Periodic Table of Elements Noble gases Alkali metals Alkaline earth metals 1A Halogens 18 8A H 2A 13 3A 14 4A 15 5A 16 6A 17 7A He Li Be B C N O F 10 Ne 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 19 K Transition metals 10 11 12 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 55 Cs 56 Ba 57 La* 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 87 Fr 88 Ra 89 Ac† 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds 111 Rg 112 Uub 113 Uut *Lanthanides 58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu †Actinides 90 Th 91 Pa 92 U 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr 118 Uuo 114 115 Uuq Uup Study Card to Accompany Zumdahl’s Introductory Chemistry Series Measurements and Calculations The Commonly Used Prefixes in the Metric System p Table 2.2 Gases Prefix Symbol Meaning Power of 10 for Scientific Notation mega M 1,000,000.000000001 106 kilo k deci d 0.1 10Ϫ1 centi c 0.01 10Ϫ2 milli m 0.001 10Ϫ3 micro m 0.000001 10Ϫ6 nano n 0.000000001 10Ϫ9 103 Types of Crystalline Solids Some Examples of Commonly Used Units A dime is mm thick A quarter is 2.5 cm in diameter The average height of an adult man is 1.8 m mass A nickel has a mass of about g A 120-lb woman has a mass of about 55 kg length volume A 12-oz can of soda has a volume of about 360 mL A half gallon of milk is equal to about L of milk cm3 ϭ mL density of H2O(l) ϭ 1.0 g/mL density ϭ mass/volume Avogadro’s number ϭ 6.022 ϫ 1023 Heat Required ϭ Q ϭ specific heat capacity ϫ mass ϫ ⌬T Specific heat capacity of H2O(l) ϭ 4.184 J/g °C Kinetic energy ϭ mv 2/2 Exothermic reactions produce heat Endothermic reactions absorb heat Kinds of Chemical Reactions 19 p Chemical reactions Oxidation–reduction reactions Ionic solids Molecular solids Atomic solids Components are ions Components are molecules Components are atoms Figure 14.13 The classes of crystalline solids Solutions Energy Precipitation reactions 45 p Crystalline solids p Table 2.6 1000 STP: °C, atm Volume of mole of ideal gas at STP ϭ 22.4 L PV ϭ nRT (Ideal Gas Law) R ϭ 0.08206 L atm/K mol Process at constant n and T: P1V1 ϭ P2V2 (Boyle’s law) Process at constant n and P: V1/T1 ϭ V2/T2 (Charles’s law) Process at constant T and P: V1/n1 ϭ V2/n2 (Avogadro’s law) Acid–base reactions Mass percent ϭ mass of solute ϫ 100% (p 481) mass of solution M ϭ molarity ϭ mol moles of solute ϭ liters of solution L (p 483) Mass of solute ϭ (molar mass of solute) ϫ (L of solution) (Molarity) number of equivalents Normality ϭ N ϭ liter of solution equiv equivalents ϭ ϭ (p 499) liter L Acids and Bases Common Strong Acids: HCl, HNO3, H2SO4, HClO4, HI Common Weak Acids: HSO4Ϫ, CH3COOH (often written HC2H3O2), HF Common Strong Bases: NaOH, KOH Common Weak Bases: NH3 Kw ϭ 10Ϫ14 ϭ [Hϩ][OHϪ] (ion-product constant for water) (p 523) Combustion reactions Synthesis reactions (Reactants are elements.) Decomposition reactions (Products are elements.) Figure 7.12 Summary of classes of reactions Atomic Structure Mass number (A) (number of protons and neutrons) 23 11Na d Element symbol Atomic number (Z) c (number of protons) A Ϫ Z ϭ #n0 #pϩ Ϫ #eϪ ϭ charge c pH ϭ Ϫlog[Hϩ] (p 526) pOH ϭ Ϫlog[OHϪ] (p 527) pH ϩ pOH ϭ 14.00 (p 529) Equilibrium Constants aA ϩ bB cC ϩ dD, K ϭ [C]c[D]d/[A]a[B]b (p 553) A2B3 (s) 2A3ϩ(aq) ϩ 3B2Ϫ(aq), Ksp ϭ [A3ϩ]2[B2Ϫ]3 [X] ϭ Molarity of X Study Card to Accompany Zumdahl’s Introductory Chemistry Series Chemical Bonding Number of Electron Pairs Bonds Electron Pair Arrangement Ball-and-Stick Model 2 Linear 3 Trigonal planar (triangular) Molecular Structure Partial Lewis Structure Linear 180˚ AOBOA Trigonal planar (triangular) 120˚ Cl Tetrahedral Tetrahedral F B F A A H 109.5˚ A Be Cl F B Ball-and-Stick Model A A 38 p Table 12.4 Arrangements of Electron Pairs and the Resulting Molecular Structures for Two, Three, and Four Electron Pairs B A H C H H H N H H A Tetrahedral Trigonal pyramid 109.5˚ A B A A Tetrahedral Bent or V-shaped 109.5˚ A B A O H H Oxidation–Reduction Reactions Common Lewis Dot Fragments Oxidation is loss of electrons (OIL) Reduction is gain of electrons (RIG) H N ؉1 O ؊1 N C C C N N O O F 58 p Rules for Assigning Oxidation States The oxidation state of an atom in an uncombined element is The oxidation state of a monatomic ion is the same as its charge Oxygen is assigned an oxidation state of Ϫ2 in most of its covalent compounds Important exception: peroxides (compounds containing the O22Ϫ group), in which each oxygen is assigned an oxidation state of Ϫ1 In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of ϩ1 In binary compounds, the element with the greater electronegativity is assigned a negative oxidation state equal to its charge as an anion in its ionic compounds For an electrically neutral compound, the sum of the oxidation states must be zero For an ionic species, the sum of the oxidation states must equal the overall charge Table of Atomic Masses* Element Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Bohrium Boron Bromine Cadmium Calcium Californium Carbon Cerium Cesium Chlorine Chromium Cobalt Copper Curium Darmstadtium Dubnium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Symbol Ac Al Am Sb Ar As At Ba Bk Be Bi Bh B Br Cd Ca Cf C Ce Cs Cl Cr Co Cu Cm Ds Db Dy Es Er Eu Fm F Fr Gd Ga Ge Atomic Number Atomic Mass Element 89 13 95 51 18 33 85 56 97 83 107 35 48 20 98 58 55 17 24 27 29 96 110 105 66 99 68 63 100 87 64 31 32 [227]§ 26.98 [243] 121.8 39.95 74.92 [210] 137.3 [247] 9.012 209.0 [264] 10.81 79.90 112.4 40.08 [251] 12.01 140.1 132.90 35.45 52.00 58.93 63.55 [247] [271] [262] 162.5 [252] 167.3 152.0 [257] 19.00 [223] 157.3 69.72 72.59 Gold Hafnium Hassium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Meitnerium Mendelevium Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium *The values given here are to four significant figures where possible §A Symbol Au Hf Hs He Ho H In I Ir Fe Kr La Lr Pb Li Lu Mg Mn Mt Md Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Atomic Number Atomic Mass Element 79 72 108 67 49 53 77 26 36 57 103 82 71 12 25 109 101 80 42 60 10 93 28 41 102 76 46 15 78 94 84 19 197.0 178.5 [265] 4.003 164.9 1.008 114.8 126.9 192.2 55.85 83.80 138.9 [260] 207.2 6.9419 175.0 24.31 54.94 [268] [258] 200.6 95.94 144.2 20.18 [237] 58.69 92.91 14.01 [259] 190.2 16.00 106.4 30.97 195.1 [244] [209] 39.10 Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Roentgenium Rubidium Ruthenium Rutherfordium Samarium Scandium Seaborgium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium value given in parentheses denotes the mass of the longest-lived isotope Symbol Pr Pm Pa Ra Rn Re Rh Rg Rb Ru Rf Sm Sc Sg Se Si Ag Na Sr S Ta Tc Te Tb Tl Th Tm Sn Ti W U V Xe Yb Y Zn Zr Atomic Number Atomic Mass 59 61 91 88 86 75 45 111 37 44 104 62 21 106 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40 140.9 [145] [231] 226 [222] 186.2 102.9 [272] 85.47 101.1 [261] 150.4 44.96 [263] 78.96 28.09 107.9 22.99 87.62 32.07 180.9 [98] 127.6 158.9 204.4 232.0 168.9 118.7 47.88 183.9 238.0 50.94 131.3 173.0 88.91 65.38 91.22 This page intentionally left blank SEVENTH EDITION Basic Chemistry Steven S Zumdahl University of Illinois Donald J DeCoste University of Illinois Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States Basic Chemistry, Seventh Edition Steven S Zumdahl and Donald J DeCoste Publisher: Charles Hartford Development Editor: Alyssa White Senior Content Project Manager: Cathy Brooks Art Directors: Jill 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permissions questions can be e-mailed to permissionrequest@cengage.com Senior Marketing Communications Manager: Linda Yip Library of Congress Control Number: 2009934046 Print Buyer: Judy Inouye ISBN-13: 978-0-538-73637-4 Permissions Editor: Roberta Broyer ISBN-10: 0-538-73637-2 Text Designer: Janet Theurer Production Service: Graphic World Inc Production Project Manager, Graphic World: Megan Greiner Copyeditor: Graphic World Inc Illustrators: Jessica Broekman, Rossi Art, and Graphic World Inc Cover Designer: Brian Salisbury Cover Image: Radius Images/Photo Library OWL Producers: Stephen Battisti, Cindy Stein, David Hart (Center for Educational Software Development; University of Massachusetts, Amherst) Compositor: Graphic World Inc Brooks/Cole 20 Davis Drive Belmont, CA 94002-3098 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at international.cengage.com/region Cengage Learning products are represented in Canada by Nelson Education, Ltd For your course and learning solutions, visit academic.cengage.com Purchase any of our products at your local college store or at our preferred online store www.ichapters.com Printed in the United States of America 13 12 11 10 09 BRIEF CONTENTS 10 11 12 13 14 15 16 Chemistry: An Introduction Measurements and Calculations 14 Matter 56 Chemical Foundations: Elements, Atoms, and Ions 74 Nomenclature 114 Chemical Reactions: An Introduction 144 Reactions in Aqueous Solutions 166 Chemical Composition 204 Chemical Quantities 248 Energy 288 Modern Atomic Theory 322 Chemical Bonding 358 Gases 402 Liquids and Solids 446 Solutions 474 Acids and Bases 514 v This page intentionally left blank 256 Chapter Chemical Quantities molar masses of Al and I2 (26.98 g/mol and 253.8 g/mol) that the mass of mol I2 is almost 10 times as great as that of mol Al We also know that we need a greater number of moles of I2 compared with Al (by a 3:2 ratio) So, we should expect to get a mass of I2 that is well over 10 times as great as 35.0 g, and we did EXAMPLE 9.4 Using Mass–Mole Conversions with Mole Ratios Propane, C3H8, when used as a fuel, reacts with oxygen to produce carbon dioxide and water according to the following unbalanced equation: C3H8(g) ϩ O2(g) → CO2(g) ϩ H2O(g) What mass of oxygen will be required to react exactly with 96.1 g of propane? SOLUTION Where Are We Going? We want to determine the mass of O2 required to react exactly with 96.1 g C3H8 What Do We Know? • The unbalanced equation for the reaction is C3H8 ϩ O2 S CO2 ϩ H2O • We start with 96.1 g mol C3H8 • We know the atomic masses of carbon, hydrogen, and oxygen from the periodic table What Do We Need To Know? • We need to know the balanced equation • We need the molar masses of O2 and C3H8 Always balance the equation for the reaction first How Do We Get There? To deal with the amounts of reactants and products, we first need the balanced equation for this reaction: C3H8(g) ϩ 5O2(g) → 3CO2(g) ϩ 4H2O(g) Our problem, in schematic form, is MATH SKILL BUILDER Remember that to show the correct significant figures in each step, we are rounding off after each calculation In doing problems, you should carry extra numbers, rounding off only at the end 96.1 g propane requires ? grams O2 Using the ideas we developed when we discussed the aluminum–iodine reaction, we will proceed as follows: We are given the number of grams of propane, so we must convert to moles of propane (C3H8) Then we can use the coefficients in the balanced equation to determine the moles of oxygen (O2) required Finally, we will use the molar mass of O2 to calculate grams of oxygen 9.3 Mass Calculations 257 We can sketch this strategy as follows: ϩ C3H8(g) 96.1 g C3H8 3CO2(g) ϩ 4H2O(g) ? moles C3H8 ? moles O2 Thus the first question we must answer is, How many moles of propane are present in 96.1 g of propane? The molar mass of propane is 44.09 g (3 ϫ 12.01 ϩ ϫ 1.008) The moles of propane present can be calculated as follows: 96.1 g C3H8 ϫ S ? grams O2 1 5O2(g) mol C3H8 ϭ 2.18 mol C3H8 44.09 g C3H8 Next we recognize that each mole of propane reacts with mol of oxygen This gives us the equivalence statement mol C3H8 ϭ mol O2 from which we construct the mole ratio mol O2 mol C3H8 that we need to convert from moles of propane molecules to moles of oxygen molecules 2.18 mol C3H8 ϫ mol O2 ϭ 10.9 mol O2 mol C3H8 Notice that the mole ratio is set up so that the moles of C3H8 cancel and the resulting units are moles of O2 Because the original question asked for the mass of oxygen needed to react with 96.1 g of propane, we must convert the 10.9 mol O2 to grams, using the molar mass of O2 (32.00 ϭ ϫ 16.00) 10.9 mol O2 ϫ 32.0 g O2 ϭ 349 g O2 mol O2 Therefore, 349 g of oxygen is required to burn 96.1 g of propane We can summarize this problem by writing out a “conversion string” that shows how the problem was done 96.1 g C3H8 ϫ mol O2 mol C3H8 ϫ ϫ 32.0 g O2 ϭ 349 g O2 mol C3H8 44.09 g C3H8 mol O2 258 Chapter Chemical Quantities MATH SKILL BUILDER Use units as a check to see that you have used the correct conversion factors (mole ratios) This is a convenient way to make sure the final units are correct The procedure we have followed is summarized below ϩ C3H8(g) 5O2(g) S 96.1 g C3H8 349 g O2 Use molar mass of C3H8 (44.09 g) Use molar mass of O2 (32.0 g) 2.18 mol C3H8 Use mole ratio: mol O2 mol C3H8 3CO2(g) ϩ 4H2O(g) 10.9 mol O2 R E A L I T Y C H E C K According to the balanced equation, more O2 is required (by moles) than C3H8 by a factor of Because the molar mass of C3H8 is not much greater than that of O2, we should expect that a greater mass of oxygen is required, and our answer confirms this Self-Check EXERCISE 9.2 What mass of carbon dioxide is produced when 96.1 g of propane reacts with sufficient oxygen? See Problems 9.23 through 9.26 ■ Self-Check EXERCISE 9.3 Calculate the mass of water formed by the complete reaction of 96.1 g of propane with oxygen See Problems 9.23 through 9.26 ■ So far in this chapter, we have spent considerable time “thinking through” the procedures for calculating the masses of reactants and products in chemical reactions We can summarize these procedures in the following steps: Steps for Calculating the Masses of Reactants and Products in Chemical Reactions Step Balance the equation for the reaction Step Convert the masses of reactants or products to moles Step Use the balanced equation to set up the appropriate mole ratio(s) Step Use the mole ratio(s) to calculate the number of moles of the desired reactant or product Step Convert from moles back to masses 9.3 Mass Calculations 259 The process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction is called stoichiometry (pronounced stoi´ ke¯-˘ om´ i-tre¯) Chemists say that the balanced equation for a chemical reaction describes the stoichiometry of the reaction We will now consider a few more examples that involve chemical stoichiometry Because real-world examples often involve very large or very small masses of chemicals that are most conveniently expressed by using scientific notation, we will deal with such a case in Example 9.5 EXAMPLE 9.5 For a review of writing formulas of ionic compounds, see Chapter Stoichiometric Calculations: Using Scientific Notation Solid lithium hydroxide has been used in space vehicles to remove exhaled carbon dioxide from the living environment The products are solid lithium carbonate and liquid water What mass of gaseous carbon dioxide can 1.00 ϫ 103 g of lithium hydroxide absorb? SOLUTION Where Are We Going? We want to determine the mass of carbon dioxide absorbed by 1.00 ϫ 103 g of lithium hydroxide What Do We Know? • The names of the reactants and products • We start with 1.00 ϫ 103 g of lithium hydroxide • We can obtain the atomic masses from the periodic table What Do We Need To Know? • We need to know the balanced equation for the reaction, but we first have to write the formulas for the reactants and products • We need the molar masses of lithium hydroxide and carbon dioxide How Do We Get There? Step Using the description of the reaction, we can write the unbalanced equation LiOH(s) ϩ CO2(g) → Li2CO3(s) ϩ H2O(l) The balanced equation is 2LiOH(s) ϩ CO2(g) → Li2CO3(s) ϩ H2O(l) Check this for yourself Step We convert the given mass of LiOH to moles, using the molar mass of LiOH, which is 6.941 g ϩ 16.00 g ϩ 1.008 g ϭ 23.95 g NASA 1.00 ϫ 103 g LiOH ϫ Astronaut Sidney M Gutierrez changes the lithium hydroxide canisters on space shuttle Columbia mol LiOH ϭ 41.8 mol LiOH 23.95 g LiOH Step The appropriate mole ratio is mol CO2 mol LiOH 260 Chapter Chemical Quantities Step Using this mole ratio, we calculate the moles of CO2 needed to react with the given mass of LiOH 41.8 mol LiOH ϫ mol CO2 ϭ 20.9 mol CO2 mol LiOH Step We calculate the mass of CO2 by using its molar mass (44.01 g) 20.9 mol CO2 ϫ 44.01 g CO2 ϭ 920 g CO2 ϭ 9.20 ϫ 102 g CO2 mol CO2 Thus 1.00 ϫ 103 g of LiOH(s) can absorb 920 g of CO2(g) We can summarize this problem as follows: 2LiOH(s) ϩ CO2(g) 1.00 ϫ 103 g LiOH Li2CO3(s) ϩ H2O(l) Grams of CO2 Use molar mass of CO Use molar mass of LiOH Moles of LiOH S Use mole ratio between CO2 and LiOH Moles of CO2 The conversion string is 1.00 ϫ 103 g LiOH ϫ MATH SKILL BUILDER Carrying extra significant figures and rounding off only at the end gives an answer of 919 g CO2 Self-Check 44.01 g CO2 mol CO2 mol LiOH ϫ ϫ 23.95 g LiOH mol LiOH mol CO2 ϭ 9.19 ϫ 102 g CO2 R E A L I T Y C H E C K According to the balanced equation, there is a 2:1 mole ratio of LiOH to CO2 There is about a 1:2 molar mass ratio of LiOH:CO2 (23.95:44.01) We should expect about the same mass of CO2 as LiOH, and our answer confirms this (1000 g compared to 920 g) EXERCISE 9.4 Hydrofluoric acid, an aqueous solution containing dissolved hydrogen fluoride, is used to etch glass by reacting with the silica, SiO2, in the glass to produce gaseous silicon tetrafluoride and liquid water The unbalanced equation is HF(aq) ϩ SiO2(s) → SiF4(g) ϩ H2O(l) a Calculate the mass of hydrogen fluoride needed to react with 5.68 g of silica Hint: Think carefully about this problem What is the balanced equation for the reaction? What is given? What you need to calculate? Sketch a map of the problem before you the calculations b Calculate the mass of water produced in the reaction described in part a See Problems 9.23 through 9.26 ■ 9.3 Mass Calculations EXAMPLE 9.6 261 Stoichiometric Calculations: Comparing Two Reactions Baking soda, NaHCO3, is often used as an antacid It neutralizes excess hydrochloric acid secreted by the stomach The balanced equation for the reaction is NaHCO3(s) ϩ HCl(aq) → NaCl(aq) ϩ H2O(l) ϩ CO2(g) Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, Mg(OH)2, is also used as an antacid The balanced equation for the reaction is Mg(OH)2(s) ϩ 2HCl(aq) → 2H2O(l) ϩ MgCl2(aq) Which antacid can consume the most stomach acid, 1.00 g of NaHCO3 or 1.00 g of Mg(OH)2? SOLUTION Where Are We Going? We want to compare the neutralizing power of two antacids, NaHCO3 and Mg(OH)2 In other words, how many moles of HCl will react with 1.00 g of each antacid? What Do We Know? • The balanced equations for the reactions • We start with 1.00 g each of NaHCO3 and Mg(OH)2 • We can obtain atomic masses from the periodic table What Do We Need to Know? • We need the molar masses of NaHCO3 and Mg(OH)2 How Do We Get There? The antacid that reacts with the larger number of moles of HCl is more effective because it will neutralize more moles of acid A schematic for this procedure is Antacid ϩ HCl → Products 1.00 g antacid Use molar mass of antacid Moles of antacid Use mole ratio from balanced equation Moles of HCl Notice that in this case we not need to calculate how many grams of HCl react; we can answer the question with moles of HCl We will now solve this problem for each antacid Both of the equations are balanced, so we can proceed with the calculations Using the molar mass of NaHCO3, which is 22.99 g ϩ 1.008 g ϩ 12.01 g ϩ 3(16.00 g) ϭ 84.01 g, we determine the moles of NaHCO3 in 1.00 g of NaHCO3 C H E M I S T R Y I N F OCUS Cars of the Future There is a great deal of concern about how we are going to sustain our personal transportation system in the face of looming petroleum shortages (and the resultant high costs) and the challenges of global warming The era of large gasoline-powered cars as the primary means of transportation in the United States seems to be drawing to a close The fact that discoveries of petroleum are not keeping up with the rapidly increasing global demand for oil has caused skyrocketing prices In addition, the combustion of gasoline produces carbon dioxide (about lb of CO2 per mile for many cars), which has been implicated in global warming So what will the car of the future in the United States be like? It seems that we are moving rapidly toward cars that have an electrical component as part of the power train Hybrid cars, which use a small gasoline motor in conjunction with a powerful battery, have been quite successful By supplementing the small gasoline engine, which would be inadequate by itself, with power from the battery, the typical hybrid gets 40 to 50 miles per gallon of gasoline In this type of hybrid car, both the battery and the engine are used to power the wheels of the car as needed Another type of system that involves both a gasoline engine and a battery is the so-called “plug-in hybrid.” In this car, the battery is the sole source of power to the car’s wheels The gasoline engine is only used to charge the battery as needed One example of this type of car is the Chevrolet Volt, which is slated for production in 2010 The Volt is being designed to run about 40 miles on each battery charge The car would be plugged into a normal household electric outlet overnight to recharge the battery For trips longer than 40 miles, the gasoline engine would turn on to charge the battery Another type of “electrical car” being tested is one powered by a hydrogen–oxygen fuel cell An example of such a car is the Honda FCX Clarity The Clarity stores hydrogen in a tank that holds 4.1 kg of H2 at a pressure of 5000 lb per square inch The H2 is sent to a fuel cell, where it reacts with oxygen from the air supplied by an air compressor About 200 of these cars are going to be tested in Southern Califor- 1.00 g NaHCO3 ϫ mol NaHCO3 ϭ 0.0119 mol NaHCO3 84.01 g NaHCO3 ϭ 1.19 ϫ 10Ϫ2 mol NaHCO3 Next we determine the moles of HCl, using the mole ratio 1.19 ϫ 10Ϫ2 mol NaHCO3 ϫ mol HCl mol NaHCO3 mol HCl ϭ 1.19 ϫ 10Ϫ2 mol HCl mol NaHCO3 Thus 1.00 g of NaHCO3 neutralizes 1.19 ϫ 10Ϫ2 mol HCl We need to compare this to the number of moles of HCl that 1.00 g of Mg(OH)2 neutralizes Using the molar mass of Mg(OH)2, which is 24.31 g ϩ 2(16.00 g) ϩ 2(1.008 g) ϭ 58.33 g, we determine the moles of Mg(OH)2 in 1.00 g of Mg(OH)2 1.00 g Mg(OH)2 ϫ 262 mol Mg(OH)2 58.33 g Mg(OH)2 ϭ 0.0171 mol Mg(OH)2 ϭ 1.71 ϫ 10Ϫ2 mol Mg(OH)2 process Intense research is now being conducted to find economically feasible ways to produce H2 from water It appears that our cars of the future will have an electrical drive component Whether it will involve a conventional battery or a fuel cell will depend on technological developments and costs Courtesy Honda Motors USA Horizon Fuel Cell Technologies Pte Ltd nia in the next years, leased to people who live near one of the three 24-hour public hydrogen stations The Clarity gets about 72 miles per kilogram of hydrogen One obvious advantage of a car powered by an H2/O2 fuel cell is that the combustion product is only H2O However, there is a catch (it seems there is always a catch) Currently, 95% of hydrogen produced is obtained by natural gas (CH4), and CO2 is a by-product of this Even model cars are becoming “green.” The H-racer from Horizon Fuel Cell Technologies uses a hydrogen–oxygen fuel cell The Honda FCX Clarity is powered by a hydrogen–oxygen fuel cell To determine the moles of HCl that react with this amount of Mg(OH)2, we mol HCl use the mole ratio mol Mg(OH)2 1.71 ϫ 10Ϫ2 mol Mg(OH)2 ϫ mol HCl ϭ 3.42 ϫ 10Ϫ2 mol HCl mol Mg(OH)2 Therefore, 1.00 g of Mg(OH)2 neutralizes 3.42 ϫ 10Ϫ2 mol HCl We have already calculated that 1.00 g of NaHCO3 neutralizes only 1.19 ϫ 10Ϫ2 mol HCl Therefore, Mg(OH)2 is a more effective antacid than NaHCO3 on a mass basis Self-Check EXERCISE 9.5 In Example 9.6 we answered one of the questions we posed in the introduction to this chapter Now let’s see if you can answer the other question posed there Determine what mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the reaction CO(g) ϩ 2H2(g) → CH3OH(l) See Problem 9.39 ■ 263 264 Chapter Chemical Quantities 9.4 The Concept of Limiting Reactants OBJECTIVE: To understand what is meant by the term “limiting reactant.” Earlier in this chapter, we discussed making sandwiches Recall that the sandwich-making process could be described as follows: pieces bread ϩ slices meat ϩ slice cheese S sandwich In our earlier discussion, we always purchased the ingredients in the correct ratios so that we used all the components, with nothing left over Now assume that you came to work one day and found the following quantities of ingredients: 20 slices of bread 24 slices of meat 12 slices of cheese How many sandwiches can you make? What will be left over? To solve this problem, let’s see how many sandwiches we can make with each component sandwich ϭ 10 sandwiches slices of bread sandwich Meat: 24 slices meat ϫ ϭ sandwiches slices of meat sandwich Cheese: 12 slices cheese ϫ ϭ 12 sandwiches slice of cheese Bread: 20 slices bread ϫ How many sandwiches can you make? The answer is When you run out of meat, you must stop making sandwiches The meat is the limiting ingredient What you have left over? Making sandwiches requires 16 pieces of bread You started with 20 pieces, so you have pieces of bread left You also used pieces of cheese for the sandwiches, so you have 12 Ϫ ϭ pieces of cheese left In this example, the ingredient present in the largest number (the meat) was actually the component that limited the number of sandwiches you could make This situation arose because each sandwich required slices of meat—more than the quantity required of any other ingredient You probably have been dealing with limiting-reactant problems for most of your life For example, suppose a lemonade recipe calls for cup of sugar for every lemons You have 12 lemons and cups of sugar Which ingredient is limiting, the lemons or the sugar?* ▲ Module 8a: Stoichiometry and Limiting Reactants (Pt 1) covers concepts in this section A Closer Look When molecules react with each other to form products, considerations very similar to those involved in making sandwiches arise We can illustrate these ideas with the reaction of N2(g) and H2(g) to form NH3(g): N2(g) ϩ 3H2(g) → 2NH3(g) *The ratio of lemons to sugar that the recipe calls for is lemons to cup of sugar We can calculate the number of lemons required to “react with” the cups of sugar as follows: cups sugar ϫ lemons ϭ 18 lemons cup sugar Thus 18 lemons would be required to use up cups of sugar However, we have only 12 lemons, so the lemons are limiting 9.4 The Concept of Limiting Reactants 265 Consider the following container of N2(g) and H2(g): = H2 = N2 What will this container look like if the reaction between N2 and H2 proceeds to completion? To answer this question, you need to remember that each N2 requires H2 molecules to form NH3 To make things more clear, we will circle groups of reactants: Before the reaction After the reaction = H2 = N2 = NH3 In this case, the mixture of N2 and H2 contained just the number of molecules needed to form NH3 with nothing left over That is, the ratio of the number of H2 molecules to N2 molecules was 15 H2 H2 ϭ N2 N2 This ratio exactly matches the numbers in the balanced equation 3H2(g) ϩ N2(g) → 2NH3(g) This type of mixture is called a stoichiometric mixture—one that contains the relative amounts of reactants that matches the numbers in the balanced equation In this case, all reactants will be consumed to form products Now consider another container of N2(g) and H2(g): = H2 = N2 266 Chapter Chemical Quantities What will the container look like if the reaction between N2(g) and H2(g) proceeds to completion? Remember that each N2 requires H2 Circling groups of reactants, we have Before the reaction After the reaction = H2 = N2 = NH3 In this case, the hydrogen (H2) is limiting That is, the H2 molecules are used up before all of the N2 molecules are consumed In this situation, the amount of hydrogen limits the amount of product (ammonia) that can form—hydrogen is the limiting reactant Some N2 molecules are left over in this case because the reaction runs out of H2 molecules first To determine how much product can be formed from a given mixture of reactants, we have to look for the reactant that is limiting—the one that runs out first and thus limits the amount of product that can form In some cases, the mixture of reactants might be stoichiometric—that is, all reactants run out at the same time In general, however, you cannot assume that a given mixture of reactants is a stoichiometric mixture, so you must determine whether one of the reactants is limiting The reactant that runs out first and thus limits the amounts of products that can form is called the limiting reactant (limiting reagent) To this point, we have considered examples where the numbers of reactant molecules could be counted In “real life” you can’t count the molecules directly—you can’t see them, and, even if you could, there would be far too many to count Instead, you must count by weighing We must therefore explore how to find the limiting reactant, given the masses of the reactants 9.5 Calculations Involving a Limiting Reactant OBJECTIVES: To learn to recognize the limiting reactant in a reaction • To learn to use the limiting reactant to stoichiometric calculations Manufacturers of cars, bicycles, and appliances order parts in the same proportion as they are used in their products For example, auto manufacturers order four times as many wheels as engines and bicycle manufacturers order twice as many pedals as seats Likewise, when chemicals are mixed together so that they can undergo a reaction, they are often mixed in stoichiometric quantities—that is, in exactly the correct amounts so that all 9.5 Calculations Involving a Limiting Reactant 267 reactants “run out” (are used up) at the same time To clarify this concept, we will consider the production of hydrogen for use in the manufacture of ammonia Ammonia, a very important fertilizer itself and a starting material for other fertilizers, is made by combining nitrogen from the air with hydrogen The hydrogen for this process is produced by the reaction of methane with water according to the balanced equation AP photo/Kevin Rivoli CH4(g) ϩ H2O(g) → 3H2(g) ϩ CO(g) Farmer Rodney Donala looks out over his corn fields in front of his 30,000-gallon tank (at right) of anhydrous ammonia, a liquid fertilizer Module 8b: Stoichiometry and Limiting Reactants (Pt 2) covers concepts in this section Let’s consider the question, What mass of water is required to react exactly with 249 g of methane? That is, how much water will just use up all of the 249 g of methane, leaving no methane or water remaining? This problem requires the same strategies we developed in the previous section Again, drawing a map of the problem is helpful CH4(g) ϩ H2O(g) 249 g CH4 3H2(g) ϩ CO(g) Grams of H2O Use molar mass of CH Moles of CH4 → Use molar mass of H 2O Use mole ratio from balanced equation Moles of H2O We first convert the mass of CH4 to moles, using the molar mass of CH4 (16.04 g/mol) 249 g CH4 ϫ mol CH4 ϭ 15.5 mol CH4 16.04 g CH4 Because in the balanced equation mol CH4 reacts with mol H2O, we have 15.5 mol CH4 ϫ mol H2O ϭ 15.5 mol H2O mol CH4 Therefore, 15.5 mol H2O will react exactly with the given mass of CH4 Converting 15.5 mol H2O to grams of H2O (molar mass ϭ 18.02 g/mol) gives 15.5 mol H2O ϫ The reactant that is consumed first limits the amounts of products that can form 18.02 g H2O ϭ 279 g H2O mol H2O This result means that if 249 g of methane is mixed with 279 g of water, both reactants will “run out” at the same time The reactants have been mixed in stoichiometric quantities If, on the other hand, 249 g of methane is mixed with 300 g of water, the methane will be consumed before the water runs out The water will be in excess In this case, the quantity of products formed will be determined by the quantity of methane present Once the methane is consumed, no more products can be formed, even though some water still remains In this situation, the amount of methane limits the amount of products that can be formed Recall from Section 9.4 that we call such a reactant the limiting reactant or the limiting reagent In any stoichiometry problem where reactants 268 Chapter Chemical Quantities Figure 9.1 A mixture of 5CH4 and 3H2O molecules undergoes the reaction CH4(g) ϩ H2O(g) S 3H2(g) ϩ CO(g) Note that the H2O molecules are used up first, leaving two CH4 molecules unreacted are not mixed in stoichiometric quantities, it is essential to determine which reactant is limiting to calculate correctly the amounts of products that will be formed This concept is illustrated in Figure 9.1 Note from this figure that because there are fewer water molecules than CH4 molecules, the water is consumed first After the water molecules are gone, no more products can form So in this case water is the limiting reactant EXAMPLE 9.7 H Group N Group Stoichiometric Calculations: Identifying the Limiting Reactant Suppose 25.0 kg (2.50 ϫ 104 g) of nitrogen gas and 5.00 kg (5.00 ϫ 103 g) of hydrogen gas are mixed and reacted to form ammonia Calculate the mass of ammonia produced when this reaction is run to completion SOLUTION Where Are We Going? We want to determine the mass of ammonia produced given the masses of both reactants What Do We Know? • The names of the reactants and products • We start with 2.50 ϫ 104 g of nitrogen gas and 5.00 ϫ 103 g of hydrogen gas • We can obtain the atomic masses from the periodic table What Do We Need To Know? • We need to know the balanced equation for the reaction, but we first have to write the formulas for the reactants and products • We need the molar masses of nitrogen gas, hydrogen gas, and ammonia • We need to determine the limiting reactant 9.5 Calculations Involving a Limiting Reactant 269 How Do We Get There? The unbalanced equation for this reaction is N2(g) ϩ H2(g) → NH3(g) which leads to the balanced equation N2(g) ϩ 3H2(g) → 2NH3(g) This problem is different from the others we have done so far in that we are mixing specified amounts of two reactants together To know how much product forms, we must determine which reactant is consumed first That is, we must determine which is the limiting reactant in this experiment To so we must add a step to our normal procedure We can map this process as follows: ϩ N2(g) → 3H2(g) 2.50 ϫ 104 g N2 5.00 ϫ 103 g H2 Use molar mass of N Use molar mass of H Moles of N2 Use mole ratios to determine limiting reactant 2NH3(g) Moles of H2 Moles of limiting reactant We will use the moles of the limiting reactant to calculate the moles and then the grams of the product N2(g) ϩ 3H2(g) → 2NH3(g) Grams of NH3 Use molar mass of NH3 Moles of limiting reactant Use mole ratios involving limiting reactant Moles of NH3 We first calculate the moles of the two reactants present: mol N2 ϭ 8.92 ϫ 102 mol N2 28.02 g N2 mol H2 5.00 ϫ 103 g H2 ϫ ϭ 2.48 ϫ 103 mol H2 2.016 g H2 2.50 ϫ 104 g N2 ϫ 270 Chapter Chemical Quantities Now we must determine which reactant is limiting (will be consumed first) We have 8.92 ϫ 102 moles of N2 Let’s determine how many moles of H2 are required to react with this much N2 Because mole of N2 reacts with moles of H2, the number of moles of H2 we need to react completely with 8.92 ϫ 102 moles of N2 is determined as follows: mol H2 mol N2 8.92 ϫ 102 mol N2 8.92 ϫ 102 mol N2 ϫ Moles of H2 required mol H2 ϭ 2.68 ϫ 103 mol H2 mol N2 Is N2 or H2 the limiting reactant? The answer comes from the comparison Moles of H2 available less than 2.48 ϫ 103 Moles of H2 required 2.68 ϫ 103 We see that 8.92 ϫ 102 mol N2 requires 2.68 ϫ 103 mol H2 to react completely However, only 2.48 ϫ 103 mol H2 is present This means that the hydrogen will be consumed before the nitrogen runs out, so hydrogen is the limiting reactant in this particular situation Note that in our effort to determine the limiting reactant, we could have started instead with the given amount of hydrogen and calculated the moles of nitrogen required mol N2 mol H2 2.48 ϫ 103 mol H2 2.48 ϫ 103 mol H2 ϫ Moles of N2 required mol N2 ϭ 8.27 ϫ 102 mol N2 mol H2 Thus 2.48 ϫ 103 mol H2 requires 8.27 ϫ 102 mol N2 Because 8.92 ϫ 102 mol N2 is actually present, the nitrogen is in excess Moles of N2 available 8.92 ϫ 102 Always check to see which, if any, reactant is limiting when you are given the amounts of two or more reactants greater than Moles of N2 required 8.27 ϫ 102 If nitrogen is in excess, hydrogen will “run out” first; again we find that hydrogen limits the amount of ammonia formed Because the moles of H2 present are limiting, we must use this quantity to determine the moles of NH3 that can form 2.48 ϫ 103 mol H2 ϫ mol NH3 ϭ 1.65 ϫ 103 mol NH3 mol H2 Next we convert moles of NH3 to mass of NH3 1.65 ϫ 103 mol NH3 ϫ 17.03 g NH3 ϭ 2.81 ϫ 104 g NH3 ϭ 28.1 kg NH3 mol NH3 Therefore, 25.0 kg of N2 and 5.00 kg of H2 can form 28.1 kg of NH3