2. 5 - 1CH104 CHAPTER 7 Chemical Reactions & Quantities Reactions & Equations Balancing Chemical Reactions Types of Reactions Oxidation-Reduction The Mole and Chemical Equations Mass Calculations Percent Yield & Limiting Reactants Energy Changes in Reactions

3. 5 - 2CH104 Examples Color Odor Taste Feel Shape Physical properties Characteristics that can be evaluated without changing the composition of the material. Density Melting / Freezing point Boiling point Compressibility Form (foil, wire, powder…)

4. 5 - 3CH104 What is Chemistry? “The study of Matter and its Changes.” Physical Changes Physical Changes = Physical Property Changes in a Physical Property Chemical Changes Chemical Changes = Chemical Property Changes in a Chemical Property Appearance: melting, freezing, evaporation…melting, freezing, evaporation… stretching, molding, cutting…stretching, molding, cutting… Chemical Composition:

5. 5 - 4CH104 Change in the Chemical Composition Burning of Magnesium Chemical Changes Rusting of Iron Decomposing of wood Souring of Milk Examples:

6. 5 - 5CH104 Examples Mulching leaves chemicalphysical Which are chemical or physical changes? Tarnishing Silver Fermentation Making ice into water Carbonated Beverage going flat Bleaching a stain

7. 5 - 6CH104 Mg + O 2  MgO + Energy Chemical Reactions Shows how the Chemical change occurs. Reactants C 3 H 8 + O 2 CO 2 + H 2 O + Energy  Fe + O 2 Fe 2 O 3 ProductsProducts

8. 5 - 7CH104 Chemical equations Chemist’s shorthand to describe a reaction. ReactantsReactants ProductsProducts The state of all substancesThe state of all substances H 2 + O 2 H 2 O + E (g)(g) (g) Any conditions used in the reactionAny conditions used in the reaction heat Same # & type atoms on each sideSame # & type atoms on each side Law of Conservation of Matter Law of Conservation of Matter 2 2

9. 5 - 8CH104 Balancing Equations WB ___W 10 + ___B 8 ___WB ReactantsReactants ProductsProducts Making Hot dogs: How many packages wieners & buns to buy so none is left over. 4540

10. 5 - 9CH104 Balancing chemical equations Reactants Products 1 Ca 1 C 3 O CaCO 3 (s) CaO (s) + CO 2 (g)  denotes heat Reactant Products

11. 5 - 10CH104 Ca H Cl Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation. 1 1 1 1 2 2

12. 5 - 11CH104 Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 Ca H Cl Ca H Cl ReactantsReactants ProductsProducts 111111 111111 122122 122122 Step 2 Step 2: Determine which atoms are not balanced. - not balanced

13. 5 - 12CH104 Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 ReactantsReactants ProductsProducts 111111 111111 122122 122122 - not balanced Step 3: Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!) 2 2 2 Ca H Cl Ca H Cl 2 2

14. 5 - 13CH104 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation. 3 1 4 1 2 1 2 8 3 1

15. 5 - 14CH104 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts 3141231412 1283112831 - not balanced Step 2 Step 2: Determine which atoms are not balanced. - not balanced

16. 5 - 15CH104 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts 3141231412 1283112831 - not balanced Step 3: Step 3: Balance elements with #’s in front of formulas until all balanced. (Never change the formulas!)

17. 5 - 16CH104 Hints: Start with a metal in a complex compound, or an element that only appears in one formula. (Like Mg here) Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts 3141231412 1283112831 - not balanced 6 6 6 6 63 3 3 2 2 6 6 1 8 8 2 6 6

18. 5 - 17CH104 Hints: Start with an element that only appears in one formula on both sides of the equation. Leave oxygen until last. Balancing Equations CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O ReactantsReactants ProductsProducts CHOCHO

19. 5 - 18CH104 Balancing Equations CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O CHOCHO CHOCHO ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation. 262262 2 6 2 123123 1 2 3

20. 5 - 19CH104 CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O Balancing Equations ReactantsReactants ProductsProducts 262262 262262 123123 123123 Step 2 Step 2: Determine which atoms are not balanced. - not balanced CHOCHO CHOCHO

21. 5 - 20CH104 262262 262262 - not balanced CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations ReactantsReactants ProductsProducts 123123 123123 - not balanced Step 3: Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!) 2 5 5 3 6 6 7 7 3.5 7 7 CHOCHO CHOCHO

22. 5 - 21CH104 CHOCHO CHOCHO CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations ReactantsReactants ProductsProducts 262262 262262 123123 123123 2 5 5 3 6 6 7 7 3.5 7 7 3.5 O 2 Can’t have 3.5 O 2, so multiply equation by 2!

23. 5 - 22CH104 3.5 CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations CHOCHO CHOCHO ReactantsReactants ProductsProducts 262262 262262 123123 123123 4 5 5 6 6 6 7 7 7 7 7 3.5 O 2 Can’t have 3.5 O 2, so multiply equation by 2! 2 4 4 4 4 12 14

24. 5 - 23CH104 (NH 2 ) 2 CO + H 2 O ______ > NH 3 + CO 2 - not balanced 1 6 2NHCONHCO1 3 1 22 Balancing Equations

25. 5 - 24CH104 (NH 2 ) 2 CO + H 2 O ______ > NH 3 + CO 2 Balancing Equations 1 6 21 3 1 2 2 6 NHCONHCO 22

26. 5 - 25CH104 Example: Decomposition of urea CH 3 OH + PCl 5  CH 3 Cl + POCl 3 + H 2 O CH 3 OH + PCl 5  CH 3 Cl + POCl 3 + H 2 O - not balanced 1 4 1CHOPClCHOPCl1 5 2 11 54

27. 5 - 26CH104 1 CH 3 OH + PCl 5  CH 3 Cl + POCl 3 + H 2 O CH 3 OH + PCl 5  CH 3 Cl + POCl 3 + H 2 O Balancing Equations 2 2 8 1 4 1CHOPClCHOPCl 5 2 11 545 2 2 8 2

28. 5 - 27CH104 Types of Chemical Reactions Combination Decomposition Single Replacement: Substitution Double Replacement: Metathesis A + BX  B + AX A + B  C C  A + B AX + BY  BX + AY

29. 5 - 28CH104 Types of Chemical Reactions Combination Decomposition Single Replacement: Substitution Double Replacement: Metathesis 2H 2 + O 2  2H 2 O CaCO 3  CaO + CO 2 Al + FeCl 3  Fe + AlCl 3 2AgNO 3 + K 2 SO 4  Ag 2 SO 4 + 2KNO 3

30. 5 - 29CH104 Types of Chemical Reactions Complete: C 3 H 8 + 5O 2  3CO 2 + 4H 2 O Combustion Incomplete: 2C 3 H 8 + 7O 2  6CO + 8H 2 O C 3 H 8 + 2O 2  3C + 4H 2 O

31. 5 - 30CH104 Combination Reactions 2H 2 + O 2  2H 2 O Formation of Acid Rain SO 3 + H 2 O  H 2 SO 4 Explosion of Hydrogen Balloon Rusting of Iron 4 Fe + 3 O 2  2 Fe 2 O 3 A + B  C

32. 5 - 31CH104 Decomposition Reactions Heating Egg Shells CaCO 3  CaO + CO 2 2 H 2 O 2 2 H 2 O + O 2 Blood with peroxide C  A + B

33. 5 - 32CH104 Single Replacement Reactions Iron Deposits on an Aluminum Pan Al + FeCl 3  Fe + AlCl 3 A + BX  B + AX

34. 5 - 33CH104 Activity series of metals potassium sodium potassium sodium calcium magnesium aluminum zinc chromium magnesium aluminum zinc chromium iron nickel tin lead iron nickel tin lead copper silver platinum gold copper silver platinum gold Hydrogen Al + Fe +3  Fe + Al +3 Fe + H +  Fe +3 + H 2 increasing reactivity Element give e’s to ion lower on list

35. 5 - 34CH104 Double Replacement Reaction BaCl 2(aq) + Na 2 SO 4(aq)  BaSO 4(s) + 2NaCl (aq) AX + BY  BX + AY Ba +2 Cl -1 Na +1 SO 4 -2 Insoluble Precipitate Formed

36. 5 - 35CH104 Predict the products: AgNO 3(aq) AlCl 3 (aq) AgNO 3(aq) + AlCl 3 (aq) Ag + NO 3 - Al +3 Cl - AgCl (s) + Al(NO 3 ) 3(aq) Double Replacement Reaction AX + BY  BX + AY Balance later as needed to get: 3AgNO 3(aq) AlCl 3 (aq) 3AgNO 3(aq) + AlCl 3 (aq) 3AgCl (s) + Al(NO 3 ) 3(aq)

37. 5 - 36CH104 Oxidation and reduction REDOX Where reactants exchange electrons - Examples: All types of batteries All types of batteries alkaline, NiCad, car batteries Rusting and corrosion Rusting and corrosion Metabolism Metabolism Antioxidants (Vit C, E prevent oxidation) Antioxidants (Vit C, E prevent oxidation)

38. 5 - 37CH104 Oxidation and reduction REDOX Where reactants exchange electrons - Oxidation Oxidation = Losing electrons LEO LEO: Lose Electrons Oxidation LEOGER LEO the lion says GER Reduction Reduction = Gaining electrons GER GER: Gain Electrons Reduction OIL OIL : Oxidation Is Losing RIG RIG : Reduction Is Gaining OIL RIG

39. 5 - 38CH104 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1-1+ Assign Oxidation States: 0 0 For simple ions, charge. Ox state = charge. For element in natural form 0 Ox State = 0.

40. 5 - 39CH104 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1- 1+ Who’s loosing or gaining electrons? 00 Loses 1 e - = LEO Gains 1 e - =GER 1 e - =GER oxidized Na loses e - (LEO) Na gets oxidized reduced Cl gains e - (GER) Cl gets reduced

41. 5 - 40CH104 Oxidation and reduction Oxidation loses Oxidation - when reactant loses e - (s). (LEO) NaNa + + e - Na (s) Na + + e - Reduction gains Reduction - when reactant gains e - (s). (GER) Cl 2 + 2 e - 2 Cl - Cl 2 (g) + 2 e - 2 Cl - half reactions These are half reactions

42. 5 - 41CH104 2 Na (s) + Cl 2 (g) + 2 e -  2 Na + + 2e - + 2 Cl - Oxidation and reduction 2 half reactions make a complete reaction 2 half reactions make a complete reaction Na (s)  Na + + e - Na (s)  Na + + e - Cl 2 (g) + 2 e -  2 Cl - Cl 2 (g) + 2 e -  2 Cl - 222 2 Na (s) + Cl 2 (g)  2 Na + + 2 Cl -

43. 5 - 42CH104 Oxidizing agent = The chemical that caused an oxidation. It is reduced. Reducing agent = The chemical that caused a reduction. It is oxidized. Oxidation and reduction

44. 5 - 43CH104 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1- 1+ Who’s loosing or gaining electrons? 00 NaNaoxidized Na loses e - (LEO) so Na gets oxidized Na caused Cl to get reduced Na caused Cl to get reduced Na is the Reducing agent Na is the Reducing agent Loses 1 e - =LEO Gains 1 e - =GER 1 e - =GER

45. 5 - 44CH104 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1- 1+ Who’s loosing or gaining electrons? 00 Loses 1 e - =LEO Gains 1 e - =GER 1 e - =GER ClClreducedCl gains e - (GER) so Cl gets reduced Cl caused Na to get oxidizedCl caused Na to get oxidized Cl is the Oxidizing agentCl is the Oxidizing agent

46. 5 - 45CH104 Oxidation and reduction 2 Na (s) + Cl 2 (g)  2 Na + + 2 Cl - NaNaoxidized Na loses e - (LEO) so Na gets oxidized Na caused Cl to get reduced Na caused Cl to get reduced Na is the Reducing agent Na is the Reducing agent ClClreducedCl gains e - (GER) so Cl gets reduced Cl caused Na to get oxidizedCl caused Na to get oxidized Cl is the Oxidizing agentCl is the Oxidizing agent

47. 5 - 46CH104 Oxidation state Rules 0 Oxidation state of element in natural form = 0. For simple monoatomic ions, charge. oxidation state = charge. Describes the charge of each element. For certain groups at certain times, group number oxidation number = group number Examples Na +1, Cl -1, Ca 2+ Examples N +5, Cl +7 Examples N 2, Na, O 2, H 2

48. 5 - 47CH104 H 1+ HydrogenH 1+ if bonded to nonmetal H 1- H 1- if bonded to metal Example HCl Oxidation states O -2 OxygenO -2 Usually O -1 O -1 in peroxides Example H 2 O Example H 2 O 2 Example NaH

49. 5 - 48CH104 Assign the oxidation states for all elements in HO H 2 O +1 -2 +1 +2-2 Oxidation states

50. 5 - 49CH104 Assign the oxidation states for all elements in HO 2 H 2 O 2 +1 +1 +2-2 Oxidation states

51. 5 - 50CH104 REDOX reactions 2 H 2 + O 2 2 H 2 O 0 Oxidation state of H is 0 0 Oxidation state of O is 0 1+ Oxidation state of H is 1+ 2- Oxidation state of O is 2- Hydrogen is oxidized and is a reducing agent. Oxygen is reduced and is an oxidizing agent. Lose 1 e - = LEO Gain 2 e - = GER

52. 5 - 51CH104 Find the oxidation state for all elements in: HNO 3 Nitrogen must be 1+ Hydrogen is 1+ 2- Oxygen is 2- 2- 2- 5+ +1+5 -6 = 0 Start with what We know Start with what We know Oxidation states

53. 5 - 52CH104 NaCl +1 +3 -2 NaClO Oxidation states of Cl NaClO 4 NaClO 2 NaClO 3 +1 +1 +1 +1 -2-2 -2-2-2 -2-2-2-2 +1 +5 +7

54. 5 - 53CH104 Types of chemical reactions Chemical Reactions NonredoxNonredox DoublereplacementDoublereplacement Combination RedoxRedox SinglereplacementSinglereplacement Decomposition

55. 5 - 54CH104 Combination reactions B + C A REDOX or NONREDOX types Non-REDOX reaction Non-REDOX reaction. Formation of Acid Rain SO 3 + H 2 O  H 2 SO 4 -2+6 -2-2-2-2 -2-2-2+1+1 +1+1 +6

56. 5 - 55CH104 Combination reactions B + C A REDOX or NONREDOX types Rusting of Iron 4 Fe + 3 O 2  2 Fe 2 O 3 0 -2-2-2 0 +3+3 REDOX reaction REDOX reaction. Fe 0+3LEO Fe goes from 0 to +3 LEO O 0 -2GER O goes from 0 to -2 GER

57. 5 - 56CH104 A B + C REDOX or NONREDOX types Decomposition reactions Non-REDOX reaction Non-REDOX reaction. Decomposition of Egg Shells CaCO 3  CaO + CO 2 -2 +2 -2-2-2-2-2 +4+4+2

58. 5 - 57CH104 Decomposition reactions 2 H 2 O 2 2 H 2 O + O 2 REDOX reaction REDOX reaction. O -10LEO Some O goes from -1 to 0 (O 2 ) LEO O -1 -2GER Some O goes from -1 to -2 (H 2 O) GER Decomposition of hydrogen peroxide +1+1+1+1-20

59. 5 - 58CH104 A + BX  B + AX Always REDOX Iron Deposits on an Aluminum Pan Al + FeCl 3  Fe + AlCl 3 00 +3+3 Single Replacement Reaction REDOX reaction REDOX reaction. Al 0+3LEO Al goes from 0 to +3 LEO Fe +3 0GER Fe goes from +3 to 0 GER

60. 5 - 59CH104 Activity series of metals potassium sodium potassium sodium calcium magnesium aluminum zinc chromium magnesium aluminum zinc chromium iron nickel tin lead iron nickel tin lead copper silver platinum gold copper silver platinum gold Hydrogen Al + Fe +3  Fe + Al +3 Fe + H +  Fe +3 + H 2 increasing reactivity Element give e’s to ion lower on list

61. 5 - 60CH104 AX + BY  BX + AY Always non-REDOX +1 +1 -2-2-2 +5+1 Double Replacement Reaction Non -REDOX reaction Non -REDOX reaction. AgNO 3 NaClAgClNaNO 3 AgNO 3(aq) + NaCl (aq)  AgCl (s) + NaNO 3(aq) +1 +5

62. 5 - 61CH104 Ionic equations Ionic substances dissociate into ions when dissolved in water. Ag + + NO 3 - Ag + + NO 3 - AgNO 3(aq) NaCl (aq) AgNO 3(aq) + NaCl (aq) Certain ions join together Others remain unchanged. Ag + NO 3 - Na + Cl - AgCl (s) + NaNO 3(aq) AgCl (s) + Na + + Cl - + Na + + NO 3 -

63. 5 - 62CH104 Ionic equations Total ionic equation Ag + + NO 3 - Na + + Cl - AgC l (s) + Na + +NO 3 - Ag + + NO 3 - + Na + + Cl - AgC l (s) + Na + +NO 3 - spectator NO 3 - and Na + are spectator ions. Net ionic equation Ag + Cl - AgCl (s) Ag + + Cl - AgCl (s)

64. 5 - 63CH104 1 pair = 1 dozen = 1 mole = 1 pair = 1 dozen = 1 mole = The Mole 1 mol eggs___ 6.02 x 10 23 eggs 1 mol Au_______ 6.02 x 10 23 Au atoms _____1 mole H 2 O_____ 6.02 x 10 23 H 2 O molecules 6.02 x 10 23 H 2 O molecules 2 12 6.02 x 10 23 602,000,000,000,000,000,000,000.

65. 5 - 64CH104 1 car ___ 4 wheels The Mole & Formulas 1 mol cars_ 4 mol wheels 4 mol wheels 1 doz cars 4 doz wheels 1 mole H 2 O 2 mol H 2 mol H 1 mole H 2 O 1 mol O 1 mol O

66. 5 - 65CH104 = doz wheels 1 car ___ 4 wheels The Mole & Formulas 1 mol cars_ 4 mol wheels 4 mol wheels 1 doz cars 4 doz wheels 1doz cars 5 doz cars 5 doz cars 1 20 2 mol H 1 mol H 2 O 5 mol H 2 O 5 mol H 2 O 1 10 = mol H

67. 5 - 66CH104 1 mole = MW in g’s The Mole & Molar Mass 1 mol Au_ 197 g Au 197 g Au 1 mol Au_ 197 g Au 197 g Au 1 mole Au = 197 g Au 197 g Au 1 mol Au 197 g Au 1 mol Au __197 g Au _ __197 g Au _ 6.02 x 10 23 atoms Au 6.02 x 10 23 atoms Au __197 g Au _ __197 g Au _ 6.02 x 10 23 atoms Au 6.02 x 10 23 atoms Au

68. 5 - 67CH104 1 mole = MW in g’s The Mole & Molar Mass 1 mol S_ 32 g S 32 g S 1 mol S_ 32 g S 32 g S 1 mole S = 32 g S 32 g S 1 mol S 32 g S 1 mol S 1 mol C 12 g C 12 g C 1 mol C 12 g C 12 g C 1 mole C = 12 g C 12 g C 1 mol C 12 g C 1 mol C

69. 5 - 68CH104 The Mole & Molar Mass 1 mol H 2 O_ 18.0 g H 2 O 1 mole H 2 O has: 1.0 g H = 1 mol H 2 mol H 2 mol H 1 2.0 g H 16.0 g O = 1 mol O 1 16.0 g O 18.0 g H 2 O 18.0 g H 2 O 1 mol H 2 O 18.0 g

70. 5 - 69CH104 Molar Mass Find the MW of Glucose; C 6 H 12 O 6 1.0 g H = 1 mol H 12 mol H 1 12.0 g H 16.0 g O = 1 mol O 6 mol O 1 96.0 g O 180.0 g C 6 H 12 O 6 C 6 H 12 O 6 1 mol C 6 H 12 O 6 12.0 g C = 1 mol C 6 mol C 1 72.0 g C

71. 5 - 70CH104 1 mol H 2 O = 18 g H 2 O Mass to Mole Conversions How many moles of water are in 36 g H 2 O? What should the answer look like? What is Unique to the problem? 36 g H 2 O 36 g H 2 O 1 mol H 2 O 2.0

72. 5 - 71CH104 1 mol H 2 O 18 g H 2 O Mass to Mole Conversions How many moles of H are in 36 g H 2 O? What should the answer look like? What is Unique to the problem? 36 g H 2 O 36 g H 2 O 1 mol H 4.0 2 mol H = 1 mol H 2 O

73. 5 - 72CH104 180 g Gluc = 1 mol Gluc Mole to Mass Conversions How many g’s of Glucose (C 6 H 12 O 6 ) are in 5 mol Glucose? What should the answer look like? What is Unique to the problem? 5 mol Gluc 1 g Glucose 900

74. 5 - 73CH104 You need a balanced equation. You need a balanced equation. H 2 + O 2 -----> H 2 O H 2 + O 2 -----> H 2 O The mole and chemical equations Stoichiometry - Calculations of quantities in a chemical rxn. 2 2

75. 5 - 74CH104 Moles in Chemical Equations 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 2 mol Na 1 mol Cl 2 1 mol Cl 2 2 mol Na 2 mol NaCl Cl 2 1 mol Cl 2 2 mol NaCl

76. 5 - 75CH104 Moles in Chemical Equations 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 2 mol Na 1 mol Cl 2 1 mol Cl 2 2 mol Na 2 mol NaCl Cl 2 1 mol Cl 2 2 mol NaCl How many moles of Cl 2 are needed to completely react with 40 moles of Na? 1 mol Cl 2 = 2 mol Na What is Unique to the problem? 40 mol Na 1 20 mol Cl 2 What should the answer look like?

77. 5 - 76CH104 1mol Na 23.0g Na 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1 mol Cl 2 2 mol Na What is Unique to the problem? 40 g Na 1 61.7 g Cl 2 What should the answer look like? g’s Cl 2 Mass in Chemical Equations mols Namols Cl 2 70.9 g Cl 2 = 1 mol Cl 2 How many g’s of Cl 2 are needed to completely react with 40 g’s of Na? 40 g’s Na

78. 5 - 77CH104 1 mol Fe 2 O 3 159.7g Fe 2 O 3 Fe 2 O 3 H 2 3H 2 O Fe 2 O 3 + 3H 2  2Fe + 3H 2 O 2 mol Fe 1 mol Fe 2 O 3 What is Unique to the problem? 20.g Fe 2 O 3 1 14 g Fe What should the answer look like? g’s Fe Mass in Chemical Equations mols Fe 2 O 3 mols Fe 55.8 g Fe = 1 mol Fe How many g’s of Fe can be produced from 20. g’s of Fe 2 O 3 ? 20.g’s Fe 2 O 3

79. 5 - 78CH104 1 mol Fe 2 O 3 159.7g Fe 2 O 3 Fe 2 O 3 H 2 3H 2 O Fe 2 O 3 + 3H 2  2Fe + 3H 2 O 2 mol Fe 1 mol Fe 2 O 3 20.g Fe 2 O 3 1 14 g Fe Percent Yield 55.8 g Fe = 1 mol Fe How many g Fe can be made from 20.g Fe 2 O 3 ? What is the percent yield if I only got 12 g’s of Fe from 20. g’s of Fe 2 O 3 ? % yield = Actual x 100 Theoretical Theoretical % yield = Actual x 100 Theoretical Theoretical % yield = 12 g x 100 = 14 g 14 g % yield = 12 g x 100 = 14 g 14 g 85.7 %

80. 5 - 79CH104 Limiting reactant 121118 1 c milk + 2 c bisquick + 1 Tbs oil +1 egg = 18 pancakes If I have 5 c. Bisquick and 2 eggs, how many pancakes can I make? From 5 c. Bisquick From 2 eggs 5 c. Bquick 2 eggs 18 p cakes = 2 c. Bquick 45 45 p cakes 36 36 p cakes 18 p cakes = 1 egg

81. 5 - 80CH104 Limiting reactant 121118 1 c milk + 2 c bisquick + 1 Tbs oil +1 egg = 18 pancakes If I have 5 c. Bisquick and 2 eggs, how many pancakes can I make? From 5 c. Bisquick From 2 eggs 45 45 p cakes 36 36 p cakes So: Eggs are the limiting reagent. I can make only 36 pancakes I’ll have bisquick left over.

82. 5 - 81CH104 Limiting reactant If I have 20.g of Fe 2 O 3 and 2.0g H 2, how many g’s Fe can I make? From 20. g Fe 2 O 3 From 2.0 g H 2 14 g Fe Fe 2 O 3 H 2 3H 2 O Fe 2 O 3 + 3H 2  2Fe + 3H 2 O 1 mol H 2 2.02 g H 2 2 mol Fe 3 mol H 2 2.g H 2 1 37 g Fe 55.8 g Fe = 1 mol Fe So: Fe 2 O 3 is the limiting reagent. I can make only 14 g Fe I’ll have H 2 left over.

83. 5 - 82CH104 31.7 18 g H 2 O = 1 molH 2 O 13.5 18 g H 2 O = 1 molH 2 O Limiting reactant If I have 20.g of NH 3 and 20.g O 2, how many g’s H 2 O can I make? From 20. g NH 3 From 20. g O 2 4NH 3 O 2 6H 2 O 4NH 3 + 5O 2  4NO + 6H 2 O 1 mol NH 3 17 g NH 3 6 mol H 2 O 4 mol NH 3 20.g NH 3 1 g H 2 O 1 mol O 2 32 g O 2 6 mol H 2 O 5 mol O 2 20.g O 2 1 g H 2 O

84. 5 - 83CH104 Limiting reactant If I have 20.g of NH 3 and 20.g O 2, how many g’s H 2 O can I make? From 20. g NH 3 From 20. g O 2 31.7 g H 2 O 4NH 3 O 2 6H 2 O 4NH 3 + 5O 2  4NO + 6H 2 O So: O 2 is the limiting reagent. We can make only 13.5 g H 2 O We’ll have NH 3 left over. 13.5 g H 2 O

85. 5 - 84CH104 13 76 g CS 2 = 1 mol CS 2 5.9 76 g CS 2 = 1 mol CS 2 Limiting reactant If I have 10.g of C and 10.g SO 2, how many g’s CS 2 can I make? From 10. g C From 10. g SO 2 5CO 2 4CO 5C + 2SO 2  1CS 2 + 4CO 1 mol C 12 g C 1 mol CS 2 5 mol C 10.g C 1 g CS 2 1 mol SO 2 64 g SO 2 1 mol CS 2 2 mol SO 2 10.g SO 2 1 g CS 2

86. 5 - 85CH104 Limiting reactant 13 g CS 2 So: SO 2 is the limiting reagent. We can make only 5.9 g CS 2 We’ll have C left over. 5.9 g CS 2 If I have 10.g of C and 10.g SO 2, how many g’s CS 2 can I make? From 10. g C From 10. g SO 2 5CO 2 4CO 5C + 2SO 2  1CS 2 + 4CO

87. 5 - 86CH104 Study of Energy changes in Reactions = used to calculate the amount of useful work produced by chemical reactions. Thermodynamics Enthalpy andEntropy  H = Change in Energy =  H  S = Change in =  S

88. 5 - 87CH104 law of Thermodynamics “Energy can’t be created or destroyed in a chemical reaction” converted Energy just gets converted from one form to another. (you can’t get something from nothing)

89. 5 - 88CH104 Exothermic Reactions Mg + O 2 MgO + Energy CH 4 + 2O 2 CO 2 + 2H 2 O + 211 kcal Breaking bonds costs E (+  H) Making bonds gives E (-  H) Old bonds break New bonds get made H products - H reactants =  H

90. 5 - 89CH104 Activation Energy (E act ) The minimum amount of energy required to produce a chemical reaction. The energy of collision must be great enough to break the old bonds and form the new ones.

91. 5 - 90CH104 H2OH2OH2OH2O H2OH2OH2OH2O more stable Energy in Chemical Reactions Exothermic reaction -  H= heat of reaction Energy Rxn Progress Reactants Products E act = Activation Energy (Gets hot) H 2 + O 2 2H 2 + O 2 2H 2 O + Energy

92. 5 - 91CH104 Exothermic (Exergonic) Rxns Energy Rxn Progress 2Mg + O 2  2MgO + Energy 2H 2 + O 2  2H 2 O + Energy CH 4 + 2O 2  CO 2 + 2H 2 O + 213 kcal

93. 5 - 92CH104 Exothermic Reactions Energy Rxn Progress Reactants(Water) Products(Ice) Energy is released Products are more stable. -H-H-H-H

94. 5 - 93CH104 more stable Energy in Chemical Reactions Endothermic reaction  H= heat of reaction Energy Rxn Progress Reactants Products E act = Activation Energy (Gets cold)

95. 5 - 94CH104 Energy Rxn Progress Endothermic (Endergonic) Rxns Energy is required Products are less stable. Products(Water) Reactants(Ice) +H+H+H+H

96. 5 - 95CH104 Examples of Energy diagrams Exothermic reaction Endothermic reaction  H > 0 reactants are more stable  H < 0 products are more stable  H  H

97. 5 - 96CH104 Examples of energy diagrams High activation energy Low heat of reaction Low activation energy High heat of reaction

98. 5 - 97CH104 1780 Energy in Reactions How many kilojoules are released when 100. g Na reacts with chlorine? 2Na (s) Cl 2(g) 819 kJ 2Na (s) + Cl 2(g)  2NaCl + 819 kJ 1 mol Na 23.0 g Na 819 kJ 2 mol Na 100.g Na 1 = kJ