Chemistry Grade 10 Student Textbook PDF

CHEMISTRY CH 3 CHEMISTRY STUDENT TEXTBOOK GRADE 10 STUDENT TEXTBOOK...

CHEMISTRY CH 3 CHEMISTRY STUDENT TEXTBOOK GRADE 10 STUDENT TEXTBOOK Surroundings Surroundings Temperature Endothermic Temperature Exothermic Heat Heat CH 3 GRADE 10 Fuel Local bear 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 FEDERAL DEMOCRATIC REPUBLIC OF ETHIOPIA MINISTRY OF EDUCATION  This textbook is the property of your school. Take good care not to damage or lose it. Here are 10 ideas to help take care of the book: 1. Cover the book with protective material, such as plastic, old newspapers or magazines. Students 2. Always keep the book in a clean dry place. 3. Be sure your hands are clean when you use the book. 4. Do not write on the cover or inside pages. 5. Use a piece of paper or cardboard as a bookmark. 6. Never tear or cut out any pictures or pages. 7. Repair any torn pages with paste or tape. 8. Pack the book carefully when you place it in your school bag. 9. Handle the book with care when passing it to another person. 10. When using a new book for the first time, lay it on its back. Open only a few pages at a time. Press lightly along the bound edge as you turn the pages. This will keep the cover in good condition.  CHEMISTRY STUDENT TEXTBOOK GRADE 10 Writers: Alemayehu Paulos (Ph.D.) Adelew Estifanos (Ph.D.) Editors: Tesfaye Semela (Ph.D.) (Curriculum Editor) Kenenisa Beresa (M.A.) (Language Editor) Ahmed Awel (M.Sc.) (Content Editor) Illustrator: Abinet Tilahun Nigusie (M.Sc.) Designer: Konno B. Hirbaye (M.Sc.) Evaluators: Amare Aregahegn Dubale (Ph.D.) (Reviewer) Nega Gichile (B.Sc., M.A.) (Evaluator) Sefiw Melesse (M.Sc.) (Evaluator) Tolessa Mergo (B.Sc., M.Sc.) (Evaluator) Federal Democratic Republic of Ethiopia Hawassa University Ministry of Education First Published August 2023 by the Federal Democratic Republic of Ethiopia, Ministry of Education, under the General Education Quality Improvement Program for Equity (GEQIP-E) supported by the World Bank, UK’s Department for International Development/DFID-now merged with the Foreign, Common wealth and Development Office/FCDO, Finland Ministry for Foreign Affairs, the Royal Norwegian Embassy, United Nations Children’s Fund/UNICEF), the Global Partnership for Education (GPE), and Danish Ministry of Foreign Affairs, through a Multi Donor Trust Fund. © 2023 by the Federal Democratic Republic of Ethiopia, Ministry of Education. All rights reserved. The moral rights of the author have been asserted. No part of this textbook reproduced, copied in a retrieval system or transmitted in any form or by any means including electronic, mechanical, magnetic, photocopying, recording or otherwise, without the prior written permission of the Ministry of Education or licensing in accordance with the Federal Democratic Republic of Ethiopia as expressed in the Federal Negarit Gazeta, Proclamation No. 410/2004 - Copyright and Neighboring Rights Protection. The Ministry of Education wishes to thank the many individuals, groups and other bodies involved – directly or indirectly – in publishing this Textbook. Special thanks are due to Hawassa University for their huge contribution in the development of this textbook in collaboration with Addis Ababa University, Bahir Dar University and Jimma University. Copyrighted materials used by permission of their owners. If you are the owner of copyrighted material not cited or improperly cited, please contact the Ministry of Education, Head Office, Arat Kilo, (P.O.Box 1367), Addis Ababa Ethiopia. Printed by: GRAVITY GROUP IND LLC 13th Industrial Area, Sharjah, UNITED ARAB EMIRATES Under Ministry of Education Contract no. : MOE/GEQIP-E/LICB/G-01/23 ISBN: 978-99990-0-017-8  CONTENTS 1 Unit : CHEMICAL REACTIONS AND STOICHIOMETRY 1 1.1 Introduction 2 1.2 Chemical Equations 3 1.3 Types of Chemical Reactions 11 1.4 Oxidation and Reduction Reactions 20 1.5 Molecular and Formula Masses, the Mole Concept and Chemical Formulas 28 1.6 Stoichiometry 34 Unit Summary 47 Review Exercises 48 Unit 2 : SOLLUTIONS 51 2.1 Heterogeneous and Homogeneous Mixtures 53 2.2 The Solution Process 59 2.3 Solubility as an Equilibrium Process 73 2.4 Ways of Expressing Concentration of Solution 81 2.5 Preparation of Solutions 94 2.6 Solution Stoichiometry 98 2.7 Describing Reactions in Solution 101 Unit Summary 103 Review Exercises 105 Unit 3 : IMPORTANT INORGANIC COMPOUNDS 109 Stomach Drain 3.1 Introduction 110 Battery Lemon Tomato Milk Blood Tablets Soap Cleaner 3.2 Oxides 111 Decreasing Increasing 3.3 Acids 122 Acidic Neutral Alkaline 3.4 Bases 140 Stomach Acid Vinegar Coffee Water Baking Ammonia Bleach 3.5 Salts 148 Soda Solution Unit Summary 162 Review Exercises 163 Unit 4 : ENERGY CHANGES AND ELECTROCHEMISTRY 166 4.1 Introduction 167 4.2 Energy Changes in Electrochemistry 174 4.3 Electrochemical Cells 182 4.4 Electrolysis 187 Unit Summary 194 Review Exercises 195 Unit 5 : METALS AND NONMETALS 198 5.1 Introduction 199 5.2 General Properties of Metals and Production of Some Metals 200 5.3 Production of Some Important Nonmetals 218 Unit Summary 230 Review Exercises 231 Unit 6 : HYDROCARBONS AND THEIR NATURAL SOURCES 234 6.1 Introduction 235 6.2 Saturated Hydrocarbons: Alkanes (CnH2n+2) 240 6.3 Unsaturated Hydrocarbons: Alkenes, Alkynes and Aromatic Hydrocarbons 262 6.4 Aromatic Hydrocarbons: Benzene 285 6.5 Natural Sources of Hydrocarbons 290 Unit Summary 295 Review Exercises 295 UNIT  CHEMICAL REACTIONS AND STOICHIOMETRY Unit Outcomes At the end of this unit, you will be able to " define the basics of chemical reaction; " describe the four major types of reactions; " develop skills in writing and balancing chemical equations; " define oxidation - reduction reactions " analyse redox reactions by specifying the oxidizing agent, the reducing agent, the substance reduced or oxidized; " explain molecular mass, formula mass, molar mass, empirical and molecular formulas; " determine molecular mass, formula mass, molar mass, empirical and molecular formulas " describe the mole concept " solve problems related to moles of substances " develop skills in solving problems based on chemical equations (mole- mole, mass - mass, volume - volume and mass - volume problems); " develop skills in determining the limiting reactant, theoretical yield, actual yield and percentage yield; " demonstrate scientific inquiry skills: observing, inferring, predicting, classifying comparing and contrasting, communicating, measuring, asking questions designing experiments, interpreting data, drawing conclusions, applying concepts, relating cause and effect and problem - solving. Chemistry Grade 10 Start-up Activity Discuss the following questions in groups of three or four students, then share your findings with the class. 1. Changes in our daily lives include fruit ripening, which involves the transformation of raw fruit (containing starting substance) into ripe fruit (containing new substance), and the fermentation of injera (new material) from teff flour (starting material). a. What are the differences between the raw and ripened fruit? b. What are the differences between the fermented injera and the teff flour? c. Are these changes reversible? Make a list of at least three more changes that occur in your day-to-day activities, and note the differences between the starting materials and new materials in each. 2. Are the following changes reversible? a. Dough is changed to bread through baking. b. Firewood changed to ash through burning. 3. Explain whether the following refer to the quality or quantity of substances. a. A farmer determines the amount of fertilizer for his farm. b. A nurse calculates the dose of a medicine for a patient. c. A chef mixes the right amounts of ingredients for cooking food. 1.1 Introduction At the end of this section, students will be able to " define chemical reaction; " give some examples of chemical reactions. " explain physical and chemical changes using examples Change is the law of nature. We observe various types of changes around us. The growth of seed in to a plant, the burning of wood in a fire place, the rusting of iron articles in moist air, the rotting of food and the evaporation of liquids are some of the changes that we see taking place around us. Scientists categorize changes as physical and chemical changes. A physical change is a change that does not involve 2  Chemical Reactions and Stoichiometry the formation of a new substance with a new chemical composition. Examples of physical changes are evaporation of liquids, powdering of sugar and melting of ice. The change which results in the formation of one or more new substances with new chemical composition is known as a chemical change. Examples of chemical changes are turning of milk into curd, photosynthesis by green plants and rotting of egg by bacteria. Chemical changes occur due to chemical reactions between substances. In general, a change is called a chemical reaction if it exhibits all or part of the following characteristics. i. Formation of new substances ii. Production of heat or light or both iii. Change in color iv. Change in temperature and etc. Chemical reaction is the process in which reacting substances, called reactants, are converted into new substances, called products. The characteristics of the products are completely different from those of the reactants. For example, if you burn sulphur with oxygen, the sulphur and oxygen are completely converted to sulphur dioxide. Sulphur dioxide is a colourless gas with a pungent odour. These characteristics of sulphur dioxide are completely different from the characteristics of the original substances, sulphur and oxygen. A chemical reaction is represented by a short hand notation called chemical equation as: Reactants Products A chemical equation uses chemical symbols to show what happens during a chemical reaction. A balanced chemical equation can be used to describe the relationships between the amounts of reactants and products. The quantitative study of reactants and products in a chemical reaction is called reaction’s stoichiometry. In this unit you will study, the types of chemical reactions, chemical equations, redox reactions, chemical formulas, masses of compounds, mole concepts, and reaction’s stoichiometry. Exercise 1.1 a. What is a chemical reaction? b. Explain the meanings of chemical and physical changes using examples. c. Give two examples of evidences for chemical reactions not mentioned in the text. 1.2 Chemical Equations At the end of this section, students will be able to " explain the conventions used to write chemical equation; " write the chemical equations of varieties of chemical reactions; 3 Chemistry Grade 10 " balance chemical equations using; The Inspection method, The Least Common Multiple (LCM) method, The Algebraic method. Form a group and discuss each of the following and present your conclusion to the rest of the class. 1. How can you show chemical symbols, physical states and quantities of substances involved in a chemical reaction? 2. What is the difference between a chemical equation and a chemical reaction? 3. Which law is applied when a chemical equation Activity 1.1 is balanced? A chemical equation is a shorthand representation of an actual chemical reaction in terms of chemical symbols and formulas. In a chemical equation the starting substances are called reactants; and the new substances produced are known as products. In a chemical equation, reactants are written on the left side and products on the right side of the equation. An arrow (→) is placed between the two sides to indicate transformation of reactants into products. Reactants → products 1.2.1 Writing Chemical Equation In writing chemical equation, instead of using words, chemical symbols and formulas are used to represent the reaction. Steps to write a chemical equation 1. Write a word equation: A word equation is stated in words. For example, the word equation for the reaction between hydrogen and nitrogen to produce ammonia is written as: Hydrogen + Nitrogen → Ammonia (Word equation) Note that we read the '+' sign as 'reacts with' and the arrow can be read as 'to produce', 'to form', 'to give' or 'to yield'. 2. Write the symbols and formulas for the reactants and products in the word equation. H2 + N2 → NH3 (Unbalanced equation) 3. Balance the equation 4  Chemical Reactions and Stoichiometry 3H2 + N2 → 2NH3 (Balanced equation) Chemically reactive substances might be solids, liquids, gases, or they can dissolve in a solvent. Particularly ionic compounds commonly react in aqueous solution, or dissolve in water. Sometimes this information is added to an equation by placing the appropriate symbols after the formulas: (s) - Solid (l) - Liquid (g) - Gas (aq) - Aqueous solution Example 1.1: Write the chemical equation for the reaction that occurs between calcium carbonate and sulfuric acid. 1. Word equation: Calcium carbonate + Sulphuric acid → Calcium sulphate + Water + Carbon dioxide 2. Chemical equation: CaCO3(s) + H2SO4(l) → CaSO4(s) + H2O(l) + CO2(g) 3. The equation is balanced Example 1.2: Write the chemical equation for the reaction of sodium chloride and silver nitrate. 1. Word equation: Silver nitrate + Sodium chloride → Silver chloride + Sodium nitrate 2. Chemical equation: AgNO3(aq) + NaCl (aq) → AgCl(s) + NaNO3 (aq) 3. The equation is balanced. Generally, any chemical equation must fulfill the following conditions: A. An equation must represent a true and possible chemical reaction. B. The symbols and formulas must be written correctly. The elements – hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine and iodine exist as diatomic molecules. For example, these elements should be written as molecules in the equation. The equation must be balanced. A chemical equation has both qualitative and quantitative meanings. Qualitatively, a chemical equation indicates the types of the reactants and products in the reaction. Quantitatively, a chemical equation expresses the relative number (amount) of moles, molecules or masses of the reactants and products. For example, the balanced chemical equation : 2H2 + O2 → 2H2O has both qualitative and quantitative meaning. Qualitatively, it shows that the reaction between hydrogen and oxygen forms water. Quantitatively, it indicates that two units of hydrogen combine with one unit of oxygen to form two units of water. The quantities may be expressed in terms of grams, number of moles or number of molecules ( see for more details in subsection 1.6.1). 5 Chemistry Grade 10 1.2.2 Balancing Chemical Equations At the end of this section, students will be able to " balance chemical equations using: * The Inspection method * The Least Common Multiple (LCM) method * The Algebraic method According to the Law of Conservation of Mass, atoms are neither created nor destroyed during a chemical reaction (refer to grade 9 chemistry). As a result, the number of atoms of each element should remain the same before and after the reaction. To balance a chemical equation means to equalize the number of atoms on both sides of the equation by putting appropriate coefficients in front of the formulas. A. Balancing Chemical Equations by Inspection Method The most common way to balance chemical equations is to use inspection method. It is exactly what the name states. You balance the equation by inspecting it. It is also known as a trial and error method or a hit and trial method. Follow four easy steps, given below, to balance a chemical equation by inspection method: 1. Write the word equation. 2. Write the correct symbols or formulas for the reactants and products (unbalanced equation). 3. Start with the most complicated compound count and tabulate the number of each type of atoms on the two sides of the unbalanced equation. 4. Make the number of each type of atoms on the left side equal to the number of corresponding atoms on the right side of the equation. Example 1.3: Balance the chemical reaction that takes place between iron and water to form iron (IV) oxide and hydrogen gas by the inspection method. Step 1: Iron + Water → Iron (IV) oxide + Hydrogen (word equation) Step 2: Fe + H2O → Fe3O4 + H2 (Unbalanced equation) Let us count and tabulate the number of various types of atoms on the reactant and product sides of the expression Fe + H2O → Fe3O4 + H2 Atom Reactant side Product side Equation Number of Fe atoms 1 3 Number of O atoms 1 4 Number of H atoms 2 2 Total no. of atoms 4 9 Unbalanced 6  Chemical Reactions and Stoichiometry Balancing Fe atoms: There is one Fe atom on the left side while there are three Fe atoms on the right side. Therefore, a suitable coefficient of Fe on the left side is 3. Thus, 3Fe + H2O → Fe3O4 + H2 (2) Balancing O atoms. There is one O atom (in H2O) on the left side and there are four O atoms on the right side. Therefore, a proper coefficient of H2O is 4 as H2O. 3Fe + 4H2O → Fe3O4 + H2 (3) Balancing H atoms. There are eight H atoms (in 4H2O) on the left side, but only two H atoms (in H2) on the right side. Therefore, an appropriate coefficient of H is 4 as 4H2. Thus, 3Fe + 4H2O → Fe3O4 + 4H2 (4) Equation 4 is a balanced chemical equation. Note that number of atoms of each element is conserved (no. of atoms reactant side = no. of atoms product side). How? These are shown below: Atom Reactant side Product side Equation Number of Fe atoms 3 3 Number of O atoms 4 4 Number of H atoms 8 8 Total number of atoms 15 15 Balanced NB: Inspection method works best for simple equations. B. Balancing chemical equations by the LCM method By taking the LCM of the total valency of the reactants and products and dividing it by the total valency of the reactants and products, the coefficients for the balanced chemical equation are derived. This is known as The LCM method. The following examples show all of the steps required to balance a chemical equation using the LCM Method. Examples: i. When aluminum reacts with oxygen, aluminum oxide is formed. Write the balanced chemical equation for the reaction Step 1: Represent the reaction by a word equation: Aluminum + Oxygen → Aluminum Oxide Step 2: Write the correct symbols or formulas for the reactants and products Al + O2 → Al2O3 Step 3: Place the total valency of each atom above it. 7 Chemistry Grade 10 3 4 6 6 Al + O2 → Al2O3 From the equation we see that the valency of aluminum is 3. The total valency of oxygen is 2 × 2 = 4. The total valency of aluminum in Al2O3 is 3 × 2 = 6. The total valency of oxygen in Al2O3 is 2 × 3 = 6, LCM is 12. Step 4: Find the LCM of each total valency and place it above the arrow, here LCM is 12. 3 4 12 6 6 Al + O2 → Al2O3 LCM Step 5: Divide the LCM by each total valency number to obtain the coefficients for each of the reactants and products. Place the obtained coefficients in front of the respective chemical formulas 4Al + 3O2 → 2Al2O3 (Balanced) Checking: There are 4 aluminum and 6 oxygen atoms on both sides of the equation. Hence, the chemical equation is correctly balanced ii. The reaction of sodium sulphate with aluminum nitrate would form aluminum sulphate and sodium nitrate. Write the balanced chemical equation. Step 1: Sodium sulphate + Aluminum nitrate → Aluminum sulphate + Sodium nitrate (word equation) Step 2: Na2SO4 + Al(NO3)3 → Al2(SO4)3 + NaNO3 2 2 3 3 6 6 1 1 Step 3: Na 2SO4 + Al(NO3)3 → Al2(SO4)3 + NaNO3 2 2 3 3 6 6 6 1 1 Step 4: Na2SO4 + Al(NO3)3 → Al2(SO4)3 + NaNO3 LCM Step 5: 3Na 2SO4 + 2Al(NO3)3 → Al2(SO4)3 + 6NaNO3 (Balanced) There are 6 nitrogen, 6 sodium, 3 sulphur, 30 oxygen and 2 aluminum atoms on both sides of the equation. Thus, the equation is correctly balanced. iii. The reaction of magnesium nitrate with ammonium phosphate would form ammonium nitrate and magnesium phosphate. Write the balanced equation. Step 1: Magnesium nitrate + Ammonium phosphate → Magnesium phosphate + Ammonium nitrate Step 2: Mg(NO3)2 + (NH4)3PO4 → NH4NO3 + Mg3(PO4)2 2 2 3 3 1 1 3 3 Step 3: Mg(NO3)2 + (NH4)3PO4 → NH4NO3 + Mg3(PO4)2 2 2 3 3 6 1 1 3 3 Step 4: Mg(NO3)2 + (NH4)3PO4 → NH4NO3 + Mg3(PO4)2 LCM 8  Chemical Reactions and Stoichiometry Step 5: 3 Mg(NO3)2 + 2 (NH4)3PO4 → 6NH4NO3 + Mg3(PO4)2 (Balanced) There are 12 nitrogen, 24 hydrogen, 3 magnesium, 2 phosphorous and 26 oxygen atoms on both sides of the equation. Thus, the equation is correctly balanced. B. Balancing Chemical Equations Using Algebraic Method This method of balancing chemical equations involves assigning algebraic variables as stoichiometric coefficients to each species in the unbalanced chemical equation. These variables are used in mathematical equations and are solved to obtain the values of each stoichiometric coefficient. Consider the formation of ammonia from hydrogen and nitrogen as an example. Step 1: Write the unbalanced equation with the correct symbols of the reactants & products: N2 + H2 → NH3 (unbalanced equation) Step 2: Assign algebraic variables to each species as coefficients (a, b, c) in the unbalanced equation: aN2 + bH2 → cNH3 Then, set equations for each element so that it is equal in both the right and left hand side. N: 2a = c (subscript of N is 2 in the left side and 1 in the right side) H: 2b = 3c (subscript of H is 2 in the left side and 3 in the right side) Step 3: Choose the smallest variable and assign arbitrary number in order to determine the remaining variables. In the above case, a is the smallest coefficient. Assuming a = 1, the values of b and c can be obtained as follows. c = 2 x 1= 2, 2b = 3c = 3 x 2 = 6; b = 6/2 = 3 Since a, b, and c have no common multiples, they can be substituted into the equation as follows. N2 + 3H2 → 2NH3 ( balanced equation) Example 1.4: Balance the reaction: Al + O2 → Al2O3 using the algebraic method. Assign variables for each element aAl + bO2 → cAl2O3 The equation for Aluminum: a = 2c (subscript of Al is 2 in the left side and 2 in the right side). The equation for oxygen: 2b = 3c (subscript of O is 2 in the left side and 3 in the right side) 9 Chemistry Grade 10 Assuming a = 1, we get: c = a/2; c = 1/2 2b = 3 x (1/2) = 3/2 ; b = 3/4 Since fractional values of b and c are obtained, the lowest common multiple denominator between the variables a, b, and c must be found and multiplied with each variable. Since the lowest common multiple denominator is 4, each of the variables must be multiplied by 4. Therefore, a = 4 x 1 = 4; b = (3/4) x 4 = 3; c = (1/2) x 4 = 2 Substituting the values of a, b, and c in the unbalanced equation, the following balanced chemical equation is obtained. 4Al + 3O2 → 2Al2O3 Example 1.5: Balance the equation: PCl5 + H2O → H3PO4 + HCl Step 1: PCl5 + H2O → H3PO4 + HCl Step 2: aPCl5 + bH2O → cH3PO4 + dHCl P: a = c (subscript of P is 1 in the left side and 1 in the right side) Cl : 5a = d (subscript of Cl is 5 in the left side and 1 in the right side) H: 2b = 3c + d (subscript of H is 2 in the left side and 3 in H3PO4 in 1 in HCl) O : b = 4c (subscript of O is 1 in the left side and 4 in the right side) Assume a =1 , therefore c=1, 5a = d ⇒ d =5, b =4. Substituting the values of a, b, c and d in the unbalanced equation: PCl5 + 4H2O → H3PO4 + 5HCl (balanced) Exercise 1.2 1. Balance the following chemical equations, using the inspection method: a. Na + H2O → NaOH + H2 b. KClO3 → KCl + O2 c. H2O2 → H2O + O2 d. Al + H3PO4 → AlPO4 + H2 e. HNO3 + H2S → NO + S + H2O 2. Write the balanced chemical equation to represent the following reactions. a. Sodium bromide reacts with chlorine to form sodium chloride and bromine 10  Chemical Reactions and Stoichiometry b. Hydrochloric acid reacts with sodium carbonate to form sodium chloride, water and carbondioxide c. Potassium chlorate when heated produces potassium chloride and oxygen d. Calcium carbonate reacts with hydrochloric acid to form water, carbon dioxide and calcium chloride e. Silver oxide decomposes to silver and oxygen gas. 3. Balance the following equations using the LCM method. a. Al + HCl → AlCl3 + H2 b. Fe2(SO4)3 + KOH → K2SO4 + Fe(OH)3 c. CaCl2 + Na3PO4 → Ca3(PO4)2 + NaCl d. FeCl3 + NH4OH → Fe(OH)3 + NH4Cl 4. Balance the following equations using the algebraic method. a. PCl5 + H2O → H3PO4 + HCl b. Mg + H2O → Mg(OH)2 + H2 c. Zn(NO3)2 → ZnO + NO2 + O2 d. H2SO4 + NaOH → Na2SO4 + H2O e. NH3 + O2 → NO + H2O f. C6H12O6 + O2 → CO2 + H2O g. FeCl3 + MgO → Fe2O3 + MgCl2 h. BaCl2 + K3PO4 → Ba3(PO4)2 + 6KCl i. P4O10 + H2O → H3PO4 1.3 Types of Chemical Reactions At the end of this section, students will be able to " list the four types of chemical reactions; " define combination reaction and give examples; " conduct experiments on combination reactions in groups; " define decomposition reaction and give examples; " conduct some experiments on decomposition reactions in groups; " define single - displacement reaction and give examples; " conduct some experiments on simple displacement reactions in groups; " define double - displacement reaction and give examples; " conduct some experiments on double - displacement reactions in group. 11 Chemistry Grade 10 In your group, discuss about the following processes and share your conclusion with the rest of the class. 1. What are the final products of digestive protein and starch? Is protein and starch digestion a building up or breaking down process ? 2. What is alcohol fermentation and its process? What are the two important chemical compounds produced during the fermentation of “Teji” or “Tella”? What is the main ingredient (starting material) for the production of “Teji” and “Tella”? 3. What are the two major chemical reactants and the two chemical products of photosynthesis? Is photosynthesis a breaking down or a building up process? Compare this process with cellular respiration. 4. What are the combustion products of kerosene? Is the combustion of kerosene a building up or a breaking down process? Activity 1.2 Different elements and compounds react in different ways to produce different kinds of new substances. Many chemical reactions can be classified as one of the four basic types of reactions. By analyzing the reactants and products of a particular reaction, you can classify them into one of these categories. Some reactions may be categorized into more than one category. A complete understanding of these reaction types will help predict the products of unknown reactions. The four basic types of chemical reactions are combinations, decompositions, single displacement, and double displacement reactions, which are described in the next section. Such reactions are generally classified as shown in Figure 1.1. Types of Chemical Reactions Direct combination Decomposition Single - displacement Double - displacement A + B → AB AB → A + B A + BC → AB + C AB + CD → AD + BC Figure 1.1 Classification of chemical reactions. 12  Chemical Reactions and Stoichiometry A. Direct Combination Reactions Combination reactions are those reactions in which two types of pure substances react directly and form a single substance. In a combination reaction, two elements, two compounds, or an element and a compound react to form a single compound. In general, a combination reaction may be represented by the equation. A + B → AB where the reactants A and B are elements or compounds, the product AB is a compound. Such type of reaction is also known as synthesis or composition reaction. Examples 2Na + Cl2 → 2NaCl Element + Element Compound CaO + CO2 → CaCO3 Compound + Compound Compound Note that in the above reactions, there is a single product. Experiment 1.1 Investigation of Combination Reaction Objective: To investigate the reaction between sulphur and iron. Apparatus: Test tube, stand, burner, watch glass Chemicals: Sulphur powder, iron filings Procedure: 1. Mix about 3 g of iron filings and 2 g of powdered sulphur in a watch glass. 2. Transfer the mixture in a glass test tube. 3. Mount the test tube in a sloping position on a stand as shown in Figure 1.2. 4. Heat the test tube until the mixture in the glass glows red hot. 5. Remove the test tube from the flame and observe the result. 13 Chemistry Grade 10 Iron fillings + sulphur Bunsen burner Figure 1.2 The reaction between iron and sulphur. Questions 1. What were the colors of iron filings and sulphur before the reaction? 2. What was the color of the resulting compound after the reaction? 3. Write a balanced chemical equation for the reaction. B. Decomposition Reactions A decomposition reaction (or analysis) is a reaction that involves the breaking down of a single compound into two or more elements or simpler compounds. A decomposition reaction can be carried out using heat (thermal decomposition), light (photo decomposition), electricity (electrical decomposition) or a catalyst. But most decomposition reactions are carried out when heat is supplied and this heat energy is indicated by a ‘delta’ (Δ) symbol above the arrow. The general form of equation for a decomposition reaction is: AB → A + B where the reactant AB must be a compound and the products A and B could be elements or compounds. The simplest kind of decomposition reaction is when a binary compound decomposes into its elements. For example, mercury (II) oxide, a red solid, decomposes when heated to produce mercury and oxygen gas, similarly a colorless potassium perchlorate decomposes to potassium chloride and oxygen gas. Δ 2HgO(s) 2Hg(l) + O2(g) Δ 2KClO3(s) 2KCl(s) + 3O2(g) 14  Chemical Reactions and Stoichiometry Experiment 1.2 Investigation of Decomposition Reaction Objective: To investigate the decomposition of copper (II) carbonate(CuCO3). Apparatus: Test tube, stand, burner, cork, delivery tube. Chemicals: CuCO3 and lime water. Procedure: Put 1g of CuCO3 powder in a glass test-tube. Mount the test tube in a sloping position on a stand as shown in Figure 1.3. Fit a cork and a delivery tube to the test tube. Put another test tube containing lime water at the end of the delivery tube. Heat the CuCO3 with a burner. Delivery tube Stand Lime water Clamp Copper(II) carbonate Bunsen burner Figure 1.3 Decomposition of copper (II) carbonate. Questions 1. What was the color of copper (II) carbonate before heating? 2. What was the color during heating and after cooling? 3. What change did you observe in the lime water? 4. Write a balanced chemical equation for the reaction. C. Single Displacement Reactions A reaction in which one element displaces another element from its compound is known as single displacement or replacement reaction. Such a reaction is represented by the following two general forms. 15 Chemistry Grade 10 A + BC → B + AC If A is a metal, it will displace B to form AC, provided A is a more active metal than B. A + BC → BA + C If A is a non-metal, it will displace C to form BA, provided A is a more active non- metal than C. In general, a more reactive element displaces a less reactive element from a compound. Examples of single-displacement reactions 1. Active metals displace hydrogen from acids Reactive metals such as potassium, calcium, sodium, and zinc displace hydrogen gas from dilute acids. For example, zinc is an active metal, and it displaces hydrogen from hydrochloric acid; but copper metal cannot do so. Zn + 2HCl → ZnCl2 + H2 Cu + HCl → No reaction 2. Reactive metals, such as potassium, calcium, and sodium react vigorously with water to displace hydrogen: 2Na + 2H2O → 2NaOH + H2 Ca + 2H2O → Ca(OH)2 + H2 3. A more active metal displaces a less active metal Zinc displaces copper from copper (II) sulphate solutions Zn + CuSO4 → ZnSO4 + Cu Iron displaces copper from copper (II) sulphate solutions Fe + CuSO4 → FeSO4 + Cu after sometime Cu deposits on Iron nail the iron nail CuSO4 solution FeSO4 solution Figure 1.4 Replacement of copper by iron metal from copper sulphate solution. 16  Chemical Reactions and Stoichiometry Experiment 1.3 Investigation of Single Displacement Reaction Objective: To investigate the displacement reaction between iron and copper (II) Sulphate (CuSO4). Apparatus: Iron rod and beaker Chemicals: CuSO4. Procedure: 1. Clean a piece of iron rod or iron knife with emery paper to remove any rust. 2. Take saturated CuSO4 solution in a beaker. 3. Dip the iron rod into the CuSO4 solution as shown in Figure 1.5 and wait for a few minutes. What did you observe on the iron rod? 4. Allow the reactants to stand for one day and observe any change on the iron rod. Iron rod Questions 1. What did you observe on the iron rod after one day? 2. Write a balanced Beaker chemical equation for the reaction. 3. Write the conclusion for CuSO4 Solution the experiment. Figure 1.5 Reaction between iron and copper (II) sulphate. Exercise 1.3 Complete the following single displacement reactions: 4. Cl2 + KI → 1. Zn + H2SO4 → 5. F2 + CaCl2→ 2. Zn + Cu(NO3)2 → 3. Cu + AgNO3 → 17 Chemistry Grade 10 D. Double Displacement Reactions A double displacement reaction is a reaction in which two compounds react together to form two new compounds by exchange of the positive and negative ions of each reactant. Such a reaction is also known as double replacement reaction or metathesis. This type of reaction can be written in the following general form of equation: AB + CD → AD + CB Experiment 1.4 Investigation of Double Displacement Reaction Objective: To observe the displacement reaction between Na2SO4 and Ba(NO3)2. Apparatus: Beaker, stirrer, filter paper, filter funnel. Chemicals: Na2SO4, Ba(NO3)2 and water. Procedure: 1. Take solution of Ba(NO3)2 into a beaker and add drop-wise Na2SO4 solution as shown in Figure 1.6. Then stir it continuously. 2. Filter the precipitate using a filter paper and funnel. Collect the filtrate or the solution in a clean beaker. Questions 1. Write the names of the compounds that are formed as a precipitate and as solution at the end of the reaction. 2. What was the colour of the precipitate. 3. Write the balanced chemical equation for the reaction. Na2SO4 solution Stirrer Ba(NO3)2 solution Figure 1.6 Double displacement reaction between Na2SO4 and Ba(NO3)2 18  Chemical Reactions and Stoichiometry Example 1.6: The two soluble compounds AgNO3 and NaCl react to produce an in soluble precipitate of AgCl and a soluble NaNO3 solution. AgNO3 + NaCl → AgCl + NaNO3 Soluble Soluble Insoluble Soluble When aqueous solutions of BaCl2 and Na2SO4 react, a precipitate of BaSO4 is formed. BaCl2 + Na2SO4 → BaSO4 ↓ + 2NaCl Soluble Soluble Insoluble Soluble When dilute hydrochloric acid is added to sodium carbonate double displacement reaction takes place. In this reaction carbondioxide gas is produced. Na2CO3 + 2HCl → 2NaCl + H2O + CO2 Exercise 1.4 Give the correct answers for the following questions. 1. What type of reaction does usually take place in each of the following reactions? a. An active metal reacting with water b. A metal reacting with a non-metal c. An acid reacting with a metal hydroxide d. Heating of a metal hydrogen carbonate 2. Classify the following reactions as combination, decomposition, single or double displacement reactions. a. FeO + C → Fe + CO b. 2NH3 + H2SO4 → (NH4)2SO4 c. CaCO3 + 2HCl →CaCl2 + CO2 + H2O d. 2Cu(NO3)2 → 2CuO + 4NO2 + O2 e. 2Na3PO4 + 3Ca(OH)2 → Ca3(PO4)2 + 6NaOH f. CuSO4.5H2O → CuSO4 + 5H2O 19 Chemistry Grade 10 1.4 Oxidation and Reduction Reactions Form a group and discuss the following phenomenon and present your conclusion to the class. When you dry your meal dishes with a towel, the towel can be termed as a drying agent and the dish as wetting agent. Assume that the meal dishes were wet with water, which one would gain water, the towel or the wetting agent and which one would lose the water, the wetting agent or the towel, after cleaning? Relate this idea with oxidizing agent and reducing agent, oxidation and reduction of substances. Activity 1.3 At the end of this section, students will be able to " define redox reactions; " define the terms oxidation and reduction in terms of electron transfer; " define oxidation number (oxidation state); " state oxidation number rules.; " determine the oxidation number of an element in a given formula. In our day to day activity, we are familiar with the chemical processes like rusting of iron, burning of substances, respiration, digestion of food and so on. All such types of processes or reactions are known as oxidation and reduction or redox reactions. Additionally, most metallic and nonmetallic elements are obtained from their ores by the process of oxidation or reduction. What are oxidation and reduction reactions? Oxidation-Reduction An oxidation is defined as the loss of one or more electrons by an atom, and a reduction is the gain of one or more electrons. Oxidation and reduction occur simultaneously. Thus, an oxidation reduction reaction, or redox reaction, is one in which electrons are transferred from one atom to another. For example, when metallic zinc is added to a solution containing copper(II) sulphate (CuSO4), zinc reduces Cu2+ by donating two electrons to it. Therefore, Zn is oxidized and copper is reduced and oxidation – reduction or redox reaction takes place as shown in the following chemical equation: Zn + CuSO4 → ZnSO4 + Cu Zn + Cu2+ → Zn2+ + Cu Oxidation and reduction reactions can also be defined in terms of oxidation number. 20  Chemical Reactions and Stoichiometry Oxidation is an increase in the oxidation number of an element and reduction is a decrease in the oxidation number. For example, Cu0 + 2Ag+ → Cu2+ + 2Ag0 The oxidation number of copper is increased from 0 to +2 and thus copper is oxidized. The oxidation number of silver is decreased from +1 to 0, and therefore silver is reduced. 1.4.1 Oxidation Number or Oxidation State At the end of this section, students will be able to " define oxidation number (oxidation state); " state oxidation number rules.; " determine the oxidation number of an element in a given formula. Oxidation number or oxidation state is the number of electrons that an atom appears to have gained or lost when it is combined with other atoms. Oxidation number could be integers including zero and fractional numbers. Rules for Assigning Oxidation Numbers Rule 1: The oxidation state of an uncombined element is zero. This applies for poly atomic molecules, S8, P4 and large structures of carbon or silicon each have an oxidation state of zero. Example 1.7: The oxidation number of Be = 0, Cu = 0, Br in Br2 = 0, O in O3 = 0, S in S8 = 0. P in P4 = 0, O in O2 = 0 Rule 2: The oxidation number of a monatomic ion is equal to the charge on the ion. Example 1.8: Na+ = +1, Mg2+ = +2, S2– = –2. Rule 3: The oxidation number of oxygen in a compound is usually –2 except in the following cases: Exceptions The oxidation number of oxygen in: A. peroxides is –1. Example: Na2O2 B. superoxides is –1/2. Example: KO2 C. oxygen diflouride is +2 Example: OF2 Rule 4: The oxidation number of hydrogen in its entire compounds is +1 except in metal hydrides, (like NaH, CaH2 and AlH3), where its oxidation number is –1. Rule 5: The sum of the oxidation number of all the atoms in a neutral compound is zero. 21 Chemistry Grade 10 Example 1.9: +1 +6 -2 H2SO4, (+2) + (+6) + (–8) = 0 Rule 6: In a polyatomic ion, the sum of the oxidation numbers of the constituent atoms equals the charge on the ion. Example 1.10: +6 –2 (SO4 )-2, (+6) + (–8) = –2 Rule 7: Elements of group IA have +1 and group IIA have +2 oxidation states in all of their compounds Rule 8: In a compound, the more electronegative element is assigned a negative oxidation number, and the less electronegative element is assigned a positive oxidation number. Example 1.11: Assign oxidation number to all elements in i. HNO3 ii. Cr2O72− iii. MnO4− iv. Ca(H2PO4)2. i. HNO3: According to rule 4 oxidation number of H = +1. Thus the other group (the nitrate ion) must have a net oxidation number of −1. Oxygen has an oxidation number of −2, and if we use x to represent the oxidation number of nitrogen, then the nitrate ion can be written as [N(x)O3(2−)]− or x + –6 = –1 ⇒ x = +5 ii. Cr2O72−: From rule 6 we see that the sum of the oxidation numbers in the dichromate ion Cr2O72− must be −2. We know that the oxidation number of O is −2, so what remains is to determine the oxidation number of Cr, which we call y. The dichromate ion can be written as [Cr2(y) O7(2−)]2−, 2y + 7 × –2 = –2 ⇒ y= +6 iii. MnO4−: Let the oxidation number of Mn be x. MnxO42−. The sum of the oxidation numbers of Mn and O in MnO–4 is –1 (Rule 6) x + (–2 × 4) = –1, x – 8 = –1 ⇒ x = +7 Therefore, the oxidation number of Mn in MnO4– is +7. iv. Ca(H2PO4)2: The oxidation number of Ca is +2. Let, the oxidation number of P be x. 22  Chemical Reactions and Stoichiometry +2 +1 x -2 Ca(H2PO4)2 +2 + (4 × (+1)) + (2 × x) + (8 × (–2)) = 0 2 + 4 + 2x – 16 = 0 2x –10 = 0 or x = +5 Hence, the oxidation number of P in Ca(H2PO4)2 is +5. Exercise 1.5 1. Find the oxidation numbers to the underlined species for the following compounds or ions. a. K4[Fe(CN)6] d. Na2S4O6 g. H2SO4 b. K2Cr2O7 e. S2O8 –2 h. HAuCl4 c. H2PtCl6 f. H2P2O7 2– i. Fe2(SO4)3 2. Determine whether the following processes are oxidation or reduction reactions a. Cu2++ 2e– → Cu d. S2–→ S + 2e– b. K → K + e + – e. Fe2+ → Fe3+ + e– c. O + 2e– → O2– f. N + 3e– → N3– 1.4.2 Oxidizing and Reducing Agents At the end of this section, students will be able to " define the terms reducing and oxidizing agent; " identify reducing and oxidizing agents from a given redox reaction; " compare and contrast oxidizing and reducing agents. Whenever one substance loses an electron (is oxidized), another substance must gain that electron (be reduced). The substance that gives up an electron and causes reduction is called a reducing agent. The substance that gains an electron and causes the oxidation is called an oxidizing agent. The following comparison shows the characteristics of reducing and oxidizing agents. Reducing agent Oxidizing agent 1. Loses one or more electrons 1. Gains one or more electrons 2. Causes reduction 2. Causes oxidation 3. Undergoes oxidation 3. Undergoes reduction 4. Becomes more positive 4. Becomes more negative Tests for an oxidizing agent are accomplished by mixing it with a substance that is easily oxidized to give a visible color change when the reaction takes place. 23 Chemistry Grade 10 Example 1.12: I. Permanganate ion (MnO4–) in acidic solution changes color from purple to colorless MnO4– → Mn2+ Figures 1.7 The reduction of Mn7+ to Mn2+ in acidic media. II. Dichromate in acidic solution changes color from orange to green Cr2O72– → Cr3+ Figures 1.8 The reduction of Cr6+ to Cr3+ in acidic media. Other common oxidizing agents are chlorine, potassium chromate, sodium chlorate and manganese (IV) oxide. Similarly, certain reducing agents undergo a visible colour change with a substance which is easily reduced. I. A moist starch solution changes potassium iodide paper to blue-black to show that iodine is formed, 2I– → I2. That is potassium iodide is a reducing agent. II. Hydrogen sulphide bubbled through a solution of an oxidizing agent forms a yellow precipitate, S2– → S. Thus, H2S is a reducing agent. Other common reducing agents are carbon, carbon monoxide, sodium thiosulphate, sodium sulphite and iron (II) salts. The oxidizing or reducing ability of substances depend on many factors. Some of these are: Electronegativity: Elements with high electronegativity such as F2, O2, N2 and Cl2 are good oxidizing agents. Elements with low electronegativity for example, metallic 24  Chemical Reactions and Stoichiometry elements like Na, K, Mg and Al are good reducing agents. Oxidation states: In a compound or ion, if one of its elements is in its higher oxidation state, then it is an oxidizing agent. Similarly, if an element of a compound or ion is in its lower oxidation state, then it is a reducing agent. Example 1.13: Oxidizing agents KMn+7O4, NaCl+7O4, K2Cr2+6O7 Reducing agents Fe+2S, C+2O, Na2S+4O3 1.4.3 Analyzing Redox Reactions At the end of this section, students will be able to explain why every reduction reaction is accompained by oxidation reaction. Oxidation and reduction or redox reactions occur simultaneously in a given reaction. Form a group and discuss the following and present your conclusion to the class. 1. How can you identify an oxidizing agent and reducing agent in a reaction? 2. Why a reducing agent undergoes oxidation and an oxidizing agent undergoes reduction? 3. Why must every redox reaction involve both an Activity 1.4 oxidizing agent and a reducing agent? Example 1.14: Identify the reducing and oxidizing agents in the following balanced redox reactions: 2Fe + 3O2 → Fe2O3 Assign oxidation states of the reactants and products, then identify the species oxidized and reduced. 0 0 +3 –2 2Fe + 3O2 → 2Fe2O3 Fe is oxidized from 0 to +3 O is reduced from 0 to –2 Therefore, Fe is reducing agent and O2 is oxidizing agent. Mg + 2HCl → MgCl2 + H2 Mg0 + 2H+1Cl-1 → Mg+2Cl2– + H20 Mg is oxidized from 0 to +2, therefore Mg is reducing agent. H is reduced from +1 to 0, therefore HCl is the oxidizing agent. 25 Chemistry Grade 10 Exercise 1.6 1. Where do the most easily reduced and oxidized elements found in the periodic table of the elements? 2. For the following substances, tell whether the oxidation number increases or decreases in a redox reaction: a. An oxidizing agent b. A reducing agent c. A substance undergoing oxidation d. A substance undergoing reduction 3. In the following reactions, label the oxidizing agent and the reducing agent. a. ZnO + C → Zn + CO b. 8Fe + S8 → 8FeS c. Fe2O3 + 3CO → 2Fe + 3CO2 d. PbS + 4H2O2 → PbSO4 + 4H2O 1.4.4 Balancing Redox Reactions: Oxidation-Number-Change Method At the end of this section, students will be able to balance redox reactions using oxidation change method.. One way to balance redox reactions is by keeping track electron transfer, by using the oxidation numbers of each of the atoms. It is based on the difference in oxidation number of oxidizing agent and the reducing agent. For the oxidation-number-change method, start with the unbalanced equation. The example below is for the reaction of iron (III) oxide with carbon monoxide. Fe2O3 + CO → Fe + CO2 Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom. +3 -2 +2 -2 0 +4 -2 Fe2O3 + CO → Fe + CO2 Step 2: Identify the atoms that are oxidized and those that are reduced. In the above equation, the C atom is being oxidized from +2 to +4. The Fe atom is being reduced from +3 to 0. Step 3: Use a line to connect the atoms that are undergoing a change in oxidation number. On the line, write the oxidation-number change. 26  Chemical Reactions and Stoichiometry -3 ← ← +3 -2 +2 -2 0 +4 -2 Fe2O3 + CO → Fe + CO2 ← ← +2 Step 4: Use coefficients to make the total increase in oxidation number equal to the total decrease in oxidation number. In this case, the least common multiple of 2 and 3 is 6. So the oxidation-number increase should be multiplied by 3, while the oxidation- number decrease should be multiplied by 2. The coefficient is also applied to the formulas in the equation. So 3 is placed in front of the CO and in front of the CO2. 2 is placed in front of the Fe on the right side of the equation. The Fe2O3 does not require a coefficient because the subscript of 2 after the Fe indicates that there are already two iron atoms. - 3*2=6 ← ← +3 -2 +2 -2 0 +4 -2 Fe2O3 + CO → Fe + CO2 ← ← +2*3=6 Step 5: Check whether each element is balanced. Occasionally, a coefficient may need to be placed in front of a molecular formula that was not involved in the redox process. In the current example, the equation is now balanced. Fe2O3 + 3CO → 2Fe + 3CO2 Example 1.15: Balance the chemical equation: HNO₃ + H₃AsO₃ → NO + H₃AsO₄ + H₂O 1. Identify the oxidation number of each atom. Reactant side: H= +1; N= +5; O = -2; As= +3 Product side: N= +2; O = -2; H= +1; As= +5 2. Determine the change in oxidation number for each atom that changes. N: +5 → +2; Change = -3 As: +3 → +5; Change = +2 3. Make the total increase in oxidation number equal to the total decrease in oxidation number. We need 2 atoms of N for every 3 atoms of As. This gives us total changes of -6 and +6. 4. Place these numbers (2 & 3) as coefficients in front of the formulas containing those atoms 2HNO₃ + 3H₃AsO₃ → 2NO + 3H₃AsO₄ + H₂O (balanced) 27 Chemistry Grade 10 Exercise 1.7 Balance the following using oxidation number change method a. Al + H2SO4 → Al2(SO4)3 + H2 b. KClO₃ → KCl + O₂ c. MnO₂ + Al → Mn + Al₂O₃ d. Cu + H2SO4 → CuSO4 + SO2 + H2O Non-redox Reactions At the end of this section, students will be able to identify redox and non redox reactions. What are non redox reactions? non redox reactions are chemical reactions where the oxidation states of chemical elements remain unchanged in reactants and products. Therefore, reactions in which neither oxidation nor reduction takes place or no species either gains or loses electrons are non-redox reactions. Neutralization and double displacement reactions are examples of non-redox reactions. Example 1.16: Na2SO4 + CaCl2 → CaSO4 + 2NaCl CaCO3 → CaO+CO2 KOH + HNO3 → KNO3 + H2O 1.5 Molecular and Formula Masses, the Mole Concept and Chemical Formulas Discuss the following in your group and present your conclusion to the class. 1. Is it possible to count 'teff' seeds for practical purposes? If not, why? How do you quantify 'teff' for practical purposes such as in markets? 2. Molecules, atoms or ions are extremely small entities. How can we express their quantities for practical purposes such as in the laboratory reactions and in Activity 1.5 industrial applications? 28  Chemical Reactions and Stoichiometry At the end of this section, students will be able to " explain and determine molecular mass, formula mass, molar mass, empirical and molecular formulas; " describe the mole concept and solve problems related to moles of substances. 1.5.1 Molecular mass (MM) and Formula Mass (FM) At the end of this topic, students will be able to explain and determine molecular mass, formula mass. The molecular mass or formula mass of a compound is used to calculate the percentage of its constituents. For example, the molecular mass of H2O is used to determine the percentage of hydrogen, and the formula mass of NaCl is used to determine the percentage of sodium. What is molecular mass or formula mass? Molecular Mass (MM) is the sum of masses of all the atoms present in a molecule. Example 1.17: MM of H2O = 2 x 1 amu (2 hydrogen atoms) + 16 amu (1 oxygen atom) = 18 amu, MM of CO2 =12 amu (1carbon atom) +2 x 16 amu (2 oxygen atoms) = 44 amu Ionic compounds are not comprised of discrete molecules, but rather, ions. The smallest unit of an ionic compound that retains the identity of an ionic compound is called a formula unit. The mass of a formula unit is called its formula mass. Fomula Mass ( FM) is the sum of the atomic masses of all atoms present in the formula unit of the compound, whether it is molecular or ionic. But FM is used mostly for ionic compounds. Example 1.18: FM of NaCl = 23 amu (1Na) + 35.5 amu (Cl) = 58.5 amu. FM of Ca(OH2)2 = 40 amu (1Ca) + 2 x [16 amu(1O) + 2 amu (2H)] = 76 amu 1.5.2 The Mole Concept At the end of this topic, students will be able to " define mole and explain the concept using examples; " define molar mass; " calculate the molar masses of substances. 29 Chemistry Grade 10 It is the concept adopted as a convenient way to deal with the tremendous amount of molecules or ions. A mole is defined as the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. 12 g of carbon-12 contains 6.022x1023 atoms. This number is known as Avogadro’s number (NA), in honor of Amedeo Avogadro (1776-1856, who first proposed the concept). The term mole, like a dozen or a gross, refers to a particular number of things. A dozen of eggs equals 12 eggs, a gross of pencils equals 144 pencils, and a mole of any substance equals 6.02x1023 units of the substance. Example 1.19: 1 mole of H2O equals 6.02 x 1023 molecules of H2O 1 mole of NaCl equals 6.02 x 1023 formula units of NaCl Molar mass is the molecular mass or formula mass expressed in grams. It is the mass of 1 mole of a substance. Molar mass of substance A = mass of one mole of A = 6.02 x1023 units of substance A Example 1.20: molar mass of H2O = 18 g = mass of 1 mole of H2O = 6.02 x 1023 molecules of H2O Molar mass of NaCl = 58.5 g = mass of one mole of NaCl =6.02 x 1023 formula units of NaCl 1.5.3 Chemical Formulas - Empirical and Molecular Formulas At the end of this topic, students will be able to " define empirical and molecular formulas; " determine empirical and molecular formulas; " calculate percentage composition of an element in a substance. Chemical formulas are used to express the composition of compounds in terms of chemical symbols. By composition we mean not only the elements present but also the ratios in which the atoms are combined. Here, you deal with two types of formulas, empirical formulas and molecular formulas. Empirical formula (or simplest formula) for a compound is the formula of a substance written with the smallest ratio (whole number ratio) subscripts. The molecular formula is the actual formula that tells you the exact number of atoms of different elements present in a molecule. Example 1.21: Empirical formula Molecular formula Benzene CH C6H6 Glucose CH2O C6H12O6 30  Chemical Reactions and Stoichiometry Note that the empirical formula, merely tells you the ratio of numbers of atoms in the compound. Percent Composition by Mass (%) Percent composition by mass tells you what percent of each element is present in a compound. Thus, it helps in chemical analysis of the given compound. The percentage composition of a given compound is defined as the ratio of the amount of individual elements present in the compound multiplied by 100. Mathematically, n × molar mass of the element x 100% Percent composition of an element = molar mass of compound where n is the number of moles of the element in 1 mole of the compound. For example, in 1 mole of hydrogen peroxide (H2O2) there are 2 moles of H atoms and 2 moles of O atoms. The molar masses of H2O2, H, and O are 34.02 g, 1.008 g, and 16.00 g, respectively. Therefore, the percent composition of H2O2 is calculated as follows: 2 x 1.00 g H %H= = 5.926% 34.02 g H2O2 2 x 16 %O= = 94.06% 34.02 H2O2 Determination of Empirical and Molecular Formulas Steps to determine empirical formula are: Step 1: Derive the number of moles of each element from its mass. Masses of elements may be given in terms of percent composition of elements or grams. Step 2: Divide each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula. Step 3: Multiply all coefficients by an integer, if necessary, to ensure that the smallest whole number ratio of subscripts is obtained Example 1.22: What is the empirical formula of a compound that contains 43.6% P and 56.4% O ? Step 1: Derivation of moles of each element, Given mass of the element Number of moles each element = Molar mass of the element Number of moles of P = 43.6 g 31.0 g/mol = 1.41 mol 56.4 g Number of moles of O = = 3.53 mol 16.0 g/mol 31 Chemistry Grade 10 Step 2: Smallest molar amount is 1.41 P: 1.41/1.41=1.00, O: 3.53/1.41 = 2.50 Step 3. Multiply both by 2 to convert into whole number, P: 2 x 1 =2, O : 2 x 2.5 = 5 Empirical formula is P2O5 Example 1.23: A compound contains 79.9% carbon and 20.1 % of hydrogen by mass. What is the empirical formula of the compound? Step 1: C= (79.9 g)/(12 g/mol) = 6.7 mol, H = 20.1g/(1g/mol) = 20.1 mol Step 2: C= 6.7/6.7 = 1, H = 20.1/6.7 = 3 Step 3: Empirical formula = CH3 Molecular formula = Empirical formula x n molar mass of molecular formula n= molar mass of empirical formula Example 1.24: What is the molecular formula of the oxide of phosphorus that has the empirical formula P2O5 if it's molecular mass is 284? Molar mass of P2O5 = 142, n = 284/142 = 2 Therefore, molecular formula = (P2O5)2 = P4O10 Example 1.25: The compound ethylene glycol is often used as antifreeze. It contains 38.7% carbon, 9.75% hydrogen, and the rest oxygen. The molecular weight of ethylene glycol is 62.07 g. What is the molecular formula of ethylene glycol? 1. Calculate the empirical formula. Assume 100 g of the compound, which will contain 38.70 g carbon, 9.75 g hydrogen and the rest oxygen. Mass of O = 100 g - (38.7 g of C + 9.75 g H) = 51.55 g 2. Calculate the moles of each element present: 38.70 g C No of moles of C = 12.01 g C = 3.22 mol C 9.75 g H No of moles of H = = 9.67 mol H 1.008 g H 51.55 g O No of moles of O = = 3.22 mol O 16.01 g O 32  Chemical Reactions and Stoichiometry Next, calculate the ratio of molecular weight to empirical formula weight. The molecular weight is given. The empirical formula is CH3O, so the empirical formula weight is 12.01 + 3(1.008) + 16.00 = 31.03. Molecular mass 62.07 = 2 n= = empirical formula mass 31.03 Therefore, the molecular formula is twice the empirical formula: C2H6O2. Exercise 1.8 1. Define and discuss each of the following terms: mole, Avogadro’s number, molar mass, molecular mass, formula mass, percentage composition 2. Distinguish between a. Formula mass and molecular mass of a compound b. Empirical and molecular formula 3. How many atoms and ions are there in one formula unit of a. K2CO3? b. Ba3(PO4)2? 4. How many moles of each element are present in a. 1.00 mole of Cu3(PO4)2 b. 2.5 mole of Na2CO3 c. 2 moles of Al2(SO4)3 5. Calculate the empirical formula for: a. 39.3% Na, 60.7% Cl b. 56.5% K, 8.7% C, 34.8% O 6. A. A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1g/mol. What is its molecular formula? B. A compound of carbon, hydrogen and chlorine has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula? 7. Calculate the molar mass of a compound if 0.372 mole of it has a mass of 152 g. 8. Calculate the molecular mass or formula mass (in amu) of each of the following substances: a. CH4 c. SO3 e. Na2SO4 b. NO2 d. C6H6 f. Ca3(PO4)2 9. Calculate the percentage composition by mass of the following compounds: a. C2H5OH c. H2SO4 e. K2Cr2O7 b. Na3PO4 d. CaCO3 33 Chemistry Grade 10 1.6 Stoichiometry Refer to grade 9 chemistry and revise the three fundamental laws of chemical reactions. Discuss in class using the following and give your own examples for each law. a. Carbon forms two types of oxides namely carbon monoxide and carbon dioxide. b. The ratio of the mass of hydrogen to the mass of oxygen in water is always 1:8. c. The total masses of calcium, carbon and oxygen are the same before and after the decomposition Activity 1.6 of calcium carbonate. At the end of this section, students will be able to " deduce mole ratios from balanced chemical equations; " solve mass-mass problems based on the given chemical equation; " define molar volume; " state Avogadro’s principle; " solve mass-mass problems based on the given chemical equation; " solve mass-volume problems based on the given chemical equation; " solve volume-volume problems based on the given chemical equation; " describe limiting and excess reactants. Stoichiometry is the study of the quantitative composition of substances and the relationship that exists between the reactants and the products involved in chemical reactions. In other words, stoichiometry is the study of the amount or ratio of moles, mass, or volumes (for gases) of reactants and products. Stoichiometric calculations are based on the following two major principles. A. The composition of any substance in the chemical equation should be expressed by a definite formula. B. The law of conservation of mass must be obeyed (the mass of reactants equals the mass of products) Form a group and discuss the following phenomenon. When wood burns, the ash weighs much less than the original wood. Where did the “lost mass” go? How can you estimate the mass of the wood that is no longer Activity 1.7 present? Present your conclusion to the class. Note that stoichiometric calculations are done for balanced chemical equations. 34  Chemical Reactions and Stoichiometry 1.6.1 Molar Ratios in Balanced Chemical Equation At the end of this section, students will be able to deduce mole ratios from balanced chemical equations. Stoichiometry problems of this type are called either mass-mass or mole-mole problems. A balanced chemical equation provides information about the nature, masses, number of moles, number of molecules/atoms of reactants and products. For example, consider the reaction of hydrogen with nitrogen to produce ammonia. The equation shows that 3 molecules of hydrogen and 1 molecule of N2 react to produce 2 molecules of ammonia. The equation can also represent the reaction of 3 moles (3x6.02x1023 molecules) of H2 with 1 mole (6.02 X 1023 molecules) of N2 to yield 2 moles (2x6.02 x1023 molecules ) of NH3. The same equation can also tell us the combination of 6 g of H2 with 28 g of N2 to yield 34 g of NH3. This can be further interpreted as follows: N2 + 3H2 → 2NH3 1 molecule N2 + 3 molecules H2 → 2 molecules NH3 (Molecular interpretation) 1mol N2 + 3 mol H2 → 2 mol NH3 (Molar interpretation) 28.0 g N2 + 3 x 2.02 g H2 → 2 x 17.0 g NH3 ( mass interpretation) Calculations based on chemical equations (stoichiometric problems) are classified into mass-mass, volume-volume and mass-volume problems. 1.6.2 Mass – Mass Relationships At the end of this section, students will be able to solve mass-mass problems based on the given chemical equation or using the conversion factor. There are two methods for solving mass-mass problems. These are mass-ratio and mole-ratio methods. A. The Mass - ratio Method In this method, the mass of one substance is determined from the given mass of the other substance using the following steps. Step 1: Write the balanced chemical equation. Step 2: Place the given mass above the corresponding formula, and x above the formula of the substance whose mass is to be determined Step 3: Write the total molar mass of the substances below the formula of each substance. (Total molar mass is the molar mass of the substance multiplied by its coefficient). Step 4: Set up the proportion. Step 5: Solve for the unknown mass, x. 35 Chemistry Grade 10 Example 1.26: How many grams of aluminum metal must be heated to produce 20.4 g of aluminium oxide? Solution: Step 1: 4Al + 3O2 → 2Al2O3 x 20.4 g Step 2: 4Al + 3O2 → 2Al2O3 x 20.4 g Step 3: 4Al + 3O2 → 2Al2O3 108 204 Step 4: x/(108 g) = (20.4 g)/(204 g) Step 5: ⟹x = (20.4 g)/(204 g) x 108 g = 10.8 g Therefore, 10.8 g of aluminum metal is needed to produce 20.4 g of aluminum oxide. Note that stoichiometric problems can also be solved using the mole ratio or molar mass as conversion factors. The numbers in a conversion factor come from the coefficients of the balanced chemical equation. In general, for mole – mole problems, mole A x mole ratio = mole B Mass-Mass problems, gram A x 1mole A x mole B x molar mass B gram B = molar mass A mole A mole B In the above example, coefficients of Al and Al2O3 are 4 and 2 respectively Therefore, conversion factor = mole ratio = 4 mol Al ∕2 mol O2. Given molar masses, 1 mole Al2O3 = 102 g, 1 mole Al = 27 g The conversion steps are summarized as follows: grams of Al2O3 → moles of Al2O3 → moles of Al → grams of Al ↓ ↓ ↓ Conversion factor: molar mass of Al2O3 mole ratio molar mass of Al 20.4 g x 1 mol Al2O3/102 g Al2O3 x 4 mol Al /2 mol Al2O3 x 27 g Al/1mol Al = 10.8 g Example 1.27: What mass of magnesium sulfate will be produced if 4.8 g of magnesium reacts with excess sulfuric acid? Step 1: Mg + H2SO4 → MgSO4 + H2 36  Chemical Reactions and Stoichiometry 4.8g x Step 2: Mg + H2SO4 → MgSO4 + H2 4.8g x Step 3: Mg + H2SO4 → MgSO4 + H2 24g 120g Step 4: ⇒x = (4.8 g)/(24 g) × 120 g = 24 g Therefore, 13.6 g MgSO4 will be produced. Check the result using conversion factor. Hint: mole ratio = 1mole Mg/1mole MgSO4 B. The Mole - ratio Method In this method, calculations are made in terms of moles; therefore the given mass is converted into mole. The obtained mole can be converted back to mass if required. Follow the steps given below to solve problems of mass-mass relationships using mole ratio method: Step 1: Write the balanced chemical equation. Step 2: Convert the given mass to moles and write the obtained moles and the required quantity, x, above the formulas of the respective substances. Step 3: Place the coefficients as the number of moles under the formula of each substance involved. Step 4: Set up the proportion. Step 5: Solve for the unknown value, x; and convert the moles obtained into mass. Example 1.28: How many grams of calcium oxide are needed to react completely with 22.0 g of carbon dioxide? Solution: Step 1: CaO + CO2 → CaCO3 Step 2: moles of CO2 = given mass/molar mass = 22 g/44 g/mol = 0.5 mol x 0.5mol Step 3: CaO + CO2 → CaCO3 1mol 1mol Step 4: x/1mol = 0.5mol/1mol ⟹ x = 0.5 mol Step 5: Convert the moles into grams of CaO Mass of CaO produced = No. of moles CaO x molar mass CaO = 0.5mol x 56 g/mol = 28 g Solve the above problem using conversion factor, gram CO2 → Mole CO2 → mole CaO → gram CaO 37 Chemistry Grade 10 1mol CO2 x 1mol CaO 56 g CaO 22 g CO2 x x = 28 g of CaO 44 g CO2 1mol CO2 1mol CaO Example 1.29: How many grams of carbon monoxide must react with excess iron oxide to produce 28 grams of iron? Step 1: Fe2O3 + 3CO → 2Fe + 3CO2 given mass Step 2: No. of moles Fe = molar mass = 28 g/ 56 g/mol = 0. 5 mol x 0.5 mol Step 3: Fe2O3 + 3CO → 2Fe + 3CO2 3 mol 2 mol Step 4: x/(3 mol) = 0.5 mol/(2 mol) ⟹ x = 0.5 x 3 mol/2 = 0.75 mol of CO Step 5: Convert the moles into grams of CO Mass of CO required = no. of moles CO x molar mass CO = 0.75 mol x 28 g/mol =21 g (check the result using the conversion factor) Note that although mass-mass and mole-mole stoichiometric problems are more common, there are also mass-mole and mole-mass problems. In mole-mass problems,the amount of one substance is given in moles and the mass of another substance is determined, usually in grams and in mass-mole problems, the mass of one substance is given, usually in grams and the amount of another substance is determined in moles. Refer to relevant sources and solve Exercise 1.9 (a,b). You may use the general sequence of conversions for a mole-mass calculation follow the following sequence: mol 1st substance → mol 2nd substance → mass 2nd substance Example: What mass of Fe2O3 must be reacted to generate 3 moles of Al2O3? Fe2O3 + 2Al → 2Fe + Al2O3 Given: Moles of Al2O3 = 3, Molar mass of Al2O3 = 102 g Required: Mass of Fe2O3 , molar mass of Fe2O3 = 160 g Solution: Follow the sequence,Gram Al2O3→ mole Al2O3→ mole Fe2O3 → Gram Fe2O3 3 moles of Al2O3 x 1 mole Fe2O3 x 160 g = 480 g of Fe2O3 1 mol Al2O3 1 mol Fe2O3 for mass- mole calculation follow the following sequence: 38  Chemical Reactions and Stoichiometry mass 1st substance → mol 2nd substance → mol 2nd substance Example: How many moles of C2H5OH are needed to generate 106.7 g of H2O? C2H5OH + 3O2 → 2CO2 + 3H2O Given: mass of H2O = 106.7 g, molar mass of H2O = 18 g Required: No. of mole of C2H5OH Solution: follow the steps: Gram H2O → moleH2O → mole C2H5OH 106.7 g x 1mole H2O x mole C2H5OH = 1.97 moles of C2H5OH 18g 3mol H2O Exercise 1.9 1. Which of the following statements is correct for the equation shown here? 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) a. 6 g of H2O are produced for every 4 g of NH3 reacted. b. 1 mole of NO is produced per mole of NH3 reacted. c. 2 moles of NO are produced for every 3 moles of O2 reacted 2. How many moles of H2O are required to produce 4.5 moles of HNO3 according to the following reaction? 3NO2 + H2O → 2HNO3 + NO 3. How many moles of CaO are needed to react with excess water to produce 370 g of calcium hydroxide? 4. In the decomposition of KClO3, how many moles of KCl are formed in the reaction that produces 0.05 moles of O2? 5. How many grams of CaCO3 are needed to react with 15.2 g of HCl according to the following equation? CaCO3 + 2HCl → CaCl2 + CO2 + H2O 6. What mass of nitrogen dioxide is produced by the decomposition of 182 g of magnesium nitrate? 7. Solve the following using either mass-mole or mole-mass relations a. What mass of hydrogen sulphide gas is burned in a reaction that produces 4 mol SO2 and water vapor: 2H2S+ 3O2 → 2SO2 + 2H2O b. How many moles of CO2 will be produced when 50 g of CaCO3 are reacted according to the chemical equation: CaCO3 + 2HCl → CaCl2 + CO2 + H2O 39 Chemistry Grade 10 1.6.3 Volume – Volume Relationships At the end of this section, students will be able to solve volume -volume problems based on the given chemical equation. At Standard Temperature (00C) and Pressure (1atm) or STP, one mole of any gas occupies 22.4 liters. This, 22.4 liters is known as molar volume at STP. The volume of a gas and its number of molecules are related and explained by Avogadro’s Law. Avogadro’s Law states that equal volumes of different gases, under the same conditions of temperature and pressure, contain equal number of molecules. According to this law, the volume of a gas is proportional to the number of molecules (moles) of the gas at STP as shown below: V ~ n; where V is the volume and n is the number of moles of gas In Volume – volume problems, the volume of one substance is given and the volume of the other substance is calculated. Example 1.30: What volume of oxygen will react with methane to produce 44.8 liters of carbon dioxide at STP? Step 1: CH4 + 2O2 → 2H2O + CO2 Step 2: Place the given volume and the required volume, x above the corresponding formulas. x 44.8 L CH4 + 2O2 → 2H2O + CO2 44.8 L 22.4 L Step 3: Write the total molar volume (22.4 L multiplied by any coefficient) below the formulas. Step 4: x/44.8 L=(44.8 L)/(22.4 L) ⟹ x = 44.8 /22.4 x 44.8 L= 89.6 L Step 5: Solve for the unknown volume, x x = 89.6 L of oxygen is needed. 40  Chemical Reactions and Stoichiometry Exercise 1.10 1. What volume of nitrogen reacts with 33.6 litres of oxygen to produce nitrogen dioxide? 2. How many litres of sulphur trioxide are formed when 4800 cm3 of sulphur dioxide is burned in air? 3. How many litres of ammonia are required to react with 145 litres of oxygen according to the following reaction? 4NH3 + 5O2 → 4NO + 6H2O 4. Calculate the volume of oxygen produced in the decomposition of 5 moles of KClO3 at STP? 5. How many moles of water vapour are formed when 10 litres of butane gas, C4H10 is burned in oxygen at STP? 1.6.4 Mass – Volume Relationships At the end of this section, students will be able to solve mass -volume problems based on the given chemical equation. In mass-volume problems, either the mass of one substance is given and the volume of the other is required or the volume of one substance is given and the mass of the other one is required. The steps to solve such type of problems are the same as the previous steps except putting the masses on one side and the volumes on the other side of the equality sign. Example 1.31: How many grams of magnesium should react with sulphuric acid to produce 5.6 L of hydrogen at STP? Solution: Step 1: Mg + H2SO4 → H2 x 5.6 L Step 2: Mg + H2SO4 → MgSO4 + H2 x 5.6 L Step 3: Mg + H2SO4 → MgSO4 + H2 24 g 22.4 L Step 4: x/24 g = 5.6 L/22.4 L ⟹x=6g Step 5: x = 6 g of magnesium is needed 41 Chemistry Grade 10 Exercise 1.11 1. How many litres of oxygen are required to react with 156 g of benzene according to the chemical equation: 2C6H6 + 15O2 → 12CO2 + 6H2O 2. What mass of iron metal would be completely oxidized by 89.6 L of oxygen to produce Fe2O3 at STP? 3. Calculate the mass of calcium carbide that is needed to produce 100 cm3 of acetylene according to the equation: CaC2 + 2H2O → C2H2 + Ca(OH)2 4. How many milliliters of sulphur dioxide are formed when 12.5 g of iron sulphide ore (pyrite) reacts with oxygen (at STP) according to the equation 4FeS2 + 11O2 → 2Fe2O3 + 8SO2 1.6.5 Limiting and Excess Reactants At the end of this section, students will be able to describe limiting and excess reactants. calculate the limiting and excess reactants of a reaction. Consider a dance contest at a club. If there are 14 men and only 9 women, then only 9 female/male pairs can compete. 1. How many men will be left without partners? 2. Who (men or women) are in excess? 3. Who (men or women) limits the number of female/ male pair that can dance in the contest? 4. How can you relate this analogue with the chemical reactions involving two or more reactants? Discuss in your group and present in class your Activity 1.8 conclusion. In a chemical reaction involving two reactants, the reaction will stop when all of one reactant has been completely consumed no matter how much of the second reactant remains. The reactant completely consumed first in a reaction is called the limiting reactant, because it limits or determines the amount of product that can be formed. When the limiting reactant is completely consumed, no more product can be formed and thus the other reactant remains excess. Excess reactants are the reactants present 42  Chemical Reactions and Stoichiometry in quantities greater than necessary to react with the quantity of the limiting reactant. Suppose you want to make some cheese sandwiches. Each is made from two slices of bread and a slice of cheese. 2 slices bread + 1 slice cheese → 1 cheese sandwich If you have 7 slices of bread and 2 slices cheese, you can make 2 cheese sandwiches. 3 slices of bread is excess. The number of sandwiches is limited by the number of slices of cheese. Hence, slices of cheese are the limiting ingredient and slices of bread are excess ingredient. Note that: Yield of the product is the one calculated using the limiting reactant and it is the smallest. Example 1.32: Magnesium metal reacts with hydrochloric acid by the following reaction: Mg + 2HCl → MgCl2 + H2 If 0.30 mol Mg is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced? Solution: Here there are 0.3 mol of Mg and 0.52 mol of HCl. Therefore, first determine the limiting reactant. Calculate the yield of H2 (product) using both reactants, the reactant that gives lower yield is the limiting reactant and the amount obtained by the limiting reactant is the yield of the product. A. Using the quantity of Mg B. Using the quantity of HCl 0.3mol/1mol = x/1mol 0.52 mol/2 mol = x/1mol 0.3 mol x 0.52 mol x Mg + 2HCl → MgCl2 + H2 Mg + 2HCl → MgCl2 + H2 1 mol 1 mol 2 mol 1mol 0.3 mol/1 mol = x/1 mol 0.52 mol/2 mol = x/1 mol ⇒ x = 0.3 mol of H2 ⇒ x = 0.26 mol of H2 H2 produced using HCl is less than H2 produced using Mg. Hence, HCl is the limiting reactant and Mg is the excess reactant. The number of moles of H2 produced is equal to the quantity of H2 obtained using HCl (limiting reactant) which is 0.26 mol. 43 Chemistry Grade 10 Example 1.33: In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde, CH3CHO, containing manganese (II) acetate (catalyst) under pressure at 60 0C. 2CH3CHO + O2 → 2HC2H3O2 In a laboratory test of this reaction, 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel. A. How many grams of acetic acid can be produced? B. How many grams of the excess reactant remain after the reaction is complete? Solution 1. Convert grams in to moles given mass No. of moles of CH3CHO = molar mass = 20 g = 0.454 mol 44 g/mol given mass No. of moles of O2 = molar mass 10 g = = 0.312 mol 32 g/mol 2. Calculate the amount of HC2H3O2 produced by 0.454 mol. CH3CHO and 0.312 mol O2 in order to decide the limiting reactant. A. Using the quantity of CH3CHO B. Using the quantity of O2 0.454 mol x 0.312 mol x 2CH3CHO +O2 → 2HC2H3O2 2CH3CHO + O2 → 2HC2H3O2 2 mol 2 mole 1 mol 2 mol 0.454 mol/2 mol = x/2 mol 0.312 mol/1 mol = x/2 mol ⟹ x = 0.454 mol of HC2H3O2 ⟹ x = 0.624 mol of HC2H3O2 Therefore, the limiting reactant is acetaldehyde, CH3CHO. It is the reactant that gives lower yield in the above reaction. Oxygen is the excess reactant in the above reaction since it gives higher yield. Number of moles of HC2H3O2 produced = 0.454 mol A. Amount of HC2H3O2 produced in grams = 0.454 mol x 60 g/mol = 27.2 g B. Calculate the amount of oxygen consumed using the product obtained by the limiting reactant x 0.454 mol 2CH3CHO + O2 → 2HC2H3O2 1 mol 2 mol 44  Chemical Reactions and Stoichiometry Amount of O2 consumed, x = 0.454 mol x 1 mol/2 mol = 0.227 mol We started with 10 g of O2 or 0.312 mol, so the excess O2 = 0.312 - 0.227 = 0.085mol or 2.7 g. Exercise 1.12 1. If 6.5 g of zinc reacts with 5.0 g of HCl, according to the following reaction. Zn + 2HCl → ZnCl2 + H2 a. Which substance is the limiting reactant? b. How many grams of the reactant remain unreacted? c. How many grams of hydrogen would be produced? 2. What mass of Na2SO4 is produced if 49 g of H2SO4 reacts with 80 g of NaOH? 3. If 20 g of CaCO3 and 25 g of HCl are mixed, what mass of CO2 is produced? CaCO3 + 2HCl → CaCl2 + CO2 + H2O 4. If 3 moles of calcium react with 3 moles of oxygen, then a. Which substance is the limiting reactant? b. How many moles of calcium oxide are formed? 5. For the reaction: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2, how many grams of hydrogen are produced if 0.8 mole of aluminum reacts with 1.0 mole of sulphuric acid? 1.6.6 Theoretical, Actual and Percentage Yields At the end of this section, students will be able to " describe theoretical, Actual and percentage yields; " calculate theoretical, Actual and percentage yields. By Considering the preparation of 'Shiro' powder from chickpeas at home, discuss in group the following questions and present your conclusion to the class. a. Have you noticed the amount of chickpeas before and after the preparation of Shiro powder? Is it possible for the amount of chickpeas to be equal after and before making the Shiro powder? If not, what is the reason? b. How could the quantity of the Shiro powder preparation be maximized? c. How can you compare this analogy with the yield of prod