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2 Pearson Education Limited Edinburgh Gate Harlow Essex CM0 JE England and Associated Companies throughout the world Visit us on the World Wide Web at: Pearson Education Limited 014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6 10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. ISBN 10: ISBN 13: British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

3 SELF STUDY ACTIVITY Stoichiometry Law of Conservation of Mass Conservation of Mass In any chemical reaction, the total amount of matter in the reactants is equal to the total amount of matter in the products. If all the reactants were weighed, they would have a total mass equal to the total mass of the products. This is known as the law of conservation of mass, which says that there is no change in the total mass of the substances reacting in a chemical reaction. Thus, no material is lost or gained when original substances are changed to new substances. For example, tarnish forms when silver reacts with sulfur to form silver sulfide. Ag(s) + S(s) Ag S(s) Ag(s) + S(s) Ag S(s) Mass of reactants = Mass of products The law of conservation of mass states that there is no matter lost or gained in a chemical reaction. In this reaction, the number of silver atoms that reacts is two times the number of sulfur atoms. When 00 silver atoms react, 100 sulfur atoms are required. However, many more atoms would actually be present in this reaction. If we are dealing with molar amounts, then the coefficients in the equation can be interpreted in terms of moles. Thus, moles of silver react with 1 mole of sulfur. Since the molar mass of each can be determined, the quantities of silver and sulfur can also be stated in terms of mass in grams of each. Therefore, an equation for a chemical equation can be interpreted several ways, as seen in Table 7. TABLE 7 Information Available from a Balanced Equation Reactants Products Equation Ag(s) S(s) Ag S(s) Atoms Ag atoms + 1 S atom 1 Ag S formula unit Avogadro s number of atoms 00 Ag atoms S atoms 100 Ag S formula units (6.0 * 10 3 ) + 1(6.0 * 10 3 ) 1(6.0 * 10 3 ) Ag S Ag atoms S atoms formula units mole of Ag S + 1 1(47.9 g) of Ag S Moles moles of Ag mole of S Mass (g) (107.9 g) of Ag (3.1 g) of S Total mass (g) 47.9 g 47.9 g 194

4 + Iron (Fe) Fe(s) + Sulfur (S) 3S(s) In the chemical reaction of Fe and S, the mass of the reactants is the same as the mass of the product, Fe S 3. Iron(III) sulfide (Fe S 3 ) Fe S 3 (s) Mole Mole Factors from an Equation When iron reacts with sulfur, the product is iron(iii) sulfide. Fe(s) + 3S(s) Fe S 3 (s) Because the equation is balanced, we know the proportions of iron and sulfur in the reaction. For this reaction, we see that moles of iron reacts with 3 moles of sulfur to form 1 mole of iron(iii) sulfide. Actually, any amount of iron or sulfur may be used, but the ratio of iron reacting with sulfur will be the same. From the coefficients, we can write mole mole factors between reactants and between reactants and products. The coefficients used in the mole mole factors are exact numbers; they do not limit the number of significant figures. moles Fe Fe and S: and 3 moles S 3 moles S moles Fe moles Fe Fe and Fe S 3 : and 1 mole Fe S 3 1 mole Fe S 3 moles Fe 3 moles S S and Fe S 3 : and 1 mole Fe S 3 1 mole Fe S 3 3 moles S CONCEPT CHECK 7 Writing Mole Mole Factors Consider the following balanced equation: Write the mole mole factors for a. Na and O b. Na and Na O ANSWER 4Na(s) + O (g) Na O(s) a. The mole mole factors for Na and O use the coefficient of Na to write 4 moles of Na, and the coefficient of 1 (understood) to write 1 mole of O. 4 moles of Na = 1 mole of O 4 moles Na and 1 mole O 1 mole O 4 moles Na b. The mole mole factors for Na and Na O use the coefficient of Na to write 4 moles of Na and the coefficient of Na O to write moles of Na O. 4 moles of Na = moles of Na O 4 moles Na and moles Na O moles Na O 4 moles Na 195

5 Moles of Reactants and Products Conversions Involving Moles Using Mole Mole Factors in Calculations Whenever you prepare a recipe, adjust an engine for the proper mixture of fuel and air, or prepare medicines in a pharmaceutical laboratory, you need to know the proper amounts of reactants to use and how much of the product will form. Earlier we wrote all the possible conversion factors that can be obtained from the balanced equation Fe(s) + 3S(s) Fe S 3 (s). Now we can use mole mole factors in chemical calculations. SAMPLE PROBLEM 7 Richard Megna Fundamental Photographs Propane fuel reacts with O in the air to produce CO, H O, and energy. Guide to Using Mole Mole Factors 1 State the given and needed quantities. Write a plan to convert the given to the needed moles. 3 Use coefficients to write relationships and mole mole factors. 4 Set up the problem using the mole mole factor that cancels given moles. Calculating Moles of a Product Propane gas (C 3 H 8 ), a fuel for camp stoves and specially equipped automobiles, reacts with oxygen to produce carbon dioxide, water, and energy. How many moles of CO can be produced when.5 moles of C 3 H 8 reacts? C 3 H 8 (g) + 5O (g) 9: 3CO (g) + 4H O(g) + energy Propane SOLUTION Step 1 State the given and needed quantities. Given moles of C 3 H 8 Need moles of CO Step Write a plan to convert the given to the needed moles. Step 3 Use coefficients to write relationships and mole mole factors. 1 mole of C 3 H 8 = 3 moles of CO 1 mole C 3 H 8 and 3 moles CO 3 moles CO 1 mole C 3 H 8 Step 4 Set up the problem using the mole mole factor that cancels given moles. STUDY CHECK 7 Mole mole moles of C 3 H 8 moles of CO factor.5 moles C 3 H 8 * 3 moles CO 1 mole C 3 H 8 = 6.75 moles of CO The answer is given with three SFs because the given quantity,.5 moles of C 3 H 8, has three SFs. The values in the mole mole factor are exact. Using the equation in Sample Problem 7, calculate the number of moles of oxygen that must react to produce mole of water. QUESTIONS AND PROBLEMS Mole Relationships in Chemical Equations 47 Write all the mole mole factors for each of the following equations: a. SO (g) + O (g) SO 3 (g) b. 4P(s) + 5O (g) P O 5 (s) 48 Write all the mole mole factors for each of the following equations: a. Al(s) + 3Cl (g) AlCl 3 (s) b. 4HCl(g) + O (g) Cl (g) + H O(g) 49 The reaction of hydrogen with oxygen produces water. H (g) + O (g) H O(g) a. How many moles of O are required to react with.0 moles of H? b. How many moles of H are needed to react with 5.0 moles of O? c. How many moles of H O form when.5 moles of O reacts? 196

6 50 Ammonia is produced by the reaction of hydrogen and nitrogen. N (g) + 3H (g) NH 3 (g) Ammonia a. How many moles of H are needed to react with 1.0 mole of N? b. How many moles of N reacted if 0.60 mole of NH 3 is produced? c. How many moles of NH 3 are produced when 1.4 moles of H reacts? 51 Carbon disulfide and carbon monoxide are produced when carbon is heated with sulfur dioxide. 5C(s) + SO (g) 9: CS (l) + 4CO(g) a. How many moles of C are needed to react with mole of SO? b. How many moles of CO are produced when 1. moles of C reacts? c. How many moles of SO are needed to produce 0.50 mole of CS? d. How many moles of CS are produced when.5 moles of C reacts? 5 In the acetylene torch, acetylene gas (C H ) burns in oxygen to produce carbon dioxide and water. C H (g) + 5O (g) 9: 4CO (g) + H O(g) a. How many moles of O are needed to react with.00 moles of C H? b. How many moles of CO are produced when 3.5 moles of C H reacts? c. How many moles of C H are required to produce 0.50 mole of H O? d. How many moles of CO are produced from mole of O? 7 Mass Calculations for Reactions When you perform a chemistry experiment in the laboratory, you measure a specific mass of the reactant. From the mass in grams, you can determine the number of moles of reactant. By using mole mole factors, you can predict the moles of product that can be produced. Then the molar mass of the product is used to convert the moles back into mass in grams as seen in Sample Problem 8. LEARNING GOAL Given the mass in grams of a substance in a reaction, calculate the mass in grams of another substance in the reaction. SAMPLE PROBLEM 8 Mass of Product from Mass of Reactant When acetylene ( C H ) burns in oxygen, high temperatures are produced that are used for welding metals. Masses of Reactants and Products SELF STUDY ACTIVITY Stoichiometry C H (g) + 5O (g) 9: 4CO (g) + H O(g) + energy How many grams of CO are produced when 54.6 g of C H is burned? SOLUTION Step 1 Use molar mass to convert grams of given to moles. 1 mole of C H = 6.0 g of C H 1 mole C H and 1 mole C H 54.6 g C H * 1 mole C H =.10 moles of C H Step Write a mole mole factor from the coefficients in the equation. moles of C H = 4 moles of CO moles C H 4 moles CO and 4 moles CO moles C H Guide to Calculating the Masses of Reactants and Products in a Chemical Reaction 1 Use molar mass to convert grams of given to moles (if necessary). Write a mole mole factor from the coefficients in the equation. 3 Convert moles of given to moles of needed substance using the mole mole factor. 4 Convert moles of needed substance to grams using molar mass. 197

7 Step 3 Convert moles of given to moles of needed substance using the mole mole factor..10 moles C H * 4 moles CO moles C H = 4.0 moles of CO Step 4 Convert moles of needed substance to grams using molar mass. 1 mole of CO = 44.0 g of CO Comstock A mixture of acetylene and oxygen undergoes combustion during the welding of metals g CO 1 mole CO and 1 mole CO 44.0 g CO 4.0 moles CO * 44.0 g CO 1 mole CO = 185 g of CO The solution can also be obtained by placing the conversion factors in sequence using the following plan: Molar Mole mole grams of C H moles of C H moles of CO mass factor Molar mass grams of CO 54.6 g C H * 1 mole C H * 4 moles CO moles C H * 44.0 g CO 1 mole CO = 185 g of CO STUDY CHECK 8 Using the equation in Sample Problem 8, calculate the grams of CO that can be produced when 5.0 g of O reacts. QUESTIONS AND PROBLEMS Mass Calculations for Reactions 53 Sodium reacts with oxygen to produce sodium oxide. 4Na(s) + O (g) Na O(s) a. How many grams of Na O are produced when 57.5 g of Na reacts? b. If you have 18.0 g of Na, how many grams of O are needed for the reaction? c. How many grams of O are needed in a reaction that produces 75.0 g of Na O? 54 Nitrogen gas reacts with hydrogen gas to produce ammonia by the following equation: N (g) + 3H (g) NH 3 (g) a. If you have 3.64 g of H, how many grams of NH 3 can be produced? b. How many grams of H are needed to react with.80 g of N? c. How many grams of NH 3 can be produced from 1.0 g of H? 55 Ammonia and oxygen react to form nitrogen and water. 4NH 3 (g) + 3O (g) N (g) + 6H O(g) a. How many grams of O are needed to react with 13.6 g of NH 3? b. How many grams of N can be produced when 6.50 g of O reacts? c. How many grams of water are formed from the reaction of 34.0 g of NH 3? 56 Iron(III) oxide reacts with carbon to give iron and carbon monoxide. Fe O 3 (s) + 3C(s) Fe(s) + 3CO(g) a. How many grams of C are required to react with 16.5 g of Fe O 3? b. How many grams of CO are produced when 36.0 g of C reacts? c. How many grams of Fe can be produced when 6.00 g of Fe O 3 reacts? 57 Nitrogen dioxide and water react to produce nitric acid, HNO 3, and nitrogen oxide. 3NO (g) + H O(l) HNO 3 (aq) + NO(g) a. How many grams of H O are needed to react with 8.0 g of NO? b. How many grams of NO are obtained from 15.8 g of NO? c. How many grams of HNO 3 are produced from 8.5 g of NO? 58 Calcium cyanamide reacts with water to form calcium carbonate and ammonia. CaCN (s) + 3H O(l) CaCO 3 (s) + NH 3 (g) a. How many grams of water are needed to react with 75.0 g of CaCN? b. How many grams of NH 3 are produced from 5.4 g of CaCN? c. How many grams of CaCO 3 form if 155 g of water reacts? 198