nwtc general chemistry ch 09

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Chapter 9 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena Calculations from Chemical Equations Accurate measurement and calculation of the correct dosage are important in dispensing the correct medicine to patients

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NWTC General Chemistry Ch 09

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  • 1. Chapter 9Calculations from Chemical EquationsAccuratemeasurementand calculationof the correctdosage areimportant indispensing thecorrect medicineto patientsIntroduction to General, Organic, and Biochemistry 10ethroughout the John Wiley & Sons, Incworld.Morris Hein, Scott Pattison, and Susan Arena

2. Chapter Outline9.1 A Short Review 9.5 Mass-Mass Calculations9.2 Introduction to Stoichiometry 9.6 Limiting Reactant and9.3 Mole-Mole CalculationsYield Calculations9.4 Mole-Mass Calculations Copyright 2012 John Wiley & Sons, Inc9-2 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 molecules6.022x1023 formula units1 mole =6.022x1023 atoms6.022x1023 ionsCopyright 2012 John Wiley & Sons, Inc 9-3 4. Molar MassWhat is the molar mass of Al(ClO3)3?1 Al 1(26.98 g)atomic mass3 Cl 3(35.45 g) Al 26.98Cl 35.459 O 9(16.00 g)O16.00 Al(ClO3)3 277.33 g/molCopyright 2012 John Wiley & Sons, Inc7-4 5. Molar MassCalculate the mass of 2.5 moles of aluminum chlorate.Plan2.5 mol Al(ClO3)3 g Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3Calculate277.33 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, Inc9-5 6. Molar MassCalculate the moles of 3.52g of aluminum chlorate.Plan3.52 g Al(ClO3)3 mol Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3Calculate1 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. Molar Mass Calculate the number of formula units contained in 12.4 g aluminum chlorate.Plan12.4 g Al(ClO3)3 formula units Al(ClO3)31 mol Al(ClO3)3 = 277.33 g = 6.022x1023 formula unitsCalculate6.022 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. 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, Inc9-8 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. Your Turn!What is the mass of 1.601023 molecules of HCl?a. 9.69 g atomic massb. 137 gH 1.01c. 0.729 gUse Cl 35.45d. 36.5 g1.601023 molecules HCl6.021023 molecules HCl /mol36.46g/mol HCl Copyright 2012 John Wiley & Sons, Inc 9-10 11. StoichiometryStoichiometry deals with the quantitative relationshipsbetween the reactants and products in a balancedchemical equation.1N2(g) + 3I2(s) 2NI3(s)1 mol N2 + 3 mol I2 2 mol NI3Mole ratios come from the coefficients in the balancedequation: 3 mol I 3 mol I 21 mol N 221 mol N 2 2 mol NI 3 2 mol NI 3The 3 other possibilities are the inverse of these ratios.Review Question 1: What is a mole ratio?2012 John Wiley & Sons, IncCopyright 9-11 12. Your Turn!Which of these statements is not true about thereaction?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. Using the mole ratioCalculate the number of moles of NI3 that can be made from5.50 mol N2 in the reaction: 1N2(g) + 3I2(s) 2NI3(s) Plan5.50 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 22 mol NI3 = 11.0 mol NI3 1 mol N 2 Copyright 2012 John Wiley & Sons, Inc9-13 14. Using the mole ratioCalculate the number of moles of I2 needed to react with5.50 mol N2 in the reaction: 1N2(g) + 3I2(s) 2NI3(s)Plan 5.50 mol N2 mol I23 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, Inc9-14 15. Your Turn!How many moles of HF will be produced by thecomplete reaction of 1.42 moles of H2 in thefollowing equation? H2 + F2 2HFa. 0.710b. 1.42c. 2.00d. 2.84Copyright 2012 John Wiley & Sons, Inc 9-15 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. Stoichiometry Copyright 2012 John Wiley & Sons, Inc 9-17 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)Plan0.0935 mol H2 mol Al2 mol AlSet-Up mole ratio =3 mol H22 mol AlCalculate 0.0935 mol H2 =.0623 mol Al3 mol H2Copyright 2012 John Wiley & Sons, Inc 9-18 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-Upmole ratio = 3 mol H2Calculate 0.0935 mol H2 6 mol HCl= 0.187 mol HCl 3 mol H2 Copyright 2012 John Wiley & Sons, Inc9-19 20. Your Turn!How many moles of H2 are made by the reaction of 1.5mol HCl with excess aluminum?2Al(s) + 6HCl (aq) 2AlCl3(aq) + 3H2(g)a. 0.75 molb. 3.0 molc. 6.0 mold. 4.5 molCopyright 2012 John Wiley & Sons, Inc 9-20 21. Your Turn!How many moles of carbon dioxide are produced when3.00 moles of oxygen react completely in thefollowing 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. Your Turn!How many moles of C3H8 are consumed when1.81x1023 molecules of CO2 are produced in thefollowing equation? C3H8 + 5O2 3CO2 + 4H2Oa. 0.100b. 0.897c. 6.03 1022d. 5.43 1023Copyright 2012 John Wiley & Sons, Inc 9-22 23. Mole-Mass CalculationsWhat 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 H2Calculate 3.0 mol HCl 3 mol H 2 1.5 mol H 26 mol HCl 1.5 mol H 2 2.02 g H 23.0 g H 2 1 mol H 2Copyright 2012 John Wiley & Sons, Inc 9-23Review Question 2: grams to moles? 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)Plan2.00 g Al mol Al mol HClCalculate 2.00 g Al1 mol Al = 0.0741 mol Al26.98 g Al6 mol HCl0.0741 mol Al0.0222 mol HCl 2 mol Al Copyright 2012 John Wiley & Sons, Inc9-24 25. Mole-Mass Calculations What mass of Al(NO3)3 (213g/mol) is needed to react with .093mol Na2CO3?3Na2CO3(aq) + 2Al(NO3)3(aq) Al2(CO3)3(s) + 6NaNO3(aq)Plan0.093 mol Na2CO3 mol Al(NO3)3 g Al(NO3)32 mol Al(NO3 )3Calculate .093 mol Na 2 CO3.062 mol Al(NO3 )33 mol Na 2CO3 213.00g Al(NO3 )3 0.062 mol Al(NO3 )313 g Al(NO3 )31 mol Al(NO3 )3 Copyright 2012 John Wiley & Sons, Inc9-25 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)Plan3.45g Na2CO3 mol Na2CO3 g Al2(CO3)3 1 mol Na 2CO3Calculate 3.45g Na 2CO3 0.0326 mol Na 2CO3105.99g Na 2CO3 1 mol Al2 (CO3 )3 0.0326 mol Na 2 CO3 0.0109 mol Na 2CO33 mol Na 2CO3 Copyright 2012 John Wiley & Sons, Inc 9-26 27. Your Turn!How many moles of oxygen are consumed when 38.0gof aluminum oxide are produced in the followingequation?atomic mass 4Al(s) + 3O2(g) 2Al2O3(s)Al 26.98a. 0.248O 16.00b. 0.559Plan 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. Your Turn!What mass of HCl is produced when 1.81x1024molecules of H2 react completely in the followingequation? atomic mass H2(g) + Cl2(g) 2HCl(g) H 1.01a. 54.7g Cl 35.45b. 72.9gc. 109gd. 219gCopyright 2012 John Wiley & Sons, Inc 9-28 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 Ca3P22 gram PH3c. 3 moles Ca(OH)2 are made for each 2 moles PH3d. Mole ratio is 2 mole PH3 / 1 mole Ca3P2Copyright 2012 John Wiley & Sons, Inc 9-29 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 H2OCa3P2 is limitingh. 200g Ca3P2 + 100g H2O57.4 g PH3Copyright 2012 John Wiley & Sons, Inc9-30 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. Mass-Mass Calculations What mass of Br2 (159.80 g/mol) is needed to completelyconsume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l) 2AlBr3(s)Plan7.00 g Al mol Al mol Br2 g Br2 1 mol AlCalculate7.00 g Al 0.259 mol Al 26.98g Al3 mol Br20.259 mol Al 0.389 mol Br22 mol Al 159.80g Br20.389 mol Br262.2 g Br21 mol Br2 Copyright 2012 John Wiley & Sons, Inc 9-32 33. Mass-Mass Calculations What mass of Fe2S3 (207.91g/mol) can be made from the reactionof 9.34 g FeCl3 (162.20 g/mol) with excess Na2S? 2FeCl3(aq) + 3Na2S(aq)Fe2S3(s) + 6NaCl(aq)Plan9.34 g FeCl3 mol FeCl3 mol Fe2S3 g Fe2S3Calculate9.34 g FeCl3 1 mol FeCl30.0576 mol FeCl3 162.20g FeCl31 mol Fe2S30.0576 mol FeCl3 0.0288 mol Fe 2S32 mol FeCl3 207.91g Fe2S3 0.0288 mol Fe 2S3 5.99 g Fe 2S31 mol Fe2S3 Copyright 2012 John Wiley & Sons, Inc 9-33 34. Your Turn!What mass of oxygen is consumed when 54.0g of wateris 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. Your Turn!What mass of H2O is produced when 12.0g of HCl reactcompletely in the following equation? 6HCl + Fe2O3 2FeCl3 + 3H2Oa. 2.97 gatomic massb. 39.4 g H 1.01c. 27.4 g O 16.00Cl 35.45d. 110. gCopyright 2012 John Wiley & Sons, Inc 9-35 36. Limiting ReactantDetermine the number of that can be madegiven these quantities of reactants and the reactionequation: + + Copyright 2012 John Wiley & Sons, Inc 9-36 37. Limiting ReactantThe limiting reactant is the reactant that limits theamount 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. Excess ReactantThe excess reactant is the reactant that remains whenthe 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. Limiting reactantFigure 9.2The number ofbicycles that canbe built from theseparts is determinedby the limitingreactant (thepedal assemblies).Copyright 2012 John Wiley & Sons, Inc9-39 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. Limiting Reactant CalculationCalculate the number of moles of water that can be madeby 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. 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 2O2 mol H2 Assuming that O2 is limiting and H2 is excess: 2 mol H 2O 0.932 mol O2 =1.86 mol H 2O1 mol 02So what is the maximum yield of H2O?Copyright 2012 John Wiley & Sons, Inc 9-42 43. Limiting Reactant Calculation continuedHow much H2 and O2 remain when the reaction stops?H2: Limiting Reactant None remains. It was used upin the reaction.O2: Excess Reactant Calculate the amount of O2 usedin the reaction with H2. Then subtract that from theoriginal amount.1 mol O21.51 mol H 2 x=0.755 mol O 22 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. Limiting Reactant CalculationCalculate the mass of copper that can be made from thecombination of 15.0 g aluminum with 25.0 g copper(II)sulfate. 2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s) Plan15 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. Limiting Reactant Calculation continued2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)1. Assume Al is limiting and CuSO4 is in excess. 1 mol Al3 mol Cu 63.55 g Cu15.0 g Al x= 53.0 g Cu26.98 g Al 2 mol Al1 mol Cu2. 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 41 mol Cu3. Compare answers.CuSO4 is the limiting reagent. The theoretical yield ofCu is 9.96 g. Copyright 2012 John Wiley & Sons, Inc9-45 46. Your Turn!True/False:You can compare the quantities of reactant when youwork a limiting reactant problem. The reactant youhave the least of is the limiting reactant.a. Trueb. False Copyright 2012 John Wiley & Sons, Inc 9-46 47. Your Turn!Which is the limiting reactant when 3.00 moles ofcopper are reacted with 3.00 moles of silver nitrate inthe following equation?Cu + 2AgNO3 Cu(NO3)2 + 2Aga. Cub. AgNO3c. Cu(NO3)2d. AgCopyright 2012 John Wiley & Sons, Inc 9-47 48. Your Turn!What is the mass of silver (107.87 g/mol) produced bythe reaction of 3.00 moles of copper with 3.00 molesof silver nitrate? Cu + 2AgNO3 Cu(NO3)2 + 2Aga. 162gb. 216gc. 324gd. 647g Copyright 2012 John Wiley & Sons, Inc 9-48 49. Percent YieldTo determine the efficiency of a process for making acompound, chemists compute the percent yield of the reaction.Actual Yield% Yield = 100Theoretical YieldThe theoretical yield is the result calculated usingstoichiometry.The actual yield of a chemical reaction is the experimentalresult, which is often less than the theoretical yield due toexperimental 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. 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 P2 mol PCl3137.33 g PCl3 5.00 g P x = 22.2g PCl330.97 g P2 mol P 1 mol PCl3Compute the % Yield.Actual Yield 17.2 g% Yield =100%=100%=77.5%Theoretical Yield22.2 gCopyright 2012 John Wiley & Sons, Inc 9-50 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. 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 when20.0g of hydrochloric acid react completely in thisreaction? 53. ContinuedWhen limestone (calcium carbonate) is reacted withhydrochloric acid the products are calcium chloride,water, and carbon dioxide.C. What mass of carbon dioxide will be produced when20.0g of hydrochloric acid react completely in thisreaction? Copyright 2012 John Wiley & Sons, Inc 9-53 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, 44Copyright 2012 John Wiley & Sons, Inc 1-54