2. Chapter 12

3. Cookies? u When baking cookies, a recipe is usually used, telling the exact amount of each ingredient If you need more, you can double or triple the amount u Thus, a recipe is much like a balanced equation

4. A. Proportional Relationships u I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar = 12.5 dozen cookies 5 eggs5 doz. 2 eggs Ratio of eggs to cookies

5. Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret balanced chemical equations several ways.

6. In terms of Particles u Element- made of atoms u Molecular compound (made of only non- metals) = molecules u Ionic Compounds (made of a metal and non-metal parts) = formula units (ions)

7. 2H 2 + O 2   2H 2 O u Two molecules of hydrogen and one molecule of oxygen form two molecules of water.  2 Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 2Na + 2H 2 O  2NaOH + H 2

8. Looking at it differently  2H 2 + O 2   2H 2 O u 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. u 2 x (6.02 x 10 23 ) molecules of hydrogen and 1 x (6.02 x 10 23 ) molecules of oxygen form 2 x (6.02 x 10 23 ) molecules of water. u 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

9. In terms of Moles  2 Al 2 O 3  Al + 3O 2  2Na + 2H 2 O  2NaOH + H 2 u The coefficients tell us how many moles of each substance are needed.

10. 36.04 g H 2 +O 2 In terms of Mass u The Law of Conservation of Mass applies u We can check using moles  2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 moles H 2 =4.04 g H 2 1 moles O 2 32.00 g O 2 1 moles O 2 =32.00 g O 2

11. In terms of Mass  2H 2 + O 2   2H 2 O 2 moles H 2 O 18.02 g H 2 O 1 mole H 2 O = 36.04 g H 2 O 2H 2 + O 2   2H 2 O 4.04 g H 2 + 32.00 O 2 = 36.04 g H 2 O

12. A. Proportional Relationships u Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio u Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

13. B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass -moles  grams Molarity - moles  liters soln. Molar volume -moles  liters gas Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

14. Mole to Mole conversions  2 Al 2 O 3  Al + 3O 2 u every time we use 2 moles of Al 2 O 3 we make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 These are two possible conversion factors

15. Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose?  2 Al 2 O 3  Al + 3O 2 3.34 mol Al 2 O 3 2 moles Al 2 O 3 3 moles O 2 =5.01 moles O 2

16. Your Turn 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O u If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.60 mol) u How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) u If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol)

17. How do you get good at this? Hydrogen Fuel Cell DVD

18. Mass in Chemical Reactions How much do you make? How much do you need?

19. Mass-Mass Calculations u We do not measure moles directly, so what can we do? u We can convert grams to moles Use the Periodic Table for mass values u Then do the math with the mole ratio Balanced equation gives mole ratio! u Then turn the moles back to grams Use Periodic table values

20. Mass-Mass Conversion u If 10.1 g of Fe +3 are added to a solution of Copper (II) Sulfate, how much solid copper would form?  Fe + CuSO 4  Fe 2 (SO 4 ) 3 + Cu  2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe 55.85 g Fe 1 mol Fe = 0.181 mol Fe

21. 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 1 mol Cu 63.55 g Cu =17.2 g Cu

22. All Together It Looks Like 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.2 g Cu

23. Examples u To make silicon for computer chips they use this reaction  SiCl 4 + 2Mg  2MgCl 2 + Si u How many grams of Mg are needed to make 9.3 g of Si? u How many grams of SiCl 4 are needed to make 9.3 g of Si? u How many grams of MgCl 2 are produced along with 9.3 g of silicon?

24. More Examples u The U. S. Space Shuttle boosters use this reaction  3 Al(s) + 3 NH 4 ClO 4  Al 2 O 3 + AlCl 3 + 3 NO + 6H 2 O u How much Al must be used to react with 652 g of NH 4 ClO 4 ? u How much water is produced? u How much AlCl 3 ?

25. Gases and Reactions

26. We Can Also Change u Movie Movie u Liters of a gas to moles If we are at STP, Standard Temperature and Pressure, which is 0ºC and 1 atmosphere pressure At STP u 22.4 L of a gas = 1 mole of any gas molecules

27. 1 mol of a gas=22.4 L at STP Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm.

28. Molar Mass (g/mol) 6.02  10 23 particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molar Volume (22.4 L/mol) LITERS OF GAS AT STP Molarity (mol/L)

29. Avogadro Told Us u Equal volumes of gas, at the same temperature and pressure contain the same number of particles. u Moles are numbers of particles u You can treat reactions as if they happen ‘liters’ at a time, as long as you keep the temperature and pressure the same.

30. Gas Stoichiometry Problems u How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol 9 mol

31. Volume-Volume Calculations u How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O2O2 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4

32. u How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O 2 22.4 L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 L Gas Stoichiometry Problems 2KClO 3  2KCl + 3O 2

33. Gas Stoichiometry Problems u How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g ? g

34. 63.55 g Cu 1 mol Cu Stoichiometry Problems u How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M

35. Shortcut for Volume-Volume u How many liters of CO 2 at STP are produced by completely burning 17.5 L of CH 4 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L CH 4 1 L CH 4 1 L CO 2 = 17.5 L CH 4 Note: This only works for Gas Volume-Volume problems.

36. Stoichiometry in the Real World Limiting Reactants and Yield

37. Limiting Reactants u Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly u Limiting Reactant bread u Excess Reactants peanut butter and jelly

38. Limiting Reactants u Limiting Reactant used up in a reaction determines the amount of product u Excess Reactant Not completely used up added to ensure that the other reactant is completely used up cheaper & easier to recycle

39. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates, the limiting reactant and the amount of product formed.

40. Limiting Reactants u 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L0.90 L 2.5M

41. Limiting Reactants 79.1 g Zn 1 mol Zn 65.39 g Zn = 27.1 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 0.90 L 2.5M

42. Limiting Reactants 22.4 L H 2 1 mol H 2 0.90 L 2.5 mol HCl 1 L = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 0.90 L 2.5M

43. Limiting Reactants Zn: 27.1 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc

44. Another Example 1. Do two stoichiometry problems. 2. The one that makes the least product is the limiting reagent. u Copper reacts with sulfur to form copper (I) sulfide. If 10.6 g of copper reacts with 3.83 g S how many grams of product will be formed?

45. u If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?  2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

46. Limiting reactants How many grams of H 2 O is produced if 20.0 grams of H 2 react with 20.0 grams of O 2 ? Which reactant is limiting? Which reactant is excess? How much excess remains? 20. 0 g H 2 Start with one of your givens 2.02 g H 2 1 mol H 2 2H 2 + O 2  2H 2 O 2 mol H 2 2 mol H 2 O 1 mol H 2 O 18.02 g H 2 O = 178 g H 2 O 20.0 g O 2 32.00 g O 2 1 mol O 2 2 mol H 2 O 1 mol H 2 O 18.02 g H 2 O = 22.5 g H 2 O Start with your other given To find the answer for how many grams of H 2 O produced compare both of your answers, the lowest number is the correct answer.

47. Limiting Reactants: How to find excess To find out how much excess is left over you take what you produce and determine excess reactant actually used. 22.5 g H 2 O Based on limiting reactant 18.02g H 2 O 1 mol H 2 O 2 mol H 2 O 2 mol H 2 1 mol H 2 2.02 g H 2 = 2.52 g H 2 This is how many grams you used up. Now take the answer and subtract from your given amount of excess. The answer that you find is how much excess is left over. 20.0g – 2.52g = 17.5g H 2

48. Your Turn u If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? u How many grams of solid will be produced? u How many atoms of Cl are in the solid? u How much excess reagent remains?

49. Your Turn II u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? u How much excess reagent will remain?

51. Yield u The amount of product made in a chemical reaction. u There are three types u Actual yield- what you get in the lab when the chemicals are mixed u Theoretical yield- what the balanced equation tells you you should make. u Percent yield-

52. Percent Yield calculated on paper measured in lab

53. Percent Yield u When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g ? g actual: 46.3 g

54. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g ? g actual: 46.3 g Theoretical Yield:

55. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g

56. Details u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %.

57. Another Example u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield? u What is the theoretical yield? u What is the percent yield? 6.78 g Cu 13.9 g Cu 48.8 %

58. Chapter 16 Energy in Chemical Reactions How Much? In or Out?

59. 58 Energy u Energy is measured in Joules or calories u Every reaction has an energy change associated with it u Exothermic reactions release energy, usually in the form of heat. u Endothermic reactions absorb energy u Energy is stored in bonds between atoms

60. 59 C + O 2  CO 2 Energy ReactantsProducts  C + O 2 395kJ + 395 kJ

61. 60 In terms of bonds C O O C O O Breaking this bond will require energy C O O O O C Making these bonds gives you energy In this case making the bonds gives you more energy than breaking them

62. 61 Exothermic u The products are lower in energy than the reactants u Releases energy

63. 62 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ CaCO 3 + 176 kJ  CaO + CO 2

64. 63 Endothermic u The products are higher in energy than the reactants u Absorbs energy

65. 64 Chemistry Happens in u MOLES u An equation that includes energy is called a thermochemical equation  CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u 1 mole of CH 4 makes 802.2 kJ of energy. u When you make 802.2 kJ you make 2 moles of water

66. 65 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u If 10. 3 grams of CH 4 are burned completely, how much heat will be produced? 10. 3 g CH 4 16.05 g CH 4 1 mol CH 4 802.2 kJ =514 kJ

67. 66 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u How many liters of O 2 at STP would be required to produce 23 kJ of heat? u How many grams of water would be produced with 506 kJ of heat?

68. Heats of Reaction

69. 68 Enthalpy u The heat content a substance has at a given temperature and pressure u Can’t be measured directly because there is no set starting point u The reactants start with a heat content u The products end up with a heat content u So we can measure how much enthalpy changes

70. 69 Enthalpy u Symbol is H  Change in enthalpy is  H u delta H u If heat is released the heat content of the products is lower   H is negative (exothermic) u If heat is absorbed the heat content of the products is higher   H is negative (endothermic)

71. 70 Energy ReactantsProducts  Change is down  H is <0

72. 71 Energy ReactantsProducts  Change is up  H is > 0

73. 72 Heat of Reaction u The heat that is released or absorbed in a chemical reaction  Equivalent to  H  C + O 2 (g)  CO 2 (g) +393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ u In thermochemical equation it is important to say what state  H 2 (g) + 1/2O 2 (g)  H 2 O(g)  H = -241.8 kJ  H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -285.8 kJ

74. 73 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance  C 2 H 4 + 3 O 2  2 CO 2 + 2 H 2 O  C 2 H 6 + O 2  CO 2 + H 2 O  2 C 2 H 6 + 5 O 2  2 CO 2 + 6 H 2 O  C 2 H 6 + (5/2) O 2  CO 2 + 3 H 2 O

75. 74 Standard Heat of Formation  The  H for a reaction that produces 1 mol of a compound from its elements at standard conditions u Standard conditions 25°C and 1 atm. u Symbol is u The standard heat of formation of an element is 0 u This includes the diatomics

76. 75 What good are they? u There are tables (pg. 190) of heats of formations u The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products

77. 76 Examples  CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = -74.86 kJO 2 (g) = 0 kJCO 2 (g) = -393.5 kJH 2 O(g) = -241.8 kJ   H= [-393.5 + 2(-241.8)]-[-74.68 +2 (0)]   H= 802.4 kJ

78. 77 Examples  2 SO 3 (g)  2SO 2 (g) + O 2 (g)

79. 78 Why Does It Work?  If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ  then H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ u If you turn an equation around, you change the sign  2 H 2 O(g)  H 2 (g) + O 2 (g)  H =+571.0 kJ u If you multiply the equation by a number, you multiply the heat by that number.

80. 79 Why does it work? u You make the products, so you need there heats of formation u You “unmake” the products so you have to subtract their heats. u How do you get good at this