1. Chapter Six Chemical Reactions: Classification and Mass Relationships Fundamentals of General, Organic, and Biological Chemistry 5th Edition James E. Mayhugh Oklahoma City University  2007 Prentice Hall, Inc.

2. Prentice Hall © 2007 Chapter Six 2 Outline ► 6.1 Chemical Equations ► 6.2 Balancing Chemical Equations ► 6.3 Avogadro’s Number and the Mole ► 6.4 Gram–Mole Conversions ► 6.5 Mole Relationships and Chemical Equations ► 6.6 Mass Relationships and Chemical Equations ► 6.7 Percent Yield ► 6.8 Classes of Chemical Reactions ► 6.9 Precipitation Reactions and Solubility Guidelines ► 6.10 Acids, Bases, and Neutralization Reactions ► 6.11 Redox Reactions ► 6.12 Recognizing Redox Reactions ► 6.13 Net Ionic Equations

3. Prentice Hall © 2007 Chapter Six 3 6.1 Chemical Equations ► Chemical equation: An expression in which symbols are used to represent a chemical reaction. ► Reactant: A substance that undergoes change in a chemical reaction and is written on the left side of the reaction arrow in a chemical equation. ► Product: A substance that is formed in a chemical reaction and is written on the right side of the reaction arrow in a chemical equation.

4. Prentice Hall © 2007 Chapter Six 4 ► The numbers and kinds of atoms must be the same on both sides of the reaction arrow. ► Numbers in front of formulas are called coefficients; they multiply all the atoms in a formula. ► The symbol 2 NaHCO 3 indicates two units of sodium bicarbonate, which contains 2 Na, 2 H, 2 C, and 6 O. ► Substances involved in chemical reactions may be solids, liquids, gases, or they may be in solution. ► This information is added to an equation by placing the appropriate symbols after the formulas: ► Solid=(s) Liquid=(l) Gas=(g) Aqueous solution=(aq)

5. Prentice Hall © 2007 Chapter Six 5 6.2 Balancing Chemical Equations ► Balancing chemical equations can be done using four basic steps: ► STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. ► STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element.

6. Prentice Hall © 2007 Chapter Six 6 ► A polyatomic ion appearing on both sides of an equation can be treated as a single unit. ► ►STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same.

7. Prentice Hall © 2007 Chapter Six 7 ► STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. ► The equation: 2 H 2 SO 4 + 4 NaOH  2 Na 2 SO 4 + 4 H 2 O is balanced, but can be simplified by dividing all coefficients by 2: H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O ► Hint: If an equation contains a pure element as a product or reactant it helps to assign that element’s coefficient last.

8. a CH 3 CH 2 OH + b O 2  c CO 2 + d H 2 O When the above equation is balanced, the coefficients a, b, c, and d are 1.a=1, b=1, c=1, d=1 2.a=1, b=2, c=2, d=3 3.a=1, b=3, c=2, d=3 4.a=2, b=7, c=4, d=6

9. a CH 3 CH 2 OH + b O 2  c CO 2 + d H 2 O When the above equation is balanced, the coefficients a, b, c, and d are 1.a=1, b=1, c=1, d=1 2.a=1, b=2, c=2, d=3 3.a=1, b=3, c=2, d=3 4.a=2, b=7, c=4, d=6

10. Determining Products ► Chapter 4 illustrated how to make molecules based on charge, we use the same procedure here. ► 3Mg(NO 3 ) 2 + 2Na 3 PO 4  Mg 3 (PO 4 ) 2 + 6NaNO 3 Prentice Hall © 2007 Chapter Six 10

11. ► Mg(NO 3 ) 2 + Na 3 PO 4  1.Identify charges, from group number or ion charge Mg 2+ (NO 3 1- ) 2 + Na 3 1+ PO 4 3-  2.Visualize charges, moving across to the product side Mg 2+ (NO 3 1- ) 2 + Na 3 1+ PO 4 3-  Determining Products Prentice Hall © 2007 Chapter Six 11 1+ 1-1+ 1-1+ 1-1+ 1- 2 + 3 -

12. 2.Visualize charges, moving across to the product side Mg 2+ (NO 3 1- ) 2 + Na 3 1+ PO 4 3-  3.Write the ion with That charge; cation first Mg 2+ (NO 3 1- ) 2 + Na 3 1+ PO 4 3-  Mg 2+ PO 4 3- + Na 1+ NO 3 1- Note, only ion and ion charges, i.e. Na 1+ not Na 3 1+ Determining Products Prentice Hall © 2007 Chapter Six 12 1+ 1-1+ 1-1+ 1-1+ 1- 2 + 3 -

13. 3.Write the ion with charge Mg 2+ (NO 3 1- ) 2 + Na 3 1+ PO 4 3-  Mg 2+ PO 4 3- + Na 1+ NO 3 1- 4.Balance the charges in the new molecules Mg(NO 3 ) 2 + Na 3 PO 4  Mg 3 (PO 4 ) 2 + NaNO 3 Determining Products Prentice Hall © 2007 Chapter Six 13

14. Determining Products 4.Balance the charges in the new molecules Mg(NO 3 ) 2 + Na 3 PO 4  Mg 3 (PO 4 ) 2 + NaNO 3 5.Balance atoms 3Mg(NO 3 ) 2 + 2Na 3 PO 4  Mg 3 (PO 4 ) 2 + 6NaNO 3 Prentice Hall © 2007 Chapter Six 14

15. Learning check Pb(NO 3 ) 2 + NaBr  Ba(OH) 2 + Li 2 SO 4  Ti 2 S 3 + Mg(H 2 PO 4 ) 2  FeCl 2 + Al(OH) 3  Prentice Hall © 2007 Chapter Six 15

16. Learning check Pb(NO 3 ) 2 + 2NaBr  PbBr 2 + 2NaNO 3 Ba(OH) 2 + Li 2 SO 4  BaSO 4 + 2LiOH Ti 2 S 3 + 3Mg(H 2 PO 4 ) 2  2Ti(H 2 PO 4 ) 3 + 3MgS 3FeCl 2 + 2Al(OH) 3  3Fe(OH) 2 + 2AlCl 3 Prentice Hall © 2007 Chapter Six 16

17. Prentice Hall © 2007 Chapter Six 17 6.3 Avogadro’s Number and the Mole ► Molecular weight: The sum of atomic weights of all atoms in a molecule. ► Formula weight: The sum of atomic weights of all atoms in one formula unit of any compound. ► Mole: One mole of any substance is the amount whose mass in grams (molar mass) is numerically equal to its molecular or formula weight. ► Avogadro’s number: The number of molecules or formula units in a mole. N A = 6.022 x 10 23

18. 18 ► A collection term indicates a specific number of items. ► For example, 1 dozen doughnuts contains 12 doughnuts. ► 1 ream of paper means 500 sheets. ► 1 case is 24 cans. Counting

19. 19 Counting In Chemistry ► Chemistry is a quantitative science — we need a “counting unit.” ► The MOLE ► 1 mole is the number of atoms in 12.0 g of 12 C. ► There are 6.02 x 10 23 carbon atoms in 12.0 g of 12 C.

20. 20 Avogadro’s Number Particles in a Mole 6.02 x 10 23 Amedeo Avogadro 1776-1856

21.   An Avogadro's number of soft drink cans would cover the surface of the earth to a depth of over 200 miles.   If you spread Avogadro's number of unpopped popcorn kernels across the USA, the entire country would be covered in popcorn to a depth of over 9 miles.   If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

22. 22 Don’t confuse a chemistry Mole with An informer / spy Rodent that burrows in the ground A tunneling machine Wave break Spicy Mexican sauce

23. 23 1 Mole “XYZ” = 6.022  10 23 “XYZ” 1 Mole Baseballs = 6.022  10 23 baseballs 1 Mole Pineapples = 6.022  10 23 Pineapples 1 Mole Calculators = 6.022  10 23 Calculators

24. 24 Just as 1 doz eggs = 12 eggs so 1 mol of eggs = 6.022  10 23 eggs A mole is a “Number” 1 mol of H = 6.022  10 23 atoms of H 1 mol of O = 6.022  10 23 atoms of O 1 mol of Al = 6.022  10 23 atoms of Al 1 mol of Cr = 6.022  10 23 atoms of Cr

25. 25 A. Calculate the number of atoms in 2.0 moles of Al. 1) 2.0 Al atoms 2) 3.0 x 10 23 Al atoms 3) 1.2 x 10 24 Al atoms B. Calculate the number of moles of S in 1.8 x 10 24 S. 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x 10 48 mole S atoms Learning Check

26. 26 A. Calculate the number of atoms in 2.0 moles of Al. 3) 1.2 x 10 24 Al atoms 2.0 moles Al x 6.02 x 10 23 Al atoms 1 mole Al B. Calculate the number of moles of S in 1.8 x 10 24 S. 2) 3.0 mole S atoms 1.8 x 10 24 S atoms x 1 mole S 6.02 x 10 23 S atoms Solution

27. Prentice Hall © 2007 Chapter Six 27 6.4 Gram – Mole Conversions ► Molar mass = Mass of 1 mole of a substance. = Mass of 6.022 x 10 23 molecules of a substance. = Molecular (formula) weight of substance in grams. ► Molar mass serves as a conversion factor between numbers of moles and mass. If you know how many moles you have, you can calculate their mass; if you know the mass of a sample, you can calculate the number of moles.

28. Chapter Seven

29. 29 Molar Mass of K 3 PO 4 ElementNumber of atoms Atomic MassTotal Mass in K 3 PO 4 K339.1 amu 117.3 amu P131.0 amu O416.0 amu 64.0 aum K 3 PO 4 8 212.3 amu amu  MW 212.3 g/mol 1 mol K 3 PO 4 = 212.3 g

30. …water, sulfur, table sugar (C 6 H 12 O 6 ), mercury, and copper each contain one mole. What are their amu or mw? Do they weigh the same? One mole of Avagadro’s number of… H2OH2O Cu S Sugar Hg

31. water or H 2 O 18 g/mol, sulfur or S 32.1 g/mol, sugar or C 6 H 12 O 6 180 g/mol, mercury or Hg 200.5g/mol, and copper or Cu 63.5g/mol One mole of Avagadro’s number of… H2OH2O Cu S Sugar Hg

32. Prentice Hall © 2007 Chapter Six 32 The molar mass of water is 18.0 g. The conversion factor between moles of water and mass of water is 18.0 g/mol and the conversion factor between mass of water and moles of water is 1 mol/18.0 g:

33. Prentice Hall © 2007 Chapter Six 33 Why do we use Molar Mass? Can you think of an instance where you weigh something to know how many you have? Is it true that 1 kg of apples = 12 apples? In chemistry we weigh things to know how many we have

34. Learning check Prentice Hall © 2007 Chapter Six 34 What is a correct method for calculating the mass of 1.9 × 10 23 potassium atoms?

35. Chapter Seven

36. Learning check ► Acetylsalicylic acid, commonly known as aspirin, has 9 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms per molecule. What is the mass, in grams, of 2.25 moles of Acetylsalicylic acid? ► A typical Aspirin is 500mg. How many moles and molecules is this. Prentice Hall © 2007 Chapter Six 36

37. Learning check ► Acetylsalicylic acid, commonly known as aspirin, has 9 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms per molecule. What is the mass, in grams, of 2.25 moles of Acetylsalicylic acid? Aspirin is C 9 H 8 O 4 and has a molar mass of: C 9 × 12 g/mol = 108 g/mol H 8 × 1 g/mol = 8 g/mol O 4 × 16 g/mol = 64 g/mol Total 180 g/mol Prentice Hall © 2007 Chapter Six 37

38. Learning check ► A typical Aspirin is 500mg. How many moles and molecules is this. Prentice Hall © 2007 Chapter Six 38

39. Prentice Hall © 2007 Chapter Six 39 6.5 Mole Relationships and Chemical Equations The coefficients in a balanced chemical equation tells how many molecules, and thus how many moles, of each reactant are needed and how many molecules, and thus moles, of each product are formed. See the example below:

40. Prentice Hall © 2007 Chapter Six 40 ► The coefficients can be put in the form of mole ratios, which act as conversion factors when setting up factor-label calculations. ► In the ammonia synthesis, the mole ratio of H 2 to N 2 is 3:1, the mole ratio of H 2 to NH 3 is 3:2, and the mole ratio of N 2 to NH 3 is 1:2 leading to the following conversion factors: (3 mol H 2 )/(1 mol N 2 ) (3 mol H 2 )/(2 mol NH 3 ) (1 mol N 2 )/(2 mol NH 3 )

41. Prentice Hall © 2007 Chapter Six 41 6.6 Mass Relationships and Chemical Equations Mole to mole conversions are carried out using mole ratios as conversion factors.

42. Prentice Hall © 2007 Chapter Six 42 ► Mole to mass and mass to mole conversions are carried out using molar mass as a conversion factor. ► Mass to mass conversions are frequently needed, but cannot be carried out directly. ► Overall, there are four steps for determining mass relationships among reactants and products.

43. Prentice Hall © 2007 Chapter Six 43 Mass to mass conversions: ► STEP 1: Write the balanced chemical equation. ► STEP 2: Choose molar masses and mole ratios to convert known information into needed information. ► STEP 3: Set up the factor- label expression, and calculate the answer. ► STEP 4: Estimate or check the answer using a ballpark solution.

44. 44  We can read the equation in “moles” by placing the word “moles” between each coefficient and formula. 4 Fe + 3 O 2 2 Fe 2 O 3 4 moles Fe + 3 moles O 2 2 moles Fe 2 O 3 Moles in Equations

45. 45 ► A mole-mole factor is a ratio of the coefficients for two substances. 4 Fe + 3 O 2 2 Fe 2 O 3 Fe and O 2 4 mole Fe and 3 mole O 2 3 mole O 2 4 mole Fe Fe and Fe 2 O 3 4 mole Fe and 2 mole Fe 2 O 3 2 mole Fe 2 O 3 4 mole Fe O 2 and Fe 2 O 3 3 mole O 2 and 2 mole Fe 2 O 3 2 mole Fe 2 O 3 3 mole O 2 Writing Mole-Mole Factors

46. 46 Consider the following equation: 3 H 2 + N 2 2 NH 3 A. A mole factor for H 2 and N 2 is 1) 3 mole N 2 2) 1 mole N 2 3) 1 mole N 2 1 mole H 2 3 mole H 2 2 mole H 2 B. A mole factor for NH 3 and H 2 is 1) 1 mole H 2 2) 2 mole NH 3 3) 3 mole N 2 2 mole NH 3 3 mole H 2 2 mole NH 3 Learning Check

47. 47 3 H 2 + N 2 2 NH 3 A. A mole factor for H 2 and N 2 is 2) 1 mole N 2 3 mole H 2 B. A mole factor for NH 3 and H 2 is 2) 2 mole NH 3 3 mole H 2 Solution

48. Chapter Seven

49. 49 Consider the following reaction: 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react? Use the appropriate mole factor to determine the moles Fe 2 O 3. 6.0 mole O 2 x 2 mole Fe 2 O 3 = 4.0 mole Fe 2 O 3 3 mole O 2 Calculations with Mole Factors

50. 50 Consider the following reaction: 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 moles of O 2 ? 1) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe Learning Check

51. 51 3) 16.0 moles Fe Consider the following reaction: 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 moles of O 2 ? 12.0 mole O 2 x 4 mole Fe = 16.0 moles Fe 3 mole O 2 Solution

52. Chapter Seven

53. 53 How many grams of O 2 are needed to produce 0.400 mole of Fe 2 O 3 ? 4 Fe + 3 O 2 2 Fe 2 O 3 1) 38.4 g O 2 2) 19.2 g O 2 3) 1.90 g O 2 Learning Check

54. 54 2) 19.2 g O 2 0.400 mole Fe 2 O 3 x 3 mole O 2 x 32.0 g O 2 2 mole Fe 2 O 3 1 mole O 2 mole factor molar mass = 19.2 g O 2 Solution

55. 55 The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? 2H 2 + O 2 2H 2 O ? g 13.1 g Plan: g H 2 O mole H 2 O mole O 2 g O 2 13.1 g H 2 O x 1 mole H 2 O x 1 mole O 2 x 32.0 g O 2 18.0 g H 2 O 2 mole H 2 O 1 mole O 2 = 11.6 g O 2 Calculating the Mass of a Reactant

56. 56 Learning Check Acetylene gas C 2 H 2 burns in the oxyactylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2 ? 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 1) 88.6 g C 2 H 2 2) 44.3 g C 2 H 2 3) 22.2 g C 2 H 2

57. 57 3) 22.2 g C 2 H 2 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 75.0 g CO 2 x 1 mole CO 2 x 2 moles C 2 H 2 x 26.0 g C 2 H 2 44.0 g CO 2 4 moles CO 2 1 mole C 2 H 2 = 22.2 g C 2 H 2 Solution

58. Prentice Hall © 2007 Chapter Six 58 6.7 Percent Yield ► The amount of product actually formed in a chemical reaction is somewhat less than the amount predicted by theory. ► Unwanted side reactions and loss of product during handling prevent one from obtaining a perfect conversion of all the reactants to desired products. ► The amount of product actually obtained in a chemical reaction is usually expressed as a percent yield.

59. Prentice Hall © 2007 Chapter Six 59 ► Percent yield is defined as: (Actual yield ÷ Theoretical yield) x 100% ► The actual yield is found by weighing the product obtained. ► The theoretical yield is found by a mass-to-mass calculation.

60. Chapter Seven

61. 61 Sample Exercise % Yield Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C + O 2 2CO What is the percent yield if 40.0 g of CO are produced from the reaction of 30.0 g O 2 ?

62. 62 Sample Exercise % Yield (cont.) 1. Calculate theoretical yield of CO. 30.0 g O 2 x 1 mole O 2 x 2 mole CO x 28.0 g CO 32.0 g O 2 1 mole O 2 1 mole CO = 52.5 g CO (theoretical) 2. Calculate the percent yield. 40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO(theoretical)

63. 63 Learning Check In the lab, N 2 and 5.0 g of H 2 are reacted and produce 16.0 g of NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 1) 31.3 % 2) 56.5 % 3) 80.0 %

64. 64 Solution 2) 56.5 % N 2 (g) + 3H 2 (g) 2NH 3 (g) 5.0 g H 2 x 1 mole H 2 x 2 moles NH 3 x 17.0 g NH 3 2.0 g H 2 3 moles H 2 1 mole NH 3 = 28.3 g NH 3 (theoretical) Percent yield = 16.0 g NH 3 x 100 = 56.5 % 28.3 g NH 3

65. Prentice Hall © 2007 Chapter Six 65 6.8 Classes of Chemical Reactions ► When learning about chemical reactions it is helpful to group the reactions of ionic compounds into three general classes: precipitation reactions, acid– base neutralization reactions, and oxidation– reduction reactions. ► Precipitation reactions are processes in which an insoluble solid called a precipitate forms when reactants are combined in aqueous solution.

66. Prentice Hall © 2007 Chapter Six 66 ► Acid–base neutralization reactions are processes in which H + ions from an acid react with OH - ions from a base to yield water. An ionic compound called a salt is also produced. The “salt” produced need not be common table salt. Any ionic compound produced in an acid–base reaction is called a salt. ► Oxidation–reduction reactions, or redox reactions, are processes in which one or more electrons are transferred between reaction partners (atoms, molecules, or ions). As a result of this transfer, the charges on atoms in the various reactants change.

67. 6.8 Classes of Chemical Reactions ► Precipitation reactions Pb(NO 3 ) 2 (aq) + 2KI (aq)  PbI 2 (s) + 2KNO 3 (aq) ► Acid–base neutralization reactions HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) ► Oxidation–reduction reactions Mg(s) + I 2 (g)  MgI 2 (s) Prentice Hall © 2007 Chapter Six 67

68. Prentice Hall © 2007 Chapter Six 68 6.9 Precipitation Reactions and Solubility Guidelines ► Reaction of aqueous Pb(NO 3 ) 2 with aqueous KI gives a yellow precipitate of PbI 2. ► To predict whether a precipitation reaction will occur on mixing aqueous solutions of two ionic compounds, you must know the solubility of the potential products.

69. Prentice Hall © 2007 Chapter Six 69 If a potential product does not contain at least one of the ions listed below, it is probably not soluble and will precipitate from solution when formed.

70. Chapter Seven

71. Predict which of the following solutions will result in a precipitation reaction. 1.Cr(NO 3 ) 3 (aq) + KCl(aq)  2.CuCl 2 (aq) + Na 2 S(aq)  3.NH 4 Br(aq) + Na 2 SO 4 (aq)  4.CsOH(aq) + RbCl(aq) 

72. Predict which of the following solutions will result in a precipitation reaction. 1.Cr(NO 3 ) 3 (aq) + KCl(aq)  2.CuCl 2 (aq) + Na 2 S(aq)  CuS (s) + 2 NaCl (aq) 3.NH 4 Br(aq) + Na 2 SO 4 (aq)  4.CsOH(aq) + RbCl(aq) 

73. Learning check Solubility Rules Predict the solubility of the products based on solubility rules AgNO 3 + Na 2 CO 3  CdCl 2 + (NH 4 ) 2 S  (NH 4 ) 3 PO 4 + LiCl  BaCl 2 + MgSO 4  Prentice Hall © 2007 Chapter Six 73

74. Prentice Hall © 2007 Chapter Six 74 6.10 Acids, Bases, and Neutralization Reactions ► When acids and bases are mixed together in correct proportion, acidic and basic properties disappear. ► A neutralization reaction produces water and a salt. HA (aq) + MOH (aq)  H 2 O (l) + MA (aq) acid + base  water + salt ► The reaction of hydrochloric acid with potassium hydroxide to produce potassium chloride is an example: ► HCl (aq) + KOH (aq)  H 2 O (l) + KCl (aq)

75. Which of the following reactions is an acid-base neutralization reaction? 1.AgNO 3 (aq) + NaBr(aq)  AgBr(s) + NaNO 3 (aq) 2.2 CH 3 OH(l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(l) 3.2 Na(s) + Br 2 (l)  2 NaBr(s) 4.H 2 SO 4 (aq) + 2 KOH(aq)  2 H 2 O(l) + K 2 SO 4 (aq) #4

76. Learning check ► Predict the products of these acid base reactions HBr + NaOH  H 3 PO 4 + Ca(OH) 2  H 2 SO 4 + Al(OH) 3  Prentice Hall © 2007 Chapter Six 76

77. Learning check ► Predict the products of these acid base reactions HBr + NaOH  NaBr + H 2 O 2H 3 PO 4 + 3Ca(OH) 2  Ca 3 (PO 4 ) 2 + 6 H 2 O 3H 2 SO 4 + 2Al(OH) 3  Al 2 (SO 4 ) 3 + 6 H 2 O Prentice Hall © 2007 Chapter Six 77

78. Prentice Hall © 2007 Chapter Six 78 6.11 Redox Reactions ► Oxidation–reduction (redox) reaction: A reaction in which electrons transfer from one atom to another. ► Oxidation: Loss of one or more electrons by an atom. ► Reduction: Gain of one or more electrons by an atom.

79. Prentice Hall © 2007 Chapter Six 79 ► Oxidation and reduction always occur together. ► A substance that is oxidized gives up an electron, causes reduction, and is called a reducing agent. ► A substance that is reduced gains an electron, causes oxidation, and is called an oxidizing agent. ► The charge on the reducing agent increases during the reaction, and the charge on the oxidizing agent decreases.

81. Prentice Hall © 2007 Chapter Six 81 Reducing agent: ► Loses one or more electrons ► Causes reduction ► Undergoes oxidation ► Becomes more positive (or less negative) Oxidizing agent: ► Gains one or more electrons ► Causes oxidation ► Undergoes reduction ► Becomes more negative (or less positive)

82. Prentice Hall © 2007 Chapter Six 82 6.12 Recognizing Redox Reactions ► One can determine whether atoms are oxidized or reduced in a reaction by keeping track of changes in electron sharing by the atoms. Each atom in a substance is assigned a value called an oxidation number or oxidation state. ► The oxidation number indicates whether the atom is neutral, electron-rich, or electron-poor. ► By comparing the oxidation state of an atom before and after reaction, we can tell whether the atom has gained or lost electrons.

83. Prentice Hall © 2007 Chapter Six 83 ► Rules for assigning oxidation numbers: ► An atom in its elemental state has an oxidation number of zero. ► ►A monatomic ion has an oxidation number equal to its charge.

84. Prentice Hall © 2007 Chapter Six 84 ► In a molecular compound, an atom usually has the same oxidation number it would have if it were a monatomic ion. ► Examples: H often has an oxidation number of +1, oxygen often has an oxidation number of -2, halogens often have an oxidation number of -1.

85. Prentice Hall © 2007 Chapter Six 85 ► For compounds with more than one nonmetal element, such as SO 2, NO, and CO 2, the more electronegative element—oxygen in these examples—has its preferred negative oxidation number. ► The less electronegative element is assigned a positive oxidation number so that the sum of the oxidation numbers in a neutral compound is 0.

86. Learning Check ► Determine the oxidation and reduction reaction as well as the oxidizing and reducing agent 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2 Prentice Hall © 2007 Chapter Six 86

87. Learning Check ► Determine the oxidation and reduction reaction as well as the oxidizing and reducing agent Prentice Hall © 2007 Chapter Six 87 Reduction: +3  0 Oxidation: 0  +4

88. Learning Check ► Determine the oxidation and reduction reaction as well as the oxidizing and reducing agent Prentice Hall © 2007 Chapter Six 88 Oxidizing Agent Reducing Agent

89. Identify the oxidized reactant, the reduced reactant, the oxidizing agent, and the reducing agent in the reaction: Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g). 1.CO(g) is oxidized and is the oxidizing agent and Fe 2 O 3 (s) is reduced and is the reducing agent. 2.CO(g) is oxidized and is the reducing agent and Fe 2 O 3 (s) is reduced and is the oxidizing agent. 3.CO(g) is reduced and is the oxidizing agent and Fe 2 O 3 (s) is oxidized and is the reducing agent. 4.CO(g) is reduced and is the reducing agent and Fe 2 O 3 (s) is oxidized and is the oxidizing agent.

90. Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g). 1.CO(g) is oxidized and is the oxidizing agent and Fe 2 O 3 (s) is reduced and is the reducing agent. 2.CO(g) is oxidized and is the reducing agent and Fe 2 O 3 (s) is reduced and is the oxidizing agent. 3.CO(g) is reduced and is the oxidizing agent and Fe 2 O 3 (s) is oxidized and is the reducing agent. 4.CO(g) is reduced and is the reducing agent and Fe 2 O 3 (s) is oxidized and is the oxidizing agent. 3+ 2- 2+ -2 0 4+ 2-

91. Learning check ► Determine the oxidation and reduction reaction as well as the oxidizing and reducing agent 2 C 2 H 2 + 5 O 2  4 CO 2 + 2 H 2 O 2NaVO 2 + 3 Zn  Na 2 VO + 3 ZnO 2 LiI + 2 H 2 O  I 2 + 2 LiOH + H 2 Prentice Hall © 2007 Chapter Six 91

92. Prentice Hall © 2007 Chapter Six 92 6.13 Net Ionic Equations ► Ionic equation: An equation in which ions are explicitly shown. ► Spectator ion: An ion that appears unchanged on both sides of a reaction arrow. ► Net ionic equation: An equation that does not include spectator ions.

93. Chapter Seven

95. 4 common gas forming reactions ► If you see the following 3 as products, replace those molecules with the appropriate gas and H 2 O (l). 1. H 2 CO 3  H 2 O(l) + CO 2 (g) 2. H 2 SO 3  H 2 O(l) + SO 2 (g) 3. NH 4 OH  NH 3 (g) + H 2 O(l) ► This is a common method for making hydrogen gas 4. H + + some metals  metal salt (aq) + H 2 (g) Prentice Hall © 2007 Chapter Six 95

96. 96 Sample Problems Write the NIE sodium sulfite + sulfuric acid Na 2 SO 3 H 2 SO 4 Na 2 SO 4 H 2 SO 3 +-->+ 2 Na + +SO 3 2- + H+H+ HSO 4 - + SO 3 2- + H + + HSO 4 - --> SO 4 2- + SO 2 + H 2 O H 2 O+SO 2 + SO 4 2- 2 Na + +

97. 97 HCl CaCO 3 CaCl 2 H 2 CO 3 +-->+ Sample Problems Write the NIE hydrochloric acid + calcium carbonate CaCO 3 CaCl 2 H 2 CO 3 +-->+ 2 H + +2 Cl - + CaCO 3 Ca 2+ +2 Cl - HOH+CO 2 + -->2 H + CaCO 3 +CO 2 HOH++Ca 2+ 2

98. 98 The two Cl - ions are SPECTATOR IONS — they do not participate. Net Ionic Equations - - Mg(s) + 2 HCl(aq) --> H 2 (g) + MgCl 2 (aq) - - Mg(s) + 2 H + (aq) + 2 Cl - (aq) - - ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq)

99. 99 - - We leave the spectator ions out in writing the NET IONIC EQUATION (NIE) - - Mg(s) + 2 H + (aq)  H 2 (g) + Mg 2+ (aq) Net Ionic Equations Mg(s) + 2 HCl(aq) --> H 2 (g) + MgCl 2 (aq) Mg(s) + 2 H + (aq) + 2 Cl - (aq) ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq)

100. 100 23 Fe 2 S 3 6 + Net Ionic Equations - - Write the NIE - - 1. iron (III) nitrate + sodium sulfide Fe(NO 3 ) 3 (aq) Na 2 S (aq) Fe 2 S 3 (s) NaNO 3 (aq) --> + - - -->2 Fe 3+ aq+6 NO 3 - aq + 6 Na + aq3 S 2- aq+ Fe 2 S 3 (s)6 NO 3 - aq+6 Na + aq+ -->2 Fe 3+ (aq) 3 S 2- (aq) + Fe 2 S 3 (s)

101. 101 Net Ionic Equations Write the NIE 4. sodium nitrate + potassium chloride NaNO 3 (aq)KCl (aq)NaCl (aq)KNO 3 (aq)+-->+ Na + + NO 3 - + K + Cl - + NO 3 - +K+K+ +Cl - Na + + no reaction (N/R)

102. 102 Learning check Write the NIE Ba + 2 NaCl  BaCl 2 + Na Ba (s) + 2 NaCl (aq)  BaCl 2 (aq) + Na (s) Ba (s) + 2 Na + (aq) + 2 Cl - (aq)  Ba 2+ (aq) + 2Cl - (aq) + 2Na (s) Ba (s) + 2 Na + (aq)  Ba 2+ (aq) + 2Na (s)

103. Learning check: Write the NIE 2Al + 6 HCl  2 AlCl 3 + 3 H 2 MgSO 4 + (NH 4 ) 2 S  MgS + (NH 4 ) 2 SO 4 barium hydroxide + hydrochloric acid  barium chloride + water KOH + HCl KOH + HCl  Prentice Hall © 2007 Chapter Six 103

104. Learning check: Write the NIE 2Al (s) + 6 H + (aq)  2 Al 3+ (aq) + 3 H 2 (g) Mg 2+ (aq) + S 2- (aq)  MgS (s) OH - (aq) + H + (aq)  H 2 O (l) Prentice Hall © 2007 Chapter Six 104

105. Prentice Hall © 2007 Chapter Six 105 Chapter Summary ► Chemical equations must be balanced; the numbers and kinds of atoms must be the same in both the reactants and the products. ► To balance an equation, coefficients are placed before formulas but the formulas themselves cannot be changed. ► A mole refers to Avogadro’s number of formula units of a substance. One mole of any substance has a mass equal to its formula weight in grams. ► Molar masses act as conversion factors between numbers of molecules and masses in grams.

106. Prentice Hall © 2007 Chapter Six 106 Chapter Summary Cont. ► The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction. ► Mole ratios relate amounts of reactants and/or products. Using molar masses and mole ratios in factor-label calculations relates unknown masses to known masses or molar amounts. ► The yield is the amount of product obtained. ► The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%.

107. Prentice Hall © 2007 Chapter Six 107 Chapter Summary Cont. ► Precipitation reactions are processes in which an insoluble solid called a precipitate is formed. ► In acid–base neutralization reactions an acid reacts with a base to yield water plus a salt. ► Oxidation–reduction (redox) reactions are processes in which one or more electrons are transferred between reaction partners. ► Oxidation is the loss of electrons by an atom, and reduction is the gain of electrons by an atom. ► Oxidation numbers are assigned to provide a measure of whether an atom is neutral, electron- rich, or electron-poor.

108. Prentice Hall © 2007 Chapter Six 108 End of Chapter Six