (a) the average molecular weight of the gas mixture, (b) weight % and (c) the mass flow rate.

Basis: 100 g moles of the gaseous mixture.

Component Weight, Molecular Weight, Weight % g mole weight g

CH₄ 10 16 160 13.56 C₂H₆ 30 30 900 76.27 H₂ 60 2 120 10.17 Total 100 — 1180 100

The average molecular weight =

1180_{= 11.8} 100

Volumetric flow rate at standard conditions = 2.5¥ ¥ (273/288) = 3.555 m3/min Moles of the gas = 3.555 = 0.156 kmole_22.414

Mass flow rate = moles ¥ average molecular weight = 0.156 ¥ 11.8 = 1.84 kg/min 2.11 In an

evaporator a dilute solution of 4% NaOH is concentrated to 25% NaOH. Calculate the evaporation of water per kg of feed. Basis: 1 kg of feed.

NaOH present is 0.04 kg, which appears as 25% in the thick liquor formed Weight of thick liquor formed = 0.04 = 0.16 kg_0.25

Weight of water evaporated = (1 – 0.16) = 0.84 kg Water evaporated per kg of feed = 0.84 kg

2.12 The average molecular weight of a flue gas sample is calculated by two different engineers. One

engineer used the correct molecular weight of N₂ as 28, while the other used an incorrect value of 14. They got the average molecular weight as 30.08 and the incorrect one as 18.74. Calculate the %

volume of N₂ in the flue gases. If the remaining gases are CO₂ and O₂ calculate their composition also.

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Basis: 100 g moles of flue gas

Component g mole I Engineer II Engineer

N₂x 28x 14x CO₂y 44y 44y O₂z 32z 32z

Total 100 3008 1874

x + y + z = 100 (i)

28x + 44y + 32z = 3008 (ii)

14x + 44y + 32z = 1874 (iii) Solving Eqs. (i), (ii) and (iii), we get

x = Moles of nitrogen = 81%

y = Moles of carbon dioxide = 11% z = Moles of oxygen = 8%

2.13 An aqueous solution contains 40% of Na₂CO₃ by weight. Express the composition in mole percent.

Basis: 100 g of solution

Component grams Molecular g mole Composition in weight mole % Na₂CO₃ 40 106 40/106 = 0.377 0.377 ´ 100/3.71 = 10.16 Water 60 18 60/18 = 3.333 3.333 ´ 100/3.71 = 89.84 Total 3.710 100.00

2.14 What is the weight of iron and water required for the production of 100 kg of hydrogen?

3Fe 4H O +→Fe O + 4H_↑234 2 (3×55.84) (4 18) (355.84) (4 16) (4 2 1)×+×××× 167.52 72 231.52 8 239.52 239.52

Method 1 (Based on absolute mass)

167.52 kg of Fe is required for producing 8 kg of H₂ \ For producing 100 kg of H₂ (by stoichiometry)

100 t167.52_{Iron (Fe) required =}

8

= 2094 kg

Similarly, for getting 100 kg of H₂ the amount of steam (H₂O) required is

= 100 t 72= 900 kg₈

The total weight of reactants is

2094 kg Fe and 900 kg H₂O = 2994 kg

The weight of Fe 231.52´ 2094 = 2894 kg₃O₄ formed is = _167.52

\ The total weight of products is (100 kg H₂ + 2894 kg Fe₃O₄) = 2994 kg The total weight of reactants is (2094 kg Fe + 900 kg H₂O) = 2994 kg

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100 kg of H₂

4 kmoles H₂ comes from

50 kmoles of H₂ comes from Weight of 37.5 katoms Fe 50 kmoles H₂ from

Weight of 50 kmoles H₂O Moles of Fe₃O₄ formed is Weight of Fe₃O₄ formed is

Total weight of reactants Total weight of products = 50 kmoles = 3 katoms of Fe (by stoichiometry)

= ¦µ50 t§¶ = 37.5 katoms Fe¨·4 = (37.5 ´ 55.84) = 2094 kg of iron = 50¦µ_{kmoles of water} 4t§¶ ¨· = (50 ´ 18) = 900 kg H₂O =37.5kmoles₃ 37.5 = ¦µ §¶´ 231.52 = 2894 kg¨· = 2994 kg = 2994 kg

2.15 How much super phosphate fertilizer can be made from one ton of calcium phosphate 93.5%

pure?

Atomic weights are: Ca : 40, P : 31, O : 16, S : 32 Ca₃(PO₄)₂ + 2H₂SO₄Æ CaH₄(PO₄)₂ + 2CaSO₄

310 (2 ¥ 98) 234 (2 ¥ 136) 506 506

One ton of raw calcium phosphate contains 0.935 tons of pure calcium phosphate Weight of super phosphate formed is = 234 ¥

0.935_{= 0.70577 tonne\}

310

2.16 SO₂ is produced by the reaction between copper and sulphuric acid. How much Cu must be used to get 10 kg of SO₂?

(24)

24

→

CuSO +SO +2H O

4 2 2 63.54 64

64 kg of sulphur dioxide is obtained from 63.54 kg of copper. 10 kg of sulphur dioxide will be obtained from 9.93 kg of copper. 2.17 How much potassium chlorate must be taken to produce the same amount of oxygen that will be produced by 2.3 g of mercuric oxide? 2KClO₃ Æ 2KCl + 3O₂

(2¥ 122.46) (2¥ 74.46) (6¥ 16) 244.92 148.92 96

2HgO Æ 2Hg + O₂

(2¥ 216.6) (2¥ 200.6) (2¥ 16)

433.2 g HgO gives 32 g of oxygen

2.3 g of HgO will give (32 ¥ 2.3/433.2) = 0.1698 g of O₂

0.1698 g O₂ is obtained from (244.92 ¥ 0.1698)/96 = 0.4332 g of KClO₃.

2.18 Ammonium phosphomolybdate is made up of the radicals NH₃, H₂O, P₂O₅ and MoO₃. What is % composition of the molecule with respect to these radicals?

The formula of ammonium phosphomolybdate is (NH₄)₃PO₄◊ 12MoO₃◊ 3H₂O

First let us form the final product from the radicals:

3NH₃ + 4.5H₂O + 12MoO₃ + ½P₂O₅Æ (NH₄)₃PO₄ 12MoO₃·3H₂O (3¥17 = 51) (4.5¥18 = 81) (12¥144 = 1728) (½¥142 = 71) 1931 % of NH₃=51 ¥ 100_{= 2.64} 1931 % of H₂O= 81 ¥100= 4.19₁₉₃₁ % of MoO₃ = 1728 ´100= 89.49₁₉₃₁ % of P₂O₅=71100= 3.68´₁₉₃₁ Total = 100.00

2.19 How many grams of salt are required to make 2500 g of salt cake? How much Glauber’s salt can

be obtained from this?

The molecular formula of Glauber’s salt is Na₂SO₄ × 10H₂O

(142 + 180 = 322) 2NaClH SO Na SO_{+ 2HCl} (2 × 58.46 =116.92) +→_{24 24} 98 142 (2×36.46)

Thus, 142 g of Na₂SO₄ is obtained from 116.92 g NaCl. 2500 g of salt cake is obtained from 116.92 ´ 2500/142 = 2058.45 g NaCl

Hence, salt needed is 2058.45 g

Glauber’s salt (Na₂SO₄ × 10H₂O) obtained is 2500 ´ 322/142 = 5669 g

2.20 (a) How many grams of K₂Cr₂O₇ are equivalent to 5 g KMnO₄? (b) How many grams of KMnO₄ are equivalent to 5 g K₂Cr₂O₇? 2KMnO₄ + 8H₂SO₄ + 10FeSO₄ ® 5Fe₂(SO₄)₃ + K₂SO₄ + 2MnSO₄

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+ 8H₂O

K₂Cr₂O₇ + 7H₂SO₄ + 6FeSO₄ ® 3Fe₂(SO₄)₃ + K₂SO₄ + (Cr₂SO₄)₃ + 7H₂O 2KMnO₄ gives 5Fe₂(SO₄)₃

(2 ´158 = 316) (5 ´400 = 2000)

K₂Cr₂O₇ gives 3Fe₂(SO₄)₃

(294) (3 ´400 = 1200) 1200 t 316_{= 189.6 g KMnO} 4\ 294 g K2Cr2O7 is equivalent to 2000 3 t189.6_{= 1.935 g KMnO} 4\ 3 g K2Cr2O7 º294 Similarly, 5 g KMnO 4 5 t 294_{= 7.75 g K} 2Cr2O7 º189.6 Alternatively, 53t_{= 7.75 g K} 2Cr2O71.935

2.21 If 45 g of iron react with H₂SO₄, how many litres of hydrogen are liberated at standard condition?

There are two possible reactions in this case: (a) Case I Fe H SO FeSO + H

(55.85)

+→

24 4 2 (i)(2)

The weight of hydrogen formed by reaction (i) is = 45 ¥2_{= 1.611 g,} 55.85 i.e. 1.611_{= 0.806 g mole} 2 0.806 g mole ∫ 0.806 ¥ 22.414 = 18.06 litres (b) Case II 2Fe3H SO_{Fe (SO ) + 3H} (111.7) +→24 2 43 2 (ii)(6)

The moles of hydrogen formed by reaction (ii) is 45 ¥6_{= 2.418 g} 111.7

2.418 g H₂ = 1.209 g moles of hydrogen ∫ 1.209 ¥ 22.414 ∫ 27.1 litres

2.22 A natural gas has the following composition by volume CH₄ : 83.5%, C₂H₆: 12.5%, and N₂: 4%. Calculate the following:

(a) composition in mole % (b) composition in weight % (c) average molecular weight (AVMWT) (d) density at standard condition (kg/m3) Basis: 100 kmoles of gas mixture

Component Molecular mole % Weight, kg Weight % weight

CH4 16 83.5 83.5¥ 16 = 1336 1336¥ 100/1823 = 73.29 C2H6 30 12.5 12.5¥30 = 375 375¥ 100/1823 = 20.57 N2 28 4.0 4.0¥28 = 112 112¥100/1823 = 6.14 Total 1823 100.0

(c) Average molecular weight =

1823_{= 18.23} 100

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(d) Density of gas at standard condition =

1823_{= 0.813 kg/m}3_2241.4

2.23 Convert 54.75 g/litre of HCl into molarity. Molarity = g moles/litre of solution

= 54.75 = 1.5_36.45

2.24 A solution of NaCl in water contains 230 g of NaCl per litre at 20 °C. The density of the

solution at this temperature is 1.148 g/cc. Find the composition in (a) weight % (b) volume % of water (c) mole % (d) atomic % (e) molality and (f) g NaCl/g water.

Basis: (a) 1 litre of solution has a weight of 1148 g

Component Molecular Weight, g Weight, %g mole mole % weight

NaCl 58.5 230 20.03 230/58.5 = 3.93 3.93/54.93 = 7.15 Water 18 918 79.97 918/18 = 51.00 51/54.93 = 92.85 Total 1148 100.00 54.93 100.00

(b) Volume % of water: 918 g is present in 1 litre of solution, i.e. 918 cc water is present in 1000 cc. of solution (density of water is 1 g/cc)

Volume % = 91.8% (d)

Element g atoms Atomic %

Na 3.93 2.443 Cl 3.93 2.443 H 102.00 63.409 O 51.00 31.705

Total 160.86 100.000

(All are based on the molecular formula)

(e) Molality = g moles of solute in 1 kg of solvent (3.93 ´ 1000/918) or, 3.93 g moles of NaCl is

present in 918 g of water (i.e.) 4.28 g moles/1000 g of solvent Molality = 4.28

Molarity = Moles of solute per litre of the solution = 3.93 (f) g NaCl/g water = 230 = 0.252₉₁₈

2.25 A benzene solution of anthracene contains 10% by weight of the solute. Find the composition in

terms of (a) molality (b) mole fraction. Basis: 100 g of solution

Component Molecular Weight, g Weight, mole weight g mole fraction Anthracene 178 10 (10/178)

0.046 0.0562 Benzene 78 90 (90/78) 0.954 1.1538 Total 1.2100 1.00 Molality = g moles of anthracene in 1000 g benzene

=0.0562 1000 = 0.624₉₀

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2.26 Calculate the weight of NaCl that should be placed in a 1 litre volumetric flask to prepare a

solution of 1.8 molality. Density of this solution is 1.06 g/cc Molality = g moles of NaCl/1000 g of water

= 1.8

or, 1.8 g moles NaCl = (1.8 ¥ 58.46) = 105.228 g

Component Weight, g Weight % NaCl 105.228 9.52

H₂O 1000.000 90.48 1105.228 100.00

Density of this solution = 1.06 g/cc

Volume of this solution, i.e. mass/density =

1105.228_{= 1042.67 cc}

1.06

or 1042.67 cc of this solution contains 105.228 g of NaCl 1000 cc of this solution will have = 1000 ¥ 105.228\_1042.67 = 100.92 g of NaCl. NaCl needed = 100.92 g

2.27 For the operation of a refrigeration plant it is desired to prepare a solution of 20% by weight of

NaCl solution.

(a) Find the weight of salt that should be added to one gallon of water at 30°C?

(b) What is the volume of this solution?

Basis: 100 lb of solution

It will have 20 lb NaCl and 80 lb water

80 lb water = 1.28 ft3 (since the density of water is 62.47 lb/ft3) We know that 1 ft3 = 7.48 gallons Therefore, 1.28 ft3 = 9.57 gallons.

20

(a) Weight of salt per gallon of water = ¦µ

§¶ = 2.09 lb.¨·

(b) Specific gravity of NaCl solution at 30 °C = 1.14

\ Density of solution = 1.14 ´ 62.4 = 71.14 lb/ft3 Weight of 1 gallon of water = 62.47 = 8.35 lb._7.48 Total weight of solution = weight of water + weight of salt = 8.35 + 2.09 = 10.44 lb.

Hence, volume of the above solution =

10.44_{= 0.147 ft}3_71.14

= 1.1 gallons.

2.28 (a) A solution has 100° Tw gravity. What is its specific gravity and °Be’?

(b) An oil has a specific gravity of 0.79. Find °API and °Be’ (a) 100 = 200 (G – 1) \ G = 1.5 °Be’ = 145 –

145 145

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= 145 – ¦µ_{= 48.3 °Be’} G §¶ ¨· 141.5 141.5 ¦µ_{– 131.5 = 47.6} ¨· – 131.5 = §¶(b) °API =¦µ §¶_¨· 140 °Be’ = §¶ ¦µ_{– 130 = 47.2} ¨·

2.29 An aqueous solution contains 15% ethyl alcohol by volume. Express the composition in weight

% and mole %. Density of ethyl alcohol and water are 790 kg/m3 and 1000 kg/m3 respectively. Basis: 1 m3 of solution.

Compound Molecular Volume, Density, Weight, Number Weight mole weight m3 kg/m3 kg of moles %%

Ethanol 46 0.15 790 118.5 2.576 12.235 5.173 Water 18 0.85 1000 850 47.222 87.765 94.827 Total 1.00 968.5 49.798 100 100

2.30 The quality of urea is expressed in terms of nitrogen content. If the nitrogen content in the sample

is only 40%, estimate the purity of sample in terms of urea content.

The molecular weight of urea (NH₂CONH₂) is 60 and that of N₂ is 28. Basis: 100 kg of sample 60 kg of urea has 28 kg of N₂

100 kg of urea will have = 28 × 100 = 46.67 kg of N₂₆₀ (Theoretically)

The given sample has 40% N₂

Hence, the % purity is = 40 × 100 = 85.71%_46.67

2.31 If the nitrogen content in ammonium nitrate sample is 28%, estimate the purity of ammonium nitrate.

Molecular weight of ammonium nitrate, NH₄NO₃ = 80 % Nitrogen in pure ammonium nitrate = 28 × 100= 35%₈₀ The % of nitrogen in the sample is 28

28

Hence, the purity of ammonium nitrate is §¶

¦µ_{× 100 = 80%} ¨·

2.32 Nitrobenzene is produced by reacting nitrating mixture with benzene. The nitrating mixture

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of nitrating mixture which sent into the reactor. If the reaction is 95%, then calculate the amount of nitrobenzene and spent acid produced. The reaction is

C₆H₆ + HNO₃ ® C₆H₅NO₂ + H₂O Feed, C₆H₆ : 663/78 = 8.5 kmole

HNO₃ : 31.5% of 1700 kg = 535.5 kg = 8.5 kmoles H₂SO₄ : 60 % of 1700 kg = 1020 kg = 10.408 kmoles H₂O : 8.5 % of 1700 kg = 144.5 kg = 8.028 kmoles Reaction is 95% complete

Hence, HNO₃ unreacted : 0.05 × 8.5 = 0.425 kmole C₆H₆ unreacted : 0.05 × 8.5 = 0.425 kmole H₂SO₄ unreacted : 10.408 kmoles

H₂O unreacted : 8.028 kmoles

H₂O formed : 8.5 × 0.95 = 8.075 kmoles Nitrobenzene formed : 8.5 × 0.95 = 8.075 kmoles

Component Weight, Molecular Weight, Weight, kmole weight kg %

HNO₃ 0.425 63 26.775 1.133 H₂SO₄ 10.408 98 1020.000 43.165 H₂O (8.075 + 8.028) = 16.103 18 289.850 12.266 Nitrobenzene 8.075 123 993.225 42.032 C₆H₆ 0.425 78 33.150 1.403 Total 2363.00 100% Nitrobenzene produced = 993.225 kg

Spent acid = 26.775 + 1020.000 + 289.85 = 1336.63 kg 2.33 A sample of caustic soda flake contains 74.6% Na₂O by weight. Estimate the purity of flakes.

Reaction is as follows: 2NaOH ® Na₂O + H₂O

Amount of Na₂O in pure flakes = 62 × 100/80 = 77.5% % Purity = 0.746/0.775 × 100 = 96.26%

2.34 Two kg of CaCO₃ and MgCO₃ was heated to a constant weight of 1.1 kg. Calculate the % amount of CaCO₃ and MgCO₃ in reacting mixture.

Reaction is as follows: CaCO₃ ® CaO + CO₂

(100) (56) (44)

MgCO₃ ® MgO + CO₂

(84) (40) (44)

Let, x be the amount of CaCO₃.

Therefore, (2 – x) be the weight of MgCO₃ 100 kg of CaCO₃gives 56 kg of CaO

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Similarly, 84 kg of MgCO₃ gives 40 kg of MgO 40 Therefore, (2 – x ) kg of MgCO 3 gives ©¸ ª¹× (2 – x) kg of MgO «º

The weight of product left behind is 1.1 kg, i.e. weight of MgO + CaO left behind 0.56x + (0.4672)(2 – x) = 1.1

0.0838x = 1.1 – 0.96524 Therefore, x = 1.761 kg

Component Weight, kg Weight, %

CaCO₃ 1.761 88.05 MgCO₃ 0.239 11.95 Total 2.000 100.00

2.35 The composition of NPK fertilizer is expressed in terms of N₂, P₂O₅ and K₂O each of about 15 weight %. Anhydrous ammonia, 100% phosphoric acid and 100% KCl are mixed to get 1 ton of fertilizer. Estimate the amount of filler in the NPK fertilizer.

Basis: 1000 kg of fertilizer Reactions are: 2NH₃ ® N₂ +3H₂ (34) (28) (6) 2H₃PO₄ ® P₂O₅ +3H₂O (196) (142) (54) 2KCl + H₂O ® K₂O + 2HCl (149) (18) (94) (73)

N₂, K₂O and P₂O₅ are each equivalent to 15 weight % = 150 kg each Ammonia reacted = 34 ×

150_{= 182.14 kg} 28 H150_{= 207.04 kg} 3PO4 needed = 196 × 142 KCl needed = 149 × 150_{= 237.77 kg} 94

The amount of inert material/filler = 1000 – 626.95 = 373.05 kg

2.36 A solution whose specific gravity is 1 contains 35% A by weight and the rest is B. If the specific

gravity of A is 0.7, find the specific gravity of B.

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Weight of A: 350 kg Weight of B: 650 kg

Volume of solution = 1 m3 (since density is 1000 kg/m3 due to specific gravity being unity) Mass/volume = density

Assuming ideal behaviour 350 650_{= 1700 SB}

r_B= 1300 kg/m3

Therefore, specific gravity of B = 1.3

2.37 An aqueous solution contains 47% of A on volume basis. If the density of A is 1250 kg/m3, express the composition of A in weight %.

Basis: 1 m3 of solution

Volume of A in solution = 1 × 0.47 = 0.47 m3 Weight of A = 0.47 × 1250 = 587.5 kg

Volume of water = (1 – 0.47) = 0.53 m3

Therefore, the weight of water = 0.53 m3 × 1000 = 530 kg Hence, weight % of A =

587.5_{= 52.57%}

530 587.5

2.38 An aqueous solution contains 43 g of K₂CO₃ in 100 g of water. The density of solution is 1.3 g/cc. Find the composition in molarity and molality.

Basis: 100 g of solvent Weight of K₂CO₃ = 43 g Weight of solution = 143 g Density of solution = 1.3 g/cc Volume of solution = 143_{= 110 cc} 1.3

Moles of solute = Weight/Molecular weight = 43_{= 312 g moles} 138

Molarity = g mole/volume of solution in lit = 0.312_{= 2.833M} 0.11

Molality = g mole/kg of solvent = 0.312= 3.12 g moles/kg solvent._0.1

2.39 A gaseous mixture contains ethylene: 30.6%, benzene: 24.5%, O₂: 1.3%, ethane: 25%, N₂: 3.1% and methane: 15.5% in volume basis. Estimate the composition in mole %, weight %, average

molecular weight and density.

Basis: 100 kmoles of mixture

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C₂H₄ 30.6 28 856.8 22.00 C₆H₆ 24.5 78 1911.0 49.07 O₂ 1.3 32 41.6 1.07 CH₄ 15.5 16 248.0 6.37 C₂H₆ 25 30 750.0 19.26 N₂ 3.1 28 86.8 2.23 Total 3894.2 100.00 Density = Weight/Volume = 3894.2_{× 22.414 = 1.737 kg/m}3 _{= 1.737 g/l.} 100

2.40 A compound has a composition of 9.76% Mg, 13.01% S, 26.01% O₂and 57.22% H₂O by weight. Find the molecular formula of this compound.

Basis: 100 g of compound

Compound Weight, Atomic weight Number g or Molecular of moles weight

Mg 9.76 24 0.410 S 13.01 32 0.410 O 26.01 16 1.615 H₂O 57.22 18 2.8738

Converting to whole numbers dividing by 0.41

1 1 3.94 6.92

Therefore, molecular formula of the compound = MgSO₄.7H₂O 2.41 A substance on analysis gave 1.978 g of Ag, 0.293 g of S and 0.587 g of O_2.Find the molecular formula of the compound.

Compound Weight, Atomic weight g or Molecular weight Number Converting to of moles whole numbers dividing by 9.156 × 10–3 Ag 1.978 108 S 0.293 32 O₂ 0.587 16 Molecular formula = Ag₂SO₄ 0.0183 2 9.156 × 10–3 1 0.0367 4.04

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2.42 Two engineers are estimating the average molecular weight of gas containing oxygen and another

gas. One uses the molecular weight as 32 and finds the average molecular weight as 39.8 and the other uses the atomic weight of oxygen as 16 and finds the average molecular weight as 33.4. Estimate the composition of the gas mixture.

By using the atomic weight of oxygen as 16, the value is 33.4 and by using the molecular weight of oxygen, the value is 39.8. Let x be the mole fraction of oxygen in the mixture and the molecular weight of the other gas be M

39.8 = (x) × (32) + (1 – x) × (M) 33.4 = (x) × (16) + (1 – x) × (M) Solving, we get x = 0.4

i.e the fraction of oxygen in the mixture is 0.4

2.43 A mixture of methane and ethane has an average molecular weight of 21.6. Find the composition.

Let the mole fraction of methane be X

21.6 = (Molecular weight of CH₄)(X) + (Molecular weight of C₂H₆) (1 – X) 21.6 = 16 × X + 30 × (1 – X)

Solving, we get X = 0.6

2.44 A mixture of FeO and Fe₃O₄ was heated in air and is found to gain 5% in mass. Find the composition of initial mixture.

Reactions involved are: 2FeO + 0.5 O₂ ® Fe₂O₃ 2Fe₃O₄ + 0.5 O₂ ® 3 Fe₂O₃

Basis: 100 kg of feed mixture

Let X be the weight of FeO in the mixture From 144 kg of FeO, Fe₂O₃formed is 160 kg Therefore, from X kg of FeO, Fe 2 O 3 formed is 160 × X 144 From 232 kg of Fe₃O_4,Fe₂O₃formed is 480 kg Therefore, from (100 – X ) of Fe 3 O 4 ,Fe 2

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O

3

formed is (100 – X ) ×

(480) 464 Since 5% gain in mass is observed, the weight of final product is 105 kg, i.e.

160X_tt(480)₁₀₅

144 464

Solving, X, the weight of FeO = 20.25 kg Fe₃O₄ = (100 – 83.45) = 79.75 kg

2.45 A sample of lime stone has 54.5% CaO. Find the weight % of lime stone.

Basis: 100 kg of lime stone

100 kg of CaCO₃will have 56% CaO

If the CaO is 54.5%, then % of CaCO₃ in the sample is 54.5 × 100/50 = 97.32%

2.46 Express the composition of magnesite in mole %.

Compound Weight %

MgCO₃ 81 SiO₂ 14 H₂O5

Compound Weight % Molecular weight moles mole % MgCO₃ 81 84 0.964 65.34 SiO₂ 14 60 0.233 15.82

H₂O 5 18 0.278 18.83

2.47 The concentration of H₃PO₄ is expressed in terms of P₂O₅ content. If 35% P₂O₅is reported, find the composition of H₃PO₄by weight. P₂O₅ + 3H₂O ® 2H₃PO₄

(142) (54) (98)

i.e. 142 kg of P₂O₅ º 196 kg of H₃PO₄

Therefore, 35 kg of P₂O₅ º 35 × 196 = 48.3 H₃PO₄₁₄₂ i.e. H₃PO₄is 48.3%

2.48 Ten kg of PbS and 3 kg of oxygen react to yield 6 kg of Pb and 1 kg of PbO₂ according to the reaction shown below:

PbS + O₂ ® Pb + SO₂ (1) PbS + 2O₂ ® PbO₂ + SO₂ (2) Estimate (i) unreacted PbS, (ii) % excess oxygen supplied, (iii) total SO₂ formed, and (iv) the % conversion of PbS to Pb.

PbS + O₂ ® Pb + SO₂ (1)

(239.2) (32) (207.2) (64)

PbS + 2O₂ ® PbO₂ + SO₂ (2)

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207.2 kg of Pb comes from 239.2 kg of PbS [from stoichiometry Eq. (1)] Therefore, 6 kg of Pb comes from 239.2 × 6 = 6.927 kg of PbS_207.2

239.2 kg of PbO₂ comes from 239.2 kg of PbS Therefore, 1 kg of PbO₂ comes from 1 kg of PbS Therefore, total PbS reacted [from Eqs. (1) and (2)] = 6.927 + 1 = 7.927 kg

Unreacted PbS = 10 – 7.927 = 2.073 kg

O₂ required for this process

From Reaction 1:

32 kg of oxygen is needed to produce 207.2 kg of Pb Therefore, to produce 6 kg of PbO₂, oxygen needed = 6 ×

32_{= 0.927 kg} 207.2

From Reaction 2:

239.2 kg of PbO₂ requires 64 kg of oxygen

Therefore, to produce 1 kg of PbO₂, oxygen required is 268 kg Therefore, total oxygen used = 0.927 + 0.268 = 1.195 kg (3 1.195) Percentage excess O 2 supplied = ©¸ ª¹× 100 = 151%«º Amount of SO₂ formed If 207.2 kg of Pb is formed, SO₂ formed is 64 kg If 6 kg of Pb is formed, SO6 = 1.853 kg₂ formed is 64 × _207.2 If 239.2 kg of PbO₂ is formed, SO₂ formed is 64 kg

If 1 kg of PbO64 = 0.268 kg₂ is formed, SO₂ formed is _239.2 Total SO₂ formed = 1.853 + 0.268 = 2.121 kg

% conversion of PbS fed to Pb = Mass of PbS converted to Pb/Total mass of PbS = 6.927 × 100= 69.27%₁₀

2.49 The composition of a liquid mixture containing A, B and C is peculiarly given as 11 kg of A, 0.5

kmole of B and 10 wt of % C. The molecular weights of A, B and C are 40, 50 and 60 respectively and their densities are 0.75 g/cc, 0.8 g/cc and 0.9 g/cc respectively. Express the composition in weight %, mole %. Also give its average molecular weight and density assuming ideal behaviours. Let the weight of mixture be W kg

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Weight of A = 11 kg Weight of C (10%) = 0.1W kg Weight of B = W – 11 – 0.1W = 0.5 kmole = 0.5 × 50 = 25 kg i.e. weight of B = W – 11 – 0.1W = 25 kg 0.9W = 36 kg W = 40 kg

i.e. total weight of mixture is 40 kg.

Component Weight, Weight, Molecular moles, mole Density, Volume, kg % weight kmole % kg/m3 m3

A 11 27.5 40 0.275 32.66 750 0.0147 B 25 62.5 50 0.500 59.38 800 0.0313 C 4 10.0 60 0.067 7.96 900 0.0044 Total 40 100.00 0.842 100.00 0.0504

40_{= 793.65 kg/m}3_{Average density = Mass/Volume =} 0.0504

Average molecular weight = Weight/Total moles = 40_{= 47.5} 0.842

(Check: Average molecular weight

= 40 × 0.3266 + 50 × 0.5938 + 60 × 0.0796 = 47.5)

EXERCISES

2.1 How many g moles are equivalent to 1.0 kg of hydrogen? 2.2 How many kilograms of charcoal is

required to reduce 3 kg of arsenic trioxide? As₂O₃+ 3C Æ 3CO + 2As

2.3 Oxygen is prepared according to the following equation: 2KClO₃Æ 2KCl + 3O₂. What is the yield of oxygen when 9.12 g

of potassium chlorate is decomposed? How many grams of potassium chlorate must be decomposed to get 5 g of oxygen?

2.4 An aqueous solution of sodium chloride contains 28 g of NaCl per 100 cc of solution at 293 K.

Express the composition in (a) percentage NaCl by weight (b) mole fraction of NaCl and (c) molality. Density of solution is 1.17 g/cc.

2.5 An aqueous solution has 20% sodium carbonate by weight. Express the composition by mole ratio

and mole percent.

2.6 A solution of caustic soda in water contains 20% NaOH by weight. The density of the solution is

1196 kg/m3. Find the molarity, normality and molality of the solution.

2.7 A saturated solution of salicylic acid in methanol contains 64 kg salicylic acid per 100 kg

methanol at 298 K. Find the composition by weight % and volume %.

2.8 A solution of sodium chloride in water contains 270 g per litre at 323 K. The density of this

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and kg of salt per kg of water.

2.9 A mixture of gases has the following composition by weight at 298 K and 740 mm Hg.

Chlorine: 60%, Bromine: 25% and Nitrogen: 15%.

Express the composition by mole % and determine the average molecular weight.

2.10 Wine making involves a series of very complex reactions most of which are performed by

microorganisms. The initial concentration of sugar determines the final alcohol content and sweetness of the wine. The general convention is to adjust the specific gravity of the starting stock to achieve a desired quality of wine. The starting solution has a specific gravity of 1.075 and contains 12.7 weight % of sugar. If all the sugar is assumed to be C₁₂H₂₂O₁₁, determine

(a) kg sugar/kg H₂O

(b) kg solution/m3 solution (c) g sugar/litre solution

2.11 The synthesis of ammonia proceeds according to the following reaction

N₂ + 3H₂ ® 2NH₃

In a given plant, 4202 lb of nitrogen and 1046 lb of hydrogen are fed to the synthesis reactor per hour. Production of pure ammonia from this reactor is 3060 lb/h.

(a) What is the limiting reactant?

(b) What is the percent excess reactant?

(c) What is the percent conversion obtained (based on the limiting reactant)?

2.12 How many grams of chromic sulphide will be formed from 0.718 g of chromic oxide according

to the following equation?

2Cr₂O₃ + 3CS₂® 2Cr₂S₃ + 3CO₂

2.13 How many kilograms of silver nitrate are there in 55.0 g mole silver nitrate?

2.14 Phosphoric acid is used in the manufacture of fertilizers and as a flavouring agent in drinks. For

a given 10 weight % phosphoric acid solution of specific gravity 1.10, determine: (a) the mole fraction composition of this mixture.

(b) the volume of this solution, which would contain 1 g mole H₃PO₄. 2.15 Hydrogen gas in the laboratory can be prepared by the reaction of sulphuric acid with zinc metal

H₂SO₄ (l) + Zn(s) ® ZnSO₄(s) + H₂ (g)

How many grams of sulphuric acid solution (97%) must act on an excess of zinc to produce 12.0 m3/h of hydrogen at standard conditions. Assume all the acid used reacts completely.

2.16 Sulphur dioxide may be produced by the reaction

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Find how much copper and how much 94% sulphuric acid must be used to obtain 32 kg of SO₂.

2.17 Aluminium sulphate is produced by reacting crushed bauxite ore with sulphuric acid as shown

below:

Al₂O₃ + 3 H₂SO₄ ® Al₂(SO₄)₃ + 3 H₂O

Bauxite ore contains 55.4% by weight Al₂O₃, the reminder being impurities. The sulphuric acid contains 77.7% H₂SO₄, the rest being water. To produce crude aluminium sulphate containing 1798 kg of pure Al₂(SO₄)₃, 1080 kg of bauxite ore and 2510 kg of sulphuric acid solution are used. Find (a) the excess reactant, (b) % of excess reactant consumed, and (c) degree of completion of the reaction.

2.18 600 kg of sodium chloride is mixed with 200 kg of KCl. Find the composition in weight % and

mole %.

2.19 What is the weight of iron and water required to produce 100 kg of hydrogen. 2.20 Cracked gas from petroleum refinery has the following composition by volume:

Methane: 42%, ethane: 13%, ethylene: 25%, propane: 6%, propylene: 9%, and rest n-butane. Find: (a) average molecular weight of mixture, (b) Composition by weight, and (c) specific gravity of the gas mixture.

2.21 A gas contains methane: 45% and carbon dioxide: 45% and rest nitrogen. Express (i) the weight

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Ideal Gases 3

3.1 RELATION BETWEEN MASS AND VOLUME FOR GASEOUS SUBSTANCES 3.1.1 Standard Conditions

1 atm. pressure or 760 mm Hg or 29.92 inches of Hg and 0 °C or 32 °F By Avogadro’s Hypothesis, 1 g mole of any gas under standard conditions will occupy 22.414 litres 1 lb mole of any gas under standard conditions will occupy 359 cu.ft.

T(K) = T °C + 273.16 T(°R) = T °F + 459.69

3.1.2 Ideal Gas Law

The ideal gas law states that,

PV = nRT

P = Pressure of gas

V = Volume of n moles of gas n = Number of moles of gas R = Gas constant

T = Absolute temperature

Using the ideal gas law (PV = nRT) and the above information one can always determine the weight of a gas if the volume is known and vice-versa.

Parameters

Pressure

Molar volume Absolute temperature Gas constant

Normal Temperature and Pressure/Standard Conditions English Metric SI 1.033 kgf/cm2 1.01325 bar 359 ft

1 atm3/lb mole 22.414 m3/kmole 22.414 m3/kmole 491.69 273.16 K 273.16 K 0.73 atm ft

oR 3/lb mole oR 0.085 kgf m3/kmole K 0.083 Bar m3/kmole K

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Pressure 1 atm = 1.033 kgf/cm2 = 1.01325 bar 1 atm = 14.67 psia = 105 N/m2 = 760 mm Hg

1 atm = 760 Torr = 29.92 inches of Hg = 76 cm Hg

R =

PV_{= 82.06 atm.cc/g mole K = 0.73 atm ft}3_{/lb mole °R}

nT

R = 10.73 lb_f ft3/in2 lb mole °R

Avogadro’s Number: 6.023 ´ 1023 molecules per g mole 2.73 ´ 1026 molecules per lb mole

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6.023 ´ 1026 molecules per kg mole

Different units are used to express pressure like atmosphere, mm of Hg, psia, kg/cm2, bar, N/m2, and Pa. Similarly, volume is expressed in cm3, m3, litre, ft3 and gallon. The temperature is expressed in °C, °F, K and °R. However, the temperature used in the application of Ideal gas law is in terms of K or °R.

Thus, the gas constant is a dimensional quantity. The following table gives the gas constant in different units.

Temperature Pressure Volume Gas constant ‘R_g’ R psia in3 18.51 R psia ft3 10.73 R atmospheres ft3 0.73 KPa m3 8314 K atmospheres m3 0.08206 K atmospheres cm3 82.06 K cm Hg cm3 6239.79 Units of gas constant

in3psia/lb mole R ft3psia/lb mole R ft3atm/lb mole R

m3Pa/kmole K m3atm/kmole K

cm3atm/g mole K (cm3cm Hg)g mole K

R_g = 8.314 J/(g mole) (K) = 1545 ft lb/lb mole °R

3.2 GASEOUS MIXTURE 3.2.1 Partial Pressure (PP)

The partial pressure of a component gas that is present in a mixture of gases is the pressure that would be exerted by that component gas if it alone were present in the same volume and at the same

temperature as the mixture.

3.2.2 Pure Component Volume (PCV)

The PCV of a component gas that is present in a mixture of gases is the volume that would be

occupied by that component gas if it alone were present at the same pressure and temperature as the mixture.

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Partial pressure p_A p_B p_C P Number of moles n_A n_B n_C n Pure component volume V_A V_B V_C V

3.2.3 Daltons Law

The total pressure (P) exerted by a gaseous mixture in a definite volume is equal to the sum of partial pressures.

p_A + p_B + p_C = P

where p_A, p_B, p_C, … represent partial pressure of components A, B, C, … .

3.2.4 Amagats Law (or) Leducs Law

The total volume (V ) occupied by a gaseous mixture is equal to the sum of the pure component volumes

V_A + V_B + V_C = V

V_A, V_B, V_C, … stand for pure component volume of components A, B, C, … Where ideal gas law is applicable;

n RT_{; p}

B = Bn RT; pC = Cn RT(a) pA = A

V V V

Adding all the partial pressures of A, B and C, we have, RT P = p A + p B + p C = ËÛ ÌÜ ´ (nA + nB + nC)ÍÝ Dividing, pRT V nA_; A/P = V ()RT n×× ++n A BC

pressure fraction = mole fraction.

(b) PV_A = n_ART; PV_B = n_BRT; PV_C = n_CRT Adding P(V_A + V_B + V_C) = RT (n_A + n_B + n_C) = nRT = PV nRTP_{= N} AV = ADividing,VA ()_{AB C}n RT ++ or, V_A = N_A × V

3.3 AVERAGE MOLECULAR WEIGHT

The weight of unit mole of the mixture is called average molecular weight, which is also equal to total weight of the gas mixture divided by the total number of moles in the mixture. This is applicable