1. Let’s make some 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.

2. Stoichiometry CH 2 (Vol. 2), page 38 I. Stoichiometry u Greek for “measuring elements” u Defined: the study of quantitative relationship between the amts of reactants used & amts of products formed by a chem. rxn. based on the Law of Conservation of Mass u (mostly) REVIEW: There are 4 ways to interpret a balanced chemical equation Mr. Mole

3. #1. In terms of Particles u An Element is made of atoms u A Molecular compound (made of only nonmetals) is made up of molecules (Don’t forget the diatomic elements!!!) u Ionic Compounds (made of a metal and nonmetal parts) are made of formula units

4. a. Example: 2H 2 + O 2 → 2H 2 O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Balanced! b. Another example: 2Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

5. #2. In terms of Moles a. The coefficients tell us how many moles of each substance 2Al 2 O 3  Al + 3O 2 2Na + 2H 2 O  2NaOH + H 2 b. A balanced equation is a Molar Ratio (more on this later)

6. #3. In terms of Mass u The Law of Conservation of Mass applies a. We can check mass by using moles (reactants). 2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 mole H 2 = 4.04 g H 2 1 mole O 2 32.00 g O 2 1 mole O 2 = 32.00 g O 2 36.04 g H 2 + O 2 + reactants

7. b. In terms of Mass (for products) 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 36.04 g H 2 + O 2 = 36.04 g H 2 O The mass of the reactants must equal the mass of the products. 36.04 grams reactant = 36.04 grams product

8. #4. In terms of Volume u At STP, 1 mol of any gas = 22.4 L 2H 2 + O 2   2H 2 O (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 )  (2 x 22.4 L H 2 O)

9. A. Mole Ratio – a ratio between the numbers of moles of any two of the substances in a balanced chemical equation 1. Example: Write all possible mole ratios for: 2HgO (s)  2Hg(l) + O 2 (g)2 mol HgO 2 mol Hg1 mol O 22 mol Hg 2 mol HgO1 mol O 2 1 mol O 2 2 mol HgO2 mol Hg

10. Practice Problem Write all the possible mole ratios for the following equation: 4 Al + 3 O 2  2 Al 2 O 3

11. Answer: 4 mol Al 3 mol O 2 2 mol Al 2 O 3 3 mol O 2 4 mol Al2 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al 3 mol O 2

12. Using Compound Masses

13. II. Stoichiometric Calculations – p. 40 u All stoichiometric calculations begin with a balanced chemical equation. u Mole ratios are needed, as well as mass-to-mole conversions (using molar mass) A. stoichiometric mole-to-mole conversion 2K(s) + 2H 2 0 (l)  2KOH (aq) + H 2 (g) From the balanced equation we know that two moles of potassium yields one mole of hydrogen. But how many moles of hydrogen is produced if we have only 0.0400 mol of K?

14. Mole to Mole conversions 2Al 2 O 3  Al + 3O 2 each time we use 2 moles of Al 2 O 3 we will also 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 the two possible conversion factors to use in the solution of the problem. Mole Ratio

15. Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2Al 2 O 3  Al + 3O 2 3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 = 5.01 mol O 2 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Conversion factor from balanced equation

16. Steps to Calculate Stoichiometric Problems 1. Balance the equation! 2. Convert the given amount into moles (if not given in moles) 3. Set up mole ratios (to use to calculate moles of desired chemical) 4. If needed, convert to grams (using Molar Mass)

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

18. Practice Problem Methane gas and sulfur (S 8 ) react to produce carbon disulfide (a liquid often used in the production of cellophane) and a colorless, poisonous gas of hydrogen sulfide (rotten egg smell). a. Write chemical equation (don’t forget to balance!) b. Calculate moles of CS 2 produced when 1.50 mol S 8 is used c. How many moles of H 2 S is produced?

19. Practice Problem - ANSWER a. 2CH 4 (g) + S 8 (s)  2CS 2 (l) + 4H 2 S(g) b. find: moles of CS 2 1.50 mol S 8 2 mol CS 2 1 mol S 8 c. Find moles of H 2 S 1.50 mol S 8 4 mol H 2 S 1 mol S 8 = 3.00 mol CS 2 = 6.00 mol H 2 S

20. Steps to Calculate Stoichiometric Problems 1. Balance the equation! 2. Convert to moles (if needed) 3. Set up mole ratios 4. If needed, convert to grams (using MM)

21. B. Stoichiometric mole-to-mass calculations Suppose you know the number of moles of a reactant or product in a reaction & you want to calculate the mass of another product or reactant. Suppose you know the number of moles of a reactant or product in a reaction & you want to calculate the mass of another product or reactant. Example:Determine the mass of NaCl produced when 1.25 mol of chlorine gas reacts with excess Na. Na(s) + Cl 2 (g)  NaCl (s) Na(s) + Cl 2 (g)  NaCl (s) 1.25 mol Cl 2 Molar Mass!!! = 146 g NaCl 58.44 g NaCl 1 mol NaCl 2 mol NaCl 1 mol Cl 2 Mole Ratio!!! 2 2

22. Practice Problem Sodium chloride is decomposed into the elements sodium and chlorine gas by means of electrical energy. How much chlorine gas, in grams, is obtained when you have 2.50 mol NaCl? Answer: 88.6 g Cl 2

23. Steps to Calculate Stoichiometric Problems 1. Balance the equation! 2. Convert to moles (if needed) 3. Set up mole ratios 4. If needed, convert to grams (using MM)

24. C. Stoichiometric mass-to-mass If you were preparing to carry out a chemical reaction in the laboratory, you would need to know how much of each reactant to use in order to produce the mass of product required.

25. Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 (6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed

26. How do you get good at this?

27. Another example: u If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

28. Practice Problem One of the reactions used to inflate automobile air bags involves solid sodium azide (NaN 3 ), which produces sodium and nitrogen gas. a. Write the chemical equation. b. Determine the mass of N 2 produced from the decomposition of 100.0 g NaN 3. Answer: 64.64 g N 2

29. D. Volume Stoichiometry At STP, 1 mol of any gas = 22.4 L 2H 2 + O 2   2H 2 O (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 )  (2 x 22.4 L H 2 O) NOTE: mass and atoms are ALWAYS conserved - however, molecules, formula units, moles, and volumes will not necessarily be conserved! 67.2 Liters of reactant ≠ 44.8 Liters of product!

30. 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. 1 mole = 22.4 L @ STP Page 135 - 139 - Chapter 3: The KMT (Section entitled: “Stoichiometry Involving Gases”)

31. Volume-Volume Calculations: 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 O 2 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 Notice anything relating these two steps?

32. Shortcut for Volume-Volume? 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 2 L O 2 1 L CH 4 = 8.75 L CH 4 Note: This only works for Volume- Volume problems.

33. molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

34. III. “Limiting” Reactants – p. 59 A. Why do reactions stop? u If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? 1. The limiting reactant is the reactant you run out of first. Determines how much product you can make! 2. The excess reactant is the one you have left over. Often referred to as limiting reagent or excess reagent

35. Limiting Reagents - Combustion

36. How do you find out which is limited? 3. The chemical that makes the least amount of product is the “limiting reactant”. u You can recognize limiting reactant problems because they will give you 2 amounts of chemical 4. Do two stoichiometry problems, one for each reactant given

37. u If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) 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 the Limiting Reagent, since it produced less product.

38. Another example: u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3 10.3 g Al 1 mol Al 3 mol Cu 63.546 g Cu 26.981 g Al 2 mol Al 1 mol Cu 51.7 g CuSO 4 1 mol CuSO 4 3 mol Cu 63.546 g Cu 159.60 g CuSO 4 3 mol CuSO 4 1 mol Cu the CuSO 4 is limited, so Cu = 20.6 g = 36.4 g Cu = 20.6 g Cu =

39. Another example: 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3 10.3 g Al 1 mol Al 3 mol Cu 63.546 g Cu 26.981 g Al 2 mol Al 1 mol Cu 51.7 g CuSO 4 1 mol CuSO 4 3 mol Cu 63.546 g Cu 159.60 g CuSO 4 3 mol CuSO 4 1 mol Cu u How much excess reactant will remain? Means how much excess Al will remain!!! 36.4 g Cu – 20.6 g Cu = 15.8 g Cu = 36.4 g Cu = 20.6 g Cu =

40. u How much excess reactant will remain? 36.4 g Cu – 20.6 g Cu = 15.8 g Cu 15.8 g Cu 63.55g Cu 1 mol Cu 3 mol Cu 2 mol Al 1 mol Al 26.981 g Al = 4.47 g Al

41. ANOTHER Example Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. 1. Determine the limiting reactant. 2. Determine the excess reactant. 3. Determine the amount of the reactant in excess. Write a balanced chemical reaction: 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (aq) + 6O 2 (g)

42. u A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. u 1. Find limiting reactant 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (aq) + 6O 2 (g) 88.0 g CO 2 43.99 g CO 2 1 mol CO 2 6 mol CO 2 6 mol O2O2 1 O2O2 31.988 g O2O2 = 64.0 g O2O2 64.O g H2OH2O 18.01 g H2OH2O 1 mol H2OH2O 6 H20H20 6 O2O2 1 O2O2 31.988 g O2O2 = 114 g O2O2 = 64.0 g O 2 Limiting Reactant is CO 2 !

43. u 2. Find excess reactant 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (aq) + 6O 2 (g) 88.0 g CO 2 43.99 g CO 2 1 mol CO 2 6 mol CO 2 6 mol O 2 1 mol O 2 31.988 g O 2 = 64.0 g O 2 64.O g H 2 O 18.01 g H 2 O 1 mol H 2 O 6 mol H 2 0 6 mol O 2 1 mol O 2 31.988 g O 2 = 113.71 g O 2 EXCESS Reactant is H 2 O!

44. 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (aq) + 6O 2 (g) 3. Determine the mass (of the reactant) in excess. 114 g O 2 – 64.0 g O 2 = 50.0 g O 2 50.O g O 2 31.998 g O 2 1 mol O 2 6 mol H 2 0 6 mol O 2 1 mol H 2 O 18.02 g H 2 O = 28.2 g H 2 O

45. Practice Problem The reaction between solid sodium and iron (III) oxide yields solid sodium oxide and iron. (Another reaction, in a series, that inflates an automobile airbag). If 100.0 g of Na and 100.0 g of iron (III) oxide are used in this reaction, determine the following: a. Limiting reactant b. Excess reactant c. Mass of solid iron produced d. Mass of excess reactant that remains after the reaction is complete

46. Practice Problem - ANSWERS 6Na(s) + Fe 2 O 3 (s)  3Na 2 O(s) + 2Fe(s) a. Fe 2 O 3 b. Na c. 69.92 g Fe d. 13.6 g Na

47. The Concept of: A little different type of yield than you had in Driver’s Education class. What is YIELD?

48. IV. Percent Yield – p. 67 A. Yield is the amount of product made in a chemical reaction. u There are three types: 1. Actual yield- what you actually get in the lab when the chemicals are mixed Or the amount of product produced when the chemical reaction is carried out in an experiment 2. Theoretical yield- what the balanced equation tells should be made The maximum amount of product that can be produced from a given amount of reactant

49. 3. Percent yield: Actual Theoretical Percent yield of the product is the ratio of the actual yield to the theoretical yield, expressed as a percent –Percent yield tells us how “efficient” a reaction is. A way of measuring the efficiency of the reaction in producing a product –Percent yield can not be bigger than 100 %. u Theoretical yield will always be larger than actual yield! –Why? Due to impure reactants, competing side reactions, loss of product in filtering or transferring between containers, measuring x 100

50. Example: u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. u Write balanced equation: 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?

51. 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? = 6.78 g Cu

52. 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? 3.92 g Al 26.982 g Al 1 mol Al 2 mol Al 3 mol Cu 1 mol Cu 63.546 g Cu = 13.8481 g Cu = 6.78 g Cu = 13.8 g Cu

53. 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 = 13.8 g Cu Percent Yield = actual yield theoretical yield X 100 6.78 g Cu 13.8 g Cu X 100 = 49.1 %

54. Your Turn: Aluminum hydroxide is often present in antacids to neutralize stomach acid (HCl). The reaction is as follows: Al(OH) 3 (s) + 3HCl(aq)  AlCl 3 (aq) + 3H 2 O(l) a. If 14.0 g of Al(OH) 3 is present in an antacid tablet, determine the theoretical yield of AlCl 3 produced when the tablet reacts with HCl. b. What is the percent yield if 20.5 g is recovered?

55. Practice Problem - ANSWER Al(OH) 3 (s) + 3HCl(aq)  AlCl 3 (aq) + 3H 2 O(l) If 14.0 g of Al(OH) 3 is present, determine the theoretical yield of AlCl 3 produced. (20.5 g / 23.9 g ) x 100% = 85.7% 14.0 g Al(OH) 3 78.004 g Al(OH) 3 1 mol Al(OH) 3 1 mol Al(OH) 3 1 mol AlCl 3 133.34 g AlCl 3 = 23.9 g AlCl 3 1 mol AlCl 3