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# CHEMICAL EQUATIONS AND REACTION STOICHIOMETRY

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### CHEMICAL EQUATIONS AND REACTION STOICHIOMETRY

1. 1.  CHEMICAL EQUATIONS & REACTION STOICHIOMETRY PROFESOR: QBA MIGUEL ÁNGEL CASTRO RAMÍREZ1
2. 2. Chapter Three Goals 1. Chemical Equations 2. Calculations Based on Chemical Equations 3. The Limiting Reactant Concept 4. Percent Yields from Chemical Reactions 5. Sequential Reactions 6. Concentrations of Solutions 7. Dilution of solutions 8. Using Solutions in Chemical Reactions 9. Synthesis Question2
3. 3. Chemical Equations  Symbolic representation of a chemical reaction that shows: 1. reactants on left side of reaction 2. products on right side of equation 3. relative amounts of each using stoichiometric coefficients3
4. 4. Chemical Equations  Attempt to show on paper what is happening at the laboratory and molecular levels.4
5. 5. Chemical Equations  Look at the information an equation provides: ∆ Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2 5
6. 6. Chemical Equations  Look at the information an equation provides: ∆ Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2  reactants yields products6
7. 7. Chemical Equations  Look at the information an equation provides: ∆ Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2  reactants yields products 1 formula unit 3 molecules 2 atoms 3 molecules7
8. 8. Chemical Equations  Look at the information an equation provides: ∆ Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2  reactants yields products 1 formula unit 3 molecules 2 atoms 3 molecules 1 mole 3 moles 2 moles 3 moles8
9. 9. Chemical Equations  Look at the information an equation provides: ∆ Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2  reactants yields products 1 formula unit 3 molecules 2 atoms 3 molecules 1 mole 3 moles 2 moles 3 moles 159.7 g 84.0 g 111.7 g 132 g9
10. 10. Chemical Equations  Law of Conservation of Matter – There is no detectable change in quantity of matter in an ordinary chemical reaction. – Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. – This law was determined by Antoine Lavoisier.  Propane,C3H8, burns in oxygen to give carbon dioxide and water. ∆ C H +5O  3 8 → 3 CO + 4 H O 2 2 210
11. 11. Law of Conservation of Matter  NH3 burns in oxygen to form NO & water You do it!11
12. 12. Law of Conservation of Matter  NH3 burns in oxygen to form NO & water ∆ 2 NH 3 + O 2  → 2 NO + 3 H 2 O 5 2 or correctly ∆ 4 NH 3 + 5 O 2  → 4 NO + 6 H 2 O12
13. 13. Law of Conservation of Matter  C7H16 burns in oxygen to form carbon dioxide and water. You do it!13
14. 14. Law of Conservation of Matter  C7H16 burns in oxygen to form carbon dioxide and water. ∆ C 7 H16 + 11 O 2 → 7 CO 2 + 8 H 2 O14
15. 15. Law of Conservation of Matter  C7H16 burns in oxygen to form carbon dioxide and water. ∆ C 7 H16 + 11 O 2 → 7 CO 2 + 8 H 2 O  Balancing equations is a skill acquired only with lots of practice – work many problems15
16. 16. Calculations Based on Chemical Equations  Can work in moles, formula units, etc.  Frequently, we work in mass or weight (grams or kg or pounds or tons). ∆ Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2 16
17. 17. Calculations Based on Chemical Equations  Example 3-1: How many CO molecules are required to react with 25 formula units of Fe2O3? 3 CO molecules ? CO molecules = 25 formula units Fe 2 O 3 × 1 Fe 2 O 3 formula unit = 75 molecules of CO17
18. 18. Calculations Based on Chemical Equations  Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide? ? Fe atoms = 2.50 × 10 formula units Fe 2 O 3 518
19. 19. Calculations Based on Chemical Equations  Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide? ? Fe atoms = 2.50 × 10 formula units Fe 2 O 3 5 2 Fe atoms × = 1 formula units Fe 2 O 319
20. 20. Calculations Based on Chemical Equations  Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide? ? Fe atoms = 2.50 × 10 formula units Fe 2 O 3 5 2 Fe atoms × = 5.00 × 10 Fe atoms 5 1 formula units Fe 2 O 320
21. 21. Calculations Based on Chemical Equations  Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide? 1 mol Fe 2 O 3 ? g CO = 146 g Fe 2 O 3 × 159.7 g Fe 2 O 321
22. 22. Calculations Based on Chemical Equations  Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide? 1 mol Fe 2 O 3 3 mol CO ? g CO = 146 g Fe 2O 3 × × 159.7 g Fe 2O 3 1 mol Fe 2O 322
23. 23. Calculations Based on Chemical Equations  Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide? 1 mol Fe 2 O 3 3 mol CO ? g CO = 146 g Fe 2O 3 × × 159.7 g Fe 2O 3 1 mol Fe 2O 3 28.0 g CO × = 76.8 g CO 1 mol CO23
24. 24. Calculations Based on Chemical Equations  Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? 3 mol CO 2 ? g CO 2 = 0.540 mol Fe 2 O3 × 1 mol Fe 2 O324
25. 25. Calculations Based on Chemical Equations  Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? 3 mol CO 2 44.0 g CO 2 ? g CO 2 = 0.540 mol Fe 2 O 3 × × 1 mol Fe 2 O 3 1 mol CO 225
26. 26. Calculations Based on Chemical Equations  Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? 3 mol CO 2 44.0 g CO 2 ? g CO 2 = 0.540 mol Fe 2 O 3 × × 1 mol Fe 2 O 3 1 mol CO 2 = 71.3 g CO 226
27. 27. Calculations Based on Chemical Equations  Example 3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? You do it!27
28. 28. Calculations Based on Chemical Equations  Example 3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? 1 molCO2 1mol Fe2O3 ? g Fe2O3 = 8.65 g CO2 × × × 44.0 g CO2 3 mol CO2 159.7 g Fe2O3 = 10.5 g Fe2O3 1 mol Fe2O328
29. 29. Calculations Based on Chemical Equations  Example 3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide? You do it!29
30. 30. Calculations Based on Chemical Equations 454 g Fe 2 O 3 ? lb CO = 125 lb Fe 2 O 3 × 1 lb Fe 2 O 3 1 mol Fe 2 O 3 3 mol CO × × × 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 28 g CO 1 lb CO × = 65.7 lb CO 1 mol CO 454 g CO YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!! Now go to your text and work the problems assigned!30
31. 31. Limiting Reactant Concept  Kitchen example of limiting reactant concept. 1 packet of muffin mix + 2 eggs + 1 cup of milk → 12 muffins  How many muffins can we make with the following amounts of mix, eggs, and milk?31
32. 32. Limiting Reactant Concept  Mix Packets Eggs Milk 1 1 dozen 1 gallon limiting reactant is the muffin mix 2 1 dozen 1 gallon 3 1 dozen 1 gallon 4 1 dozen 1 gallon 5 1 dozen 1 gallon 6 1 dozen 1 gallon 7 1 dozen 1 gallon limiting reactant is the dozen eggs32
33. 33. Limiting Reactant Concept  Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box? ( 87 bolts 1 set ) 1 bolt = 87 sets ( 110 washers 1 set ) 2 washers = 55 sets ( 99 nuts 1 set ) 1 nut = 99 sets the maximum number we can make is 55 sets determined by the smallest number33
34. 34. Limiting Reactant Concept  Look at a chemical limiting reactant situation. Zn + 2 HCl→ ZnCl2 + H234
35. 35. Limiting Reactant Concept  Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? CS2 + 3 O 2 → CO 2 + 2 SO 235
36. 36. Limiting Reactant Concept  Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? CS2 + 3 O 2 → CO 2 + 2 SO 2 1 mol 3 mol 1 mol 2 mol36
37. 37. Limiting Reactant Concept  Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? CS2 + 3 O 2 → CO 2 + 2 SO 2 1 mol 3 mol 1 mol 2 mol 76.2 g 3(32.0 g) 44.0 g 2(64.1 g)37
38. 38. Limiting Reactant Concept  Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? CS2 + 3 O 2 → CO 2 + 2 SO 2 1 mol CS2 ? mol SO 2 = 95.6 g CS2 × 76.2 g38
39. 39. Limiting Reactant Concept CS2 + 3 O 2 → CO 2 + 2 SO 2 1 mol CS 2 ? mol SO 2 = 95.6 g CS2 × 76.2 g 2 mol SO 2 64.1 g SO 2 × × =161 g SO 2 1 mol CS2 1 mol SO 2 What do we do next? You do it!39
40. 40. Limiting Reactant Concept CS2 + 3 O 2 → CO 2 + 2 SO 2 1 mol CS2 2 mol SO 2 64.1 g SO 2 ? mol SO 2 = 95.6 g CS2 × × × = 161 g SO 2 76.2 g 1 mol CS2 1 mol SO 2 1 mol O 2 2 mol SO 2 64.1 g SO 2 ? mol SO 2 = 110 g O 2 × × × = 147 g SO 2 32.0 g O 2 3 mol O 2 1 mol SO 2  Which is limiting reactant?  Limiting reactant is O2.  What is maximum mass of sulfur dioxide?  Maximum mass is 147 g.40
41. 41. Percent Yields from Reactions  Theoretical yield is calculated by assuming that the reaction goes to completion. – Determined from the limiting reactant calculation.  Actual yield is the amount of a specified pure product made in a given reaction. – In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried.  Percent yield indicates how much of the product is obtained from a reaction. actual yield % yield = × 100% theoretical yield41
42. 42. Percent Yields from Reactions  Example3-9: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?42
43. 43. Percent Yields from Reactions CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O 1. Calculate the theoretical yield43
44. 44. Percent Yields from Reactions CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O 1. Calculate the theoretical yield 88.0 g CH 3COOC 2 H 5 ? g CH 3COOC 2 H 5 = 10.0 g C 2 H 5OH × 46.0 g C 2 H 5OH = 19.1 g CH 3COOC 2 H 544
45. 45. Percent Yields from Reactions CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O 1. Calculate the theoretical yield 88.0 g CH 3COOC 2 H 5 ? g CH 3COOC 2 H 5 = 10.0 g C 2 H 5OH × 46.0 g C 2 H 5OH = 19.1 g CH 3COOC 2 H 5 2. Calculate the percent yield.45
46. 46. Percent Yields from Reactions CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O 1. Calculate the theoretical yield 88.0 g CH 3COOC 2 H 5 ? g CH 3COOC 2 H 5 = 10.0 g C 2 H 5OH × 46.0 g C 2 H 5OH = 19.1 g CH 3COOC 2 H 5 2. Calculate the percent yield. 14.8 g CH 3COOC 2 H 5 % yield = × 100% = 77.5% 19.1 g CH 3COOC 2 H 546
47. 47. Sequential Reactions N O2 NH2 HNO3 Sn H2SO4 Conc. HCl benzene nitrobenzene aniline  Example 3-10: Starting with 10.0 g of benzene (C6H6), calculate the theoretical yield of nitrobenzene (C6H5NO2) and of aniline (C6H5NH2).47
48. 48. Sequential Reactions 1 mol benzene ? g nitrobenzene = 10.0 g benzene × × 78.0 g benzene48
49. 49. Sequential Reactions 1 mol benzene ? g nitrobenzene = 10.0 g benzene × × 78.0 g benzene 1 mol nitrobenzene 123.0 g nitrobenzene × = 15.8 g nitrobenzene 1 mol benzene 1 mol nitrobenzene  Next calculate the mass of aniline produced. You do it!49
50. 50. Sequential Reactions N O2 NH2 HNO3 Sn H2SO4 Conc. HCl benzene nitrobenzene aniline 1 mol nitrobenzene ? g aniline = 15.8 g nitrobenzene × × 123.0 g nitrobenzene50
51. 51. Sequential Reactions N O2 NH2 HNO3 Sn H2SO4 Conc. HCl benzene nitrobenzene aniline 1 mol nitrobenzene ? g aniline = 15.8 g nitrobenzene × × 123.0 g nitrobenzene 1 mol aniline 93.0 g aniline × = 11.9 g aniline 1 mol nitrobenzene 1 mol aniline51
52. 52. Sequential Reactions  If6.7 g of aniline is prepared from 10.0 g of benzene, what is the percentage yield? You do it! 6.7 g aniline % yield = ×100% = 56% 11.9 g aniline52
53. 53. Concentration of Solutions  Solution is a mixture of two or more substances dissolved in another. – Solute is the substance present in the smaller amount. – Solvent is the substance present in the larger amount. – In aqueous solutions, the solvent is water.  The concentration of a solution defines the amount of solute dissolved in the solvent. – The amount of sugar in sweet tea can be defined by its concentration.  One common unit of concentration is: mass of solute % by mass of solute = ×100% mass of solution mass of solution = mass of solute + mass of solvent53 % by mass of solute has the symbol % w/w
54. 54. Concentration of Solutions  Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH. 100.0 g solution ? g solution = 32.0 g NaOH × 8.00 g NaOH = 400. g sol n54
55. 55. Concentration of Solutions  Example 3-11: What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% w/w NaOH?  8.00 g NaOH  250.0 g solution  100.0 g solution  = 20.0 g NaOH   55
56. 56. Concentration of Solutions  Example 3-13: Calculate the mass of NaOH in 300.0 mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL. You do it! 1.09 g sol n ? g NaOH = 300.0 mL sol n × × 1 mL sol n 8.00 g NaOH = 26.2 g NaOH 100 g sol n56
57. 57. Concentrations of Solutions  Example 3-14: What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL. You do it!57
58. 58. Concentrations of Solutions  Example 3-14: What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL. 100.0 g solution 1 mL solution ? mL solution = 40.0 g KOH × × 12.0 g KOH 1.11 g solution = 300. mL solution58
59. 59. Concentrations of Solutions  Second common unit of concentration: number of moles of solute molarity = number of liters of solution moles M= L mmol M= mL59
60. 60. Concentrations of Solutions  Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. You do it!60
61. 61. Concentrations of Solutions  Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. ? mol H 2SO 4 12.5 g H 2SO 4 1 mol H 2SO 4 = × L sol n 1.75 L sol n 98.1 g H 2SO 461
62. 62. Concentrations of Solutions  Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. ? mol H 2SO 4 12.5 g H 2SO 4 1 mol H 2SO 4 = × L sol n 1.75 L sol n 98.1 g H 2SO 4 0.0728 mol H 2SO 4 = L = 0.0728 M H 2SO 462
63. 63. Concentrations of Solutions  Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . You do it!63
64. 64. Concentrations of Solutions  Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . 0.800 mol Ca(NO 3 ) 2 ? g Ca(NO 3 ) 2 = 3.50 L × × L 164 g Ca(NO 3 ) 2 = 459 g Ca(NO 3 ) 2 1 mol Ca(NO 3 ) 264
65. 65. Concentrations of Solutions  Oneof the reasons that molarity is commonly used is because: M x L = moles solute and M x mL = mmol solute65
66. 66. Concentrations of Solutions  Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity? specific gravity = 1.185 tells us density =1.185 g/mL or 1185 g/L66
67. 67. Concentrations of Solutions  Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity? specific gravity = 1.185 tells us density =1.185 g/mL or 1185g/L 1185 g solution 36.31 g HCl ? mol HCl/L = × × L solution 100 g sol n67
68. 68. Concentrations of Solutions  Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity? specific gravity = 1.185 tells us density =1.185 g/mL or 1185g/L 1185 g solution 36.31 g HCl ? mol HCl/L = × × L solution 100 g sol n 1 mol HCl = 11.80 M HCl 36.46 g HCl68
69. 69. Dilution of Solutions  To dilute a solution, add solvent to a concentrated solution. – One method to make tea “less sweet.” – How fountain drinks are made from syrup.  The number of moles of solute in the two solutions remains constant.  The relationship M1V1 = M2V2 is appropriate for dilutions, but not for chemical reactions.69
70. 70. Dilution of Solutions  Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.70
71. 71. Dilution of Solutions  Example 3-18: If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution? M 1V1 = M 2 V2 12.0 M ×10.0 mL = M 2 ×100.0 mL 12.0 M ×10.0 mL M2 = 100.0 mL =1.20 M71
72. 72. Dilution of Solutions  Example 3-19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? You do it!72
73. 73. Dilution of Solutions  Example 3-19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? M 1 V1 = M 2 V2 M 2 × V2 V1 = M1 2.50 L × 2.40 M V1 = 18.0 M73 = 0.333 L or 333 mL
74. 74. Using Solutions in Chemical Reactions  Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.74
75. 75. Using Solutions in Chemical Reactions  Example 3-20: What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4? Na 2SO 4 + BaCl 2 → BaSO 4 + 2 NaCl75
76. 76. Using Solutions in Chemical Reactions  Example 3-20: What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4? Na 2SO 4 + BaCl 2 → BaSO 4 + 2 NaCl 1 mol Na 2SO 4 ? L BaCl 2 = 4.32 gNa 2SO 4 × × 142 g Na 2SO 476
77. 77. Using Solutions in Chemical Reactions  Example 3-20: What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4? Na 2SO 4 + BaCl 2 → BaSO 4 + 2 NaCl 1 mol Na 2SO 4 ? L BaCl 2 = 4.32 gNa 2SO 4 × × 142 g Na 2SO 4 1 mol BaCl 2 1 L BaCl 2 × = 0.0608 L 1 mol Na 2SO 4 0.500 mol BaCl 277
78. 78. Using Solutions in Chemical Reactions  Example3-21: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO3)3? Al ( NO3 ) 3 + 3 NaOH → Al ( OH ) 3 + 3 NaNO3 You do it!78
79. 79. Using Solutions in Chemical Reactions  Example3-20: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate? Al( NO 3 ) 3 + 3 NaOH →Al(OH)3 + 3 NaNO3 1L ? mL NaOH = 50.0 mL Al(NO3 ) 3 sol n × 1000 mL 0.200 mol Al(NO3 ) 3 sol n 3 mol NaOH × × 1 L Al(NO3 ) 3 sol n 1 mol Al(NO3 ) 3 1 L NaOH = 0.150 L or 150 mL NaOH sol n79 0.200 mol NaOH
80. 80. Using Solutions in Chemical Reactions  (b)What mass of Al(OH)3 precipitates in (a)? You do it!80
81. 81. Using Solutions in Chemical Reactions  (b) What mass of Al(OH)3 precipitates in (a)? 1L ? g Al(OH) 3 = 50.0 mL Al(NO 3 )3 sol n × 1000 mL 0.200 mol Al(NO 3 )3 1 mol Al(OH) 3 78.0 g Al(OH) 3 × × 1 L Al(NO 3 )3 sol n 1 mol Al(NO 3 )3 1 mol Al(OH) 3 = 0.780 g Al(OH) 381
82. 82. Using Solutions in Chemical Reactions  Titrationsare a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. – Requires special lab glassware  Buret, pipet, and flasks – Must have an an indicator also82
83. 83. Using Solutions in Chemical Reactions  Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl? KOH + HCl → KCl + H 2 O83
84. 84. Using Solutions in Chemical Reactions  Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl? KOH + HCl → KCl + H 2 O 43.2 mL × 0.223 M HCl = 9.63 mmol HCl84
85. 85. Using Solutions in Chemical Reactions  Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl? KOH + HCl → KCl + H 2 O 43.2 mL × 0.223 M HCl = 9.63 mmol HCl 1 mmol KOH 9.63 mmol HCl × = 9.63 mmol KOH 1 mmol HCl85
86. 86. Using Solutions in Chemical Reactions  Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl? KOH + HCl → KCl + H 2 O 43.2 mL × 0.223 M HCl = 9.63 mmol HCl 1 mmol KOH 9.63 mmol HCl × = 9.63 mmol KOH 1 mmol HCl 9.63 mmol KOH = 0.249 M KOH 38.7 mL KOH86
87. 87. Using Solutions in Chemical Reactions  Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution? Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O (44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl87
88. 88. Using Solutions in Chemical Reactions  Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution? Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O (44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl 1 mmol Ba(OH)2 4.54 mmol HCl × 2 mmol HCl88
89. 89. Using Solutions in Chemical Reactions  Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution? Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O (44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl 1 mmol Ba(OH)2 4.54 mmol HCl × 2 mmol HCl = 2.27 mmol Ba(OH)289
90. 90. Using Solutions in Chemical Reactions  Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution? Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O (44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl 1 mmol Ba(OH)2 4.54 mmol HCl × 2 mmol HCl = 2.27 mmol Ba(OH)2 2.27 mL Ba(OH)2 = 0.0593M Ba(OH)290 38.3 mL Ba(OH) 2
91. 91. Synthesis Question  Nylonis made by the reaction of hexamethylene diamine CH2 CH2 CH2 NH2 H2N CH2 CH2 CH2 with adipic acid. HO H2 H2 C O C C C C C O H2 H2 OH91
92. 92. Synthesis Question in a 1 to 1 mole ratio. The structure of nylon is: H2 H2 O H2 H2 H2 C C C C C C * C C C N C C N n * H2 H2 H H2 H2 H O where the value of n is typically 450,000. On a daily basis, a DuPont factory makes 1.5 million pounds of nylon. How many pounds of hexamethylene diamine and adipic acid must they have available in the plant each day?92
93. 93. Synthesis Question Molar mass of 1 nylon molecule = [(12 × 12) + (1× 22) + ( 2 × 14) + (2 × 16)] 450,000             C atoms H atoms N atoms O atoms # of units = [226 g/mol] 450,000 = 1.02 × 108 g/mol nylon molecules93
94. 94. Synthesis Question Molar mass of 1 nylon molecule = [(12 × 12) + (1× 22) + ( 2 × 14) + (2 × 16)] 450,000             C atoms H atoms N atoms O atoms # of units = [226 g/mol] 450,000 = 1.02 × 108 g/mol nylon molecules  454 g  1.5 million pounds = (1.5 × 106 lb)   lb  = 6.81× 108 g94
95. 95. Synthesis Question Molar mass of 1 nylon molecule = [(12 × 12) + (1× 22) + ( 2 ×14) + (2 ×16)] 450,000             C atoms H atoms N atoms O atoms # of units = [226 g/mol] 450,000 = 1.02 ×108 g/mol nylon molecules  454 g  1.5 million pounds = (1.5 ×10 lb) 6   lb  = 6.81×108 g  1 mol nylon  ( # mol of nylon molecules = 6.81×108 g  )  1.02 ×108 g     = 6.68 mol of nylon95
96. 96. Synthesis Question Because the nylon formation reaction uses 1 mole of adipic acid × 450,000 plus 1 mole of hexamethylene diamine × 450,000 per mole of nylon formed, to make 6.68 mol of nylon requires :96
97. 97. Synthesis Question Because the nylon formation reaction uses 1 mole of adipic acid × 450,000 plus 1 mole of hexamethylene diamine × 450,000 per mole of nylon formed, to make 6.68 mol of nylon requires :  1 lb  ( ) adipic acid - 6.68 × 450,000 ×146 g/mol = 4.39 ×108 g   = 9.66 ×105 lb  454 g   97
98. 98. Synthesis Question Because the nylon formation reaction uses 1 mole of adipic acid × 450,000 plus 1 mole of hexamethylene diamine × 450,000 per mole of nylon formed, to make 6.68 mol of nylon requires :  1 lb  ( ) adipic acid - 6.68 × 450,000 ×146 g/mol = 4.39 ×108 g   454 g  = 9.66 ×10 lb  5    1 lb  ( hexamethylene diamine - 6.68 × 450,000 ×116 g/mol = 3.49 ×108 g  ) 454 g  = 7.68 ×10 lb  5  98
99. 99. Group Activity Manganese dioxide, potassium hydroxide and oxygen react in the following fashion: 4 MnO 2 + 4 KOH + 3 O 2 → 4 KMnO 4 + 2 H 2 O A mixture of 272.9 g of MnO2, 26.6 L of 0.250 M KOH, and 41.92 g of O2 is allowed to react as shown above. After the reaction is finished, 234.6 g of KMnO4 is separated from the reaction mixture. What is the per cent yield of99 this reaction?
100. 100. End of Chapter 3100