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NWTC General Chemistry Ch 09

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NWTC General Chemistry Ch 09

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NWTC General Chemistry Ch 09

1. 1. Chapter 9 Calculations from Chemical EquationsAccuratemeasurementand calculationof the correctdosage areimportant indispensing thecorrect medicineto patients Introduction to General, Organic, and Biochemistry 10ethroughout the John Wiley & Sons, Incworld. Morris Hein, Scott Pattison, and Susan Arena
2. 2. Chapter Outline9.1 A Short Review 9.5 Mass-Mass Calculations9.2 Introduction to Stoichiometry 9.6 Limiting Reactant and9.3 Mole-Mole Calculations Yield Calculations9.4 Mole-Mass Calculations Copyright 2012 John Wiley & Sons, Inc 9-2
3. 3. Molar MassMolar Mass – sum of atomic masses of all atoms in 1 mole of an element or compound ; the units are g/mol. 6.022x1023 molecules 6.022x1023 formula units 1 mole = 6.022x1023 atoms 6.022x1023 ions Copyright 2012 John Wiley & Sons, Inc 9-3
4. 4. Molar MassWhat is the molar mass of Al(ClO3)3? 1 Al 1(26.98 g) atomic mass 3 Cl 3(35.45 g) Al 26.98 Cl 35.45 9 O 9(16.00 g) O 16.00 Al(ClO3)3 277.33 g/mol Copyright 2012 John Wiley & Sons, Inc 7-4
5. 5. Molar MassCalculate the mass of 2.5 moles of aluminum chlorate.Plan 2.5 mol Al(ClO3)3  g Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3Calculate 277.33 g Al(ClO3 )3 2.5 mol Al(ClO3 )3 = 690 g Al(ClO3 )3 1 mol Al(ClO3 )3 Copyright 2012 John Wiley & Sons, Inc 9-5
6. 6. Molar MassCalculate the moles of 3.52g of aluminum chlorate.Plan 3.52 g Al(ClO3)3  mol Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3Calculate 1 mol Al(ClO3 )3 3.52 g Al(ClO3 )3 = 1.27 10 2 mol Al(ClO 3 ) 3 277.33 g Al(ClO3 )3 Copyright 2012 John Wiley & Sons, Inc 9-6
7. 7. Molar Mass Calculate the number of formula units contained in 12.4 g aluminum chlorate. Plan 12.4 g Al(ClO3)3  formula units Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g = 6.022x1023 formula units Calculate 6.022 1023 formula units12.4 g Al(ClO3 )3 = 2.69 1022 formula units 277.33 g Al(ClO3 )3 Copyright 2012 John Wiley & Sons, Inc 9-7
8. 8. Your Turn!What is the mass of 3.61 moles of CaCl2?a. 3.61 g atomic massb. 272 g Ca 40.08 24 g Cl 35.45c. 2.17 × 10d. 401 g 1 Ca 1(40.08 g) 2 Cl 2(35.45 g) CaCl2 401 g/mol Copyright 2012 John Wiley & Sons, Inc 9-8
9. 9. You Turn!How many moles of HCl are contained in 18.2 g HCl?a. 1.00 mol atomic massb. 0.500 mol 1 H 1(1.01 g) H 1.01c. 0.250 mol 1 Cl 1(35.45 g) Cl 35.45d. 0.125 mol HCl 36.46g/mol 18.2g HCl / HCl 36.46g/mol= Copyright 2012 John Wiley & Sons, Inc 9-9
10. 10. Your Turn!What is the mass of 1.60×1023 molecules of HCl?a. 9.69 g atomic massb. 137 g H 1.01c. 0.729 g Use Cl 35.45d. 36.5 g 1.60×1023 molecules HCl 6.02×1023 molecules HCl /mol 36.46g/mol HCl Copyright 2012 John Wiley & Sons, Inc 9-10
11. 11. Stoichiometry Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation. 1N2(g) + 3I2(s)  2NI3(s) 1 mol N2 + 3 mol I2  2 mol NI3 Mole ratios come from the coefficients in the balanced equation: 3 mol I 3 mol I 2 1 mol N 2 2 1 mol N 2 2 mol NI 3 2 mol NI 3 The 3 other possibilities are the inverse of these ratios.Review Question 1: What is a mole ratio?2012 John Wiley & Sons, Inc Copyright 9-11
12. 12. Your Turn!Which of these statements is not true about the reaction? 1N2(g) + 3I2(s)  2NI3(s)a. 1 mole of nitrogen is needed for every 3 moles of iodineb. 1 gram of nitrogen is needed for every 3 grams of iodinec. Both statements are true Copyright 2012 John Wiley & Sons, Inc 9-12
13. 13. Using the mole ratioCalculate the number of moles of NI3 that can be made from 5.50 mol N2 in the reaction: 1N2(g) + 3I2(s)  2NI3(s) Plan 5.50 mol N2  mol NI3 moles of desired substance in equation Set-Up mole ratio = moles of starting substance in equation 2 mol NI 3 mol ratio = 1 mol N 2 Calculate 5.50 mol N 2 2 mol NI3 = 11.0 mol NI3 1 mol N 2 Copyright 2012 John Wiley & Sons, Inc 9-13
14. 14. Using the mole ratioCalculate the number of moles of I2 needed to react with 5.50 mol N2 in the reaction: 1N2(g) + 3I2(s)  2NI3(s)Plan 5.50 mol N2  mol I2 3 mol I 2Set-Up mole ratio = 1 mol N 2Calculate 5.50 mol N 2 3 mol I2 = 16.5 mol I 2 1 mol N2 Copyright 2012 John Wiley & Sons, Inc 9-14
15. 15. Your Turn!How many moles of HF will be produced by the complete reaction of 1.42 moles of H2 in the following equation? H2 + F2 2HFa. 0.710b. 1.42c. 2.00d. 2.84 Copyright 2012 John Wiley & Sons, Inc 9-15
16. 16. StoichiometryProblem Solving Strategy for stoichiometry problems:1. Convert starting substance to moles.2. Convert the moles of starting substance to moles of desired substance.3. Convert the moles of desired substance to the units specified in the problem. Copyright 2012 John Wiley & Sons, Inc 9-16
17. 17. Stoichiometry Copyright 2012 John Wiley & Sons, Inc 9-17
18. 18. Mole-Mole CalculationsHow many moles of Al are needed to make 0.0935 mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)Plan 0.0935 mol H2  mol Al 2 mol AlSet-Up mole ratio = 3 mol H2 2 mol AlCalculate 0.0935 mol H2 =.0623 mol Al 3 mol H2 Copyright 2012 John Wiley & Sons, Inc 9-18
19. 19. Mole-Mole CalculationsHow many moles of HCl are needed to make 0.0935 mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)Plan 0.0935 mol H2  mol HCl 6 mol HClSet-Up mole ratio = 3 mol H2Calculate 0.0935 mol H2 6 mol HCl = 0.187 mol HCl 3 mol H2 Copyright 2012 John Wiley & Sons, Inc 9-19
20. 20. Your Turn!How many moles of H2 are made by the reaction of 1.5 mol HCl with excess aluminum? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)a. 0.75 molb. 3.0 molc. 6.0 mold. 4.5 mol Copyright 2012 John Wiley & Sons, Inc 9-20
21. 21. Your Turn!How many moles of carbon dioxide are produced when 3.00 moles of oxygen react completely in the following equation? C3H8 + 5O2 3CO2 + 9H2Oa. 5.00 molb. 3.00 molc. 1.80 mold. 1.50 mol Copyright 2012 John Wiley & Sons, Inc 9-21
22. 22. Your Turn!How many moles of C3H8 are consumed when 1.81x1023 molecules of CO2 are produced in the following equation? C3H8 + 5O2 3CO2 + 4H2Oa. 0.100b. 0.897c. 6.03 × 1022d. 5.43 × 1023 Copyright 2012 John Wiley & Sons, Inc 9-22
23. 23. Mole-Mass Calculations What mass of H2 (2.02 g/mol) is made by the reaction of 3.0 mol HCl with excess aluminum? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) Plan 3.0 mol HCl mol H2  g H2 Calculate 3.0 mol HCl 3 mol H 2 1.5 mol H 2 6 mol HCl 1.5 mol H 2 2.02 g H 2 3.0 g H 2 1 mol H 2 Copyright 2012 John Wiley & Sons, Inc 9-23Review Question 2: grams to moles?
24. 24. Mole-Mass Calculations How many moles of HCl are needed to completely consume 2.00 g Al (26.98g/mol)? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)Plan 2.00 g Al  mol Al  mol HClCalculate 2.00 g Al 1 mol Al = 0.0741 mol Al 26.98 g Al 6 mol HCl 0.0741 mol Al 0.0222 mol HCl 2 mol Al Copyright 2012 John Wiley & Sons, Inc 9-24
25. 25. Mole-Mass Calculations What mass of Al(NO3)3 (213g/mol) is needed to react with .093 mol Na2CO3? 3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)Plan 0.093 mol Na2CO3  mol Al(NO3)3  g Al(NO3)3 2 mol Al(NO3 )3Calculate .093 mol Na 2 CO3 .062 mol Al(NO3 )3 3 mol Na 2CO3 213.00g Al(NO3 )3 0.062 mol Al(NO3 )3 13 g Al(NO3 )3 1 mol Al(NO3 )3 Copyright 2012 John Wiley & Sons, Inc 9-25
26. 26. Mole-Mass Calculations How many moles of Al2(CO3)3 are made by the reaction of 3.45g Na2CO3 (105.99 g/mol) with excess Al(NO3)3? 3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)Plan 3.45g Na2CO3  mol Na2CO3  g Al2(CO3)3 1 mol Na 2CO3Calculate 3.45g Na 2CO3 0.0326 mol Na 2CO3 105.99g Na 2CO3 1 mol Al2 (CO3 )3 0.0326 mol Na 2 CO3 0.0109 mol Na 2CO3 3 mol Na 2CO3 Copyright 2012 John Wiley & Sons, Inc 9-26
27. 27. Your Turn!How many moles of oxygen are consumed when 38.0g of aluminum oxide are produced in the following equation? atomic mass 4Al(s) + 3O2(g) 2Al2O3(s) Al 26.98a. 0.248 O 16.00b. 0.559 Plan g to mol, mol to molc. 1.50 2 Al 2(26.98 g)d. 3.00 3 O 3(16.00 g) Al2O3 101.96 g/mol Copyright 2012 John Wiley & Sons, Inc 9-27
28. 28. Your Turn!What mass of HCl is produced when 1.81x1024 molecules of H2 react completely in the following equation? atomic mass H2(g) + Cl2(g) 2HCl(g) H 1.01a. 54.7g Cl 35.45b. 72.9gc. 109gd. 219g Copyright 2012 John Wiley & Sons, Inc 9-28
29. 29. Review Question 3 Part 1Answer the following given the reaction? Show evidence. Ca3P2 + 6H2O 3Ca(OH)2 + 2PH3a. 1 mole Ca3P2 2 mole PH3b. 1 gram Ca3P2 2 gram PH3c. 3 moles Ca(OH)2 are made for each 2 moles PH3d. Mole ratio is 2 mole PH3 / 1 mole Ca3P2 Copyright 2012 John Wiley & Sons, Inc 9-29
30. 30. Review Question 3 Part 2Answer the following given the reaction? Show evidence. Ca3P2 + 6H2O 3Ca(OH)2 + 2PH3e. 2 mole Ca3P2 + 3 mole H2O 4 mole PH3f. 2 mole Ca3P2 + 15 mole H2O 6 mole Ca(OH)2g. 200g Ca3P2 + 100g H2O Ca3P2 is limitingh. 200g Ca3P2 + 100g H2O 57.4 g PH3 Copyright 2012 John Wiley & Sons, Inc 9-30
31. 31. Mass-Mass CalculationsNow we will put it all together.What mass of Br2 (159.80 g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l)  2AlBr3(s) 7.00 g Al mol Al  mol Br2  g Br2 Copyright 2012 John Wiley & Sons, Inc 9-31
32. 32. Mass-Mass Calculations What mass of Br2 (159.80 g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l)  2AlBr3(s)Plan 7.00 g Al mol Al  mol Br2  g Br2 1 mol AlCalculate 7.00 g Al 0.259 mol Al 26.98g Al 3 mol Br2 0.259 mol Al 0.389 mol Br2 2 mol Al 159.80g Br2 0.389 mol Br2 62.2 g Br2 1 mol Br2 Copyright 2012 John Wiley & Sons, Inc 9-32
33. 33. Mass-Mass Calculations What mass of Fe2S3 (207.91g/mol) can be made from the reaction of 9.34 g FeCl3 (162.20 g/mol) with excess Na2S? 2FeCl3(aq) + 3Na2S(aq) Fe2S3(s) + 6NaCl(aq)Plan 9.34 g FeCl3  mol FeCl3  mol Fe2S3  g Fe2S3Calculate 9.34 g FeCl3 1 mol FeCl3 0.0576 mol FeCl3 162.20g FeCl3 1 mol Fe2S3 0.0576 mol FeCl3 0.0288 mol Fe 2S3 2 mol FeCl3 207.91g Fe2S3 0.0288 mol Fe 2S3 5.99 g Fe 2S3 1 mol Fe2S3 Copyright 2012 John Wiley & Sons, Inc 9-33
34. 34. Your Turn!What mass of oxygen is consumed when 54.0g of water is produced in the following equation? 2H2 + O2 2H2O atomic massa. 0.167 g H 1.01 O 16.00b. 0.667 gc. 1.50 g Plan g to mol, mol to mol, mol to gd. 47.9 g H2O 18.02 g/mol Copyright 2012 John Wiley & Sons, Inc 9-34
35. 35. Your Turn!What mass of H2O is produced when 12.0g of HCl react completely in the following equation? 6HCl + Fe2O3 2FeCl3 + 3H2Oa. 2.97 g atomic massb. 39.4 g H 1.01c. 27.4 g O 16.00 Cl 35.45d. 110. g Copyright 2012 John Wiley & Sons, Inc 9-35
36. 36. Limiting ReactantDetermine the number of that can be made given these quantities of reactants and the reaction equation: +  +  Copyright 2012 John Wiley & Sons, Inc 9-36
37. 37. Limiting ReactantThe limiting reactant is the reactant that limits the amount of product that can be made.The reaction stops when the limiting reactant is used up.What was the limiting reactant in the reaction: + The small blue balls. Copyright 2012 John Wiley & Sons, Inc 9-37
38. 38. Excess ReactantThe excess reactant is the reactant that remains when the reaction stops.There is always left over excess reactant.What was the excess reactant in the reaction: + The excess reactant was the larger blue ball. Copyright 2012 John Wiley & Sons, Inc 9-38
39. 39. Limiting reactant Figure 9.2 The number of bicycles that can be built from these parts is determined by the “limiting reactant” (the pedal assemblies). Copyright 2012 John Wiley & Sons, Inc 9-39
40. 40. Limiting Reactant CalculationsTechnique for solving limiting reactant problems:1. Convert reactant 1 to moles or mass of product2. Convert reactant 2 to moles or mass of product3. Compare answers. The smaller answer is the maximum theoretical yield. Copyright 2012 John Wiley & Sons, Inc 9-40
41. 41. Limiting Reactant CalculationCalculate the number of moles of water that can be made by the reaction of 1.51 mol H2 with 0.932 mol O2. 2H2(g) + O2(g)  2H2O(g)1. Calculate the theoretical yield of H2O assuming H2 is the limiting reactant and that O2 is the excess reactant.2. Calculate the theoretical yield of H2O assuming that O2 is the limiting reactant and that H2 is the excess reactant. Copyright 2012 John Wiley & Sons, Inc 9-41
42. 42. Limiting Reactant Calculation continuedAssuming that H2 is limiting and O2 is excess: 2 mol H 2O 1.51 mol H 2 =1.51 mol H 2O 2 mol H2 Assuming that O2 is limiting and H2 is excess: 2 mol H 2O 0.932 mol O2 =1.86 mol H 2O 1 mol 02 So what is the maximum yield of H2O? Copyright 2012 John Wiley & Sons, Inc 9-42
43. 43. Limiting Reactant Calculation continuedHow much H2 and O2 remain when the reaction stops?H2: Limiting Reactant – None remains. It was used up in the reaction.O2: Excess Reactant – Calculate the amount of O2 used in the reaction with H2. Then subtract that from the original amount. 1 mol O2 1.51 mol H 2 x =0.755 mol O 2 2 mol H 2 0.932 mol O2 to start - 0.755 mol O2 = 0.177 mol of excess O2 Copyright 2012 John Wiley & Sons, Inc 9-43
44. 44. Limiting Reactant CalculationCalculate the mass of copper that can be made from the combination of 15.0 g aluminum with 25.0 g copper(II) sulfate. 2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s) Plan 15 g Al  mol Al  mol Cu  g Cu 25 g CuSO4  mol CuSO4  mol Cu  g Cu Compare answers. The smaller number is the right answer. Copyright 2012 John Wiley & Sons, Inc 9-44
45. 45. Limiting Reactant Calculation continued 2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s) 1. Assume Al is limiting and CuSO4 is in excess. 1 mol Al 3 mol Cu 63.55 g Cu 15.0 g Al x = 53.0 g Cu 26.98 g Al 2 mol Al 1 mol Cu 2. Assume CuSO4 is limiting and Al is in excess. 1 mol CuSO4 3 mol Cu 63.55 g Cu25.0 g CuSO4 = 9.96 g Cu 159.58 g CuSO4 3 mol CuSO 4 1 mol Cu 3. Compare answers. CuSO4 is the limiting reagent. The theoretical yield of Cu is 9.96 g. Copyright 2012 John Wiley & Sons, Inc 9-45
46. 46. Your Turn!True/False:You can compare the quantities of reactant when you work a limiting reactant problem. The reactant you have the least of is the limiting reactant.a. Trueb. False Copyright 2012 John Wiley & Sons, Inc 9-46
47. 47. Your Turn!Which is the limiting reactant when 3.00 moles of copper are reacted with 3.00 moles of silver nitrate in the following equation? Cu + 2AgNO3 Cu(NO3)2 + 2Aga. Cub. AgNO3c. Cu(NO3)2d. Ag Copyright 2012 John Wiley & Sons, Inc 9-47
48. 48. Your Turn!What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate? Cu + 2AgNO3 Cu(NO3)2 + 2Aga. 162gb. 216gc. 324gd. 647g Copyright 2012 John Wiley & Sons, Inc 9-48
49. 49. Percent YieldTo determine the efficiency of a process for making acompound, chemists compute the percent yield of the reaction. Actual Yield % Yield = 100 Theoretical Yield The theoretical yield is the result calculated using stoichiometry. The actual yield of a chemical reaction is the experimental result, which is often less than the theoretical yield due to experimental losses and errors along the way. Copyright 2012 John Wiley & Sons, Inc 9-49Review Question 5 & 6: Theatrical vs actual yield? And, how to calculate?
50. 50. Percent Yield Calculate the % yield of PCl3 that results from reacting 5.00 g P with excess Cl2 if only 17.2 g of PCl3 were recovered. 2P + 3Cl2 2PCl3Compute the expected yield of PCl3 from 5.00 g P with excess Cl2. 1 mol P 2 mol PCl3 137.33 g PCl3 5.00 g P x = 22.2g PCl3 30.97 g P 2 mol P 1 mol PCl3Compute the % Yield. Actual Yield 17.2 g % Yield = 100%= 100%=77.5% Theoretical Yield 22.2 g Copyright 2012 John Wiley & Sons, Inc 9-50
51. 51. Your Turn!In a reaction to produce ammonia, the theoretical yield is 420. g. What is the percent yield if the actual yield is 350. g? A. 83.3% B. 20.0% C. 16.7% D. 120.% Copyright 2012 John Wiley & Sons, Inc 9-51
52. 52. Putting it togetherWhen limestone (calcium carbonate) is reacted with hydrochloric acid the products are calcium chloride, water, and carbon dioxide.A. Write a balanced chemical equation for this reaction.B. What mass of calcium carbonate will be consumed when 20.0g of hydrochloric acid react completely in this reaction?
53. 53. ContinuedWhen limestone (calcium carbonate) is reacted with hydrochloric acid the products are calcium chloride, water, and carbon dioxide.C. What mass of carbon dioxide will be produced when 20.0g of hydrochloric acid react completely in this reaction? Copyright 2012 John Wiley & Sons, Inc 9-53
54. 54. QuestionsReview Questions – #4Paired Questions (pg 186) – Do 1, 3, 7, 9, 11, 15, 21, 27, 29, 31, 35 , 39, 43, 47 – Practice later 2, 6, 12, 16, 20, 24, 28, 32, 36, 40, 44 Copyright 2012 John Wiley & Sons, Inc 1-54