Chapter9 stoichiometry-100707061730-phpapp01

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Chemicals reactions, Limit reactant, Yield.

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Chapter9 stoichiometry-100707061730-phpapp01

  1. 1. Chapter 9Stoichiometry
  2. 2. Introduction to StoichiometrySection 1
  3. 3.  Much knowledge of chemistry based on quantitative analysis of substances in chemical reactions Composition stoichiometry  deals with mass relationships of elements in compounds Reaction stoichiometry  deals with mass relationships between reactants and products in a chemical reaction
  4. 4.  Based on chemical equations and law of conservation of matter All calculations start with balanced chemical equation This gives numbers of moles of reactants and products
  5. 5. Reaction-Stoichiometry Problems Can be classified according to information given in problem and info you are expected to find, the unknown May both be reactants, may both be products, or both Masses usually expressed in grams Solved by using ratios to convert given quantities by following methods:
  6. 6. Problem Type 1: Given and unknown quantities are amounts in moles General plan: Amt of given (mol)  amt of unknown (mol)
  7. 7. Problem Type 2: Given is in moles and unknown is a mass in gramsAmt given (mol)  amt unknown (mol)  amt unknown (g)
  8. 8. Problem Type 3: Given is mass (g) and unknown is in molesMass given (g)  amt given (mol)  amount unknown (mol)
  9. 9. Problem Type 4: Given is mass (g) and unknown is mass (g)Mass given  amt given (mol)  amt unknown (mol)  mass unknown
  10. 10. Mole Ratio Solving any reaction-stoichiometry problem requires use of mole ratio to convert from moles to grams Mole ratio  conversion factor that relates amounts in moles of any two substances involved in a chemical reaction Get directly from balanced equation
  11. 11. Example: 2Al2O3(l)  4Al(s) + 3O2(g)  Moles ratios:
  12. 12. 2Al2O3(l)  4Al(s) + 3O2(g) Use mole ratios to convert from amount in moles of one substance to amount in moles of another Ex. 13.0 mol Al2O3
  13. 13. Molar Mass Mass of one mole of a substance It is the conversion factor that relates mass of substance to amount in moles To solve stoichiometry problems, you need to determine molar mass of substances
  14. 14.  Molar masses of substances: Al2O3 = 101.96 g/mol O2 = 32.00 g/mol Al = 26.98 g/mol Can use these as conversion factors
  15. 15.  Find number of grams of Al equal to 26.0 mol of Al
  16. 16. Ideal Stoichiometric CalculationsSection 2
  17. 17. Ideal Stoichiometric Calculations Equations are very important because you get the mole ratio directly from it First thing to solving these types of problems is balancing the equation Chemical equations help make predictions about reactions without having to run experiments (and waste resources) in lab
  18. 18.  Calculations in this book are theoretical They tell amounts of reactants and products under ideal conditions (where all reactants completely converted into products) Hardly ever happens in real life
  19. 19. Conversions of Quantities inMoles If asked for moles of product made from specific number of moles of reactant:
  20. 20.  Plan requires 1 conversion factor Mole ratio
  21. 21. Sample Problem
  22. 22. 1. Analyze Given: Amount of CO2 = 20 mol Unknown: Amount of LiOH in moles
  23. 23. 2. Plan Amount CO2 (mol)  amount LiOH (mol) Requires a mole to mole ratio between CO2 and LiOH
  24. 24. 3. Compute
  25. 25.  How many moles of sodium will react with water to produce 4.0 mol of hydrogen in the following reaction? 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) 8.0 mol Na
  26. 26.  How many moles of lithium chloride will be formed by the reaction of chlorine with 0.046 mol of lithium bromide in the following reaction? 2LiBr(aq) + Cl2(g)  2LiCl(aq) + Br2(l) 0.046 mol LiCl
  27. 27. Conversions of Amounts in Moles toMass Usually asked to find mass in grams of product formed
  28. 28. Practice Problem In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, is produced when 3.00 mol of water react with carbon dioxide?
  29. 29. 1. Analyze Given: Amount H2O = 3.00 mol Unknown: Mass of glucose produced (in g)
  30. 30. 2. Plan Chemical equation is Need 2 conversion factors: Mole ratio of CO2 to H2O Molar mass of CO2
  31. 31. 3. Compute
  32. 32. Practice Problem 1 Phosphorous burns in air to produce a phosphorous oxide in the following reaction. 4P(s) + 5O2(g)  P4O10(s)What mass of phosphorous will be needed to produce 3.25 mol P4O10? 403 g
  33. 33. Practice Problem 2 Hydrogen peroxide breaks down, releasing oxygen, in the following reaction. 2H2O2(aq)  2H2O(l) + O2(g)What mass of oxygen is produced when 1.840 mol of H2O2 decompose? 29.44 g
  34. 34. Conversions of Mass to Amounts inMoles In this type of problem you are starting with mass of some substance Plan:
  35. 35. Sample Problem
  36. 36. 1. Analyze Given: Mass of NH3 = 824 g Unknown: a. amount of NO produced (in mol) b. amount of H2O produced (in mol)
  37. 37. 2. Plan First write balanced equation
  38. 38.  Mole ratios needed:
  39. 39. 3. Compute
  40. 40. Mass-Mass Calculations Since you can never measure moles directly, mass-mass calculations are more common
  41. 41. Sample Problem
  42. 42. 1. Analyze Given: Amount of HF = 30.00 g Unknown: Mass of SnF2 produced in grams
  43. 43. 2. Plan
  44. 44. 3. Compute
  45. 45. Practice Problem 1 Calculate the mass of silver bromide produced from 22.5 g of silver nitrate in the following reaction: 2AgNO3 + MgBr2  2AgBr + Mg(NO3)2 24.9 g AgBr
  46. 46. Practice Problem 2 What mass of acetylene, C2H2, will be produced from the reaction of 90. g of calcium carbide, CaC2, with water in the following reaction?CaC2(s) + 2H2O(l)  C2H2(s) + Ca(OH)2(aq) 37 g
  47. 47. Limiting Reactants and PercentYieldSection 3
  48. 48.  In experiments, a reaction is rarely ever done with exact amounts of reactants Usually one or more reactants is in excess (too much) Once one of the reactants is used up, the reaction stops Substance used up first is the limiting reactant
  49. 49.  Limiting reactant  the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction Excess reactant  the substance that is not used up completely in reaction Consider the following reaction C(s) + O2(g)  CO2(g) According to equation, 1 mol C reacts with 1 mol oxygen to form 1 mol carbon dioxide Supposed you could mix 5 mol C with 10 mol O2
  50. 50.  Consider the following reaction C(s) + O2(g)  CO2(g) According to equation, 1 mol C reacts with 1 mol oxygen to form 1 mol carbon dioxide Supposed you could mix 5 mol C with 10 mol O2
  51. 51. C(s) + O2(g)  CO2(g)
  52. 52. Problem Limit ReactantProblem LR :Aluminum oxidizes accordingto the following equation: 4Al + 3O2 2Al2O3Powdered Al (0.048 mol) is placed into acontainer containing 0.030 mol O2. What isthe limiting reactant?O2
  53. 53. 1. Analyze Given: Amount of HF = 2.0 mol Amount of SiO2 = 4.5 mol Unknown: Limiting reactant
  54. 54. 2. Plan The given amount of either reactant is used to calculate required amount of other reactant Calculated amount compared with amount you actually have Limiting reactant can be identified
  55. 55.  Under ideal conditions, 2.0 mol HF requires 0.50 mol SiO2 for complete reaction Because 4.5 mol SiO2 is available, that is more than is required so…. HF is limiting reactant
  56. 56. Practice Problem 1 Aluminum oxidizes according to the following equation: 4Al + 3O2  2Al2O3 Powdered Al (0.048 mol) is placed into a container containing 0.030 mol O2. What is the limiting reactant? O2
  57. 57. Practice Problem 2 Heating zinc sulfide in the presence of oxygen yields the following: ZnS + O2  ZnO + SO2 If 1.72 mol ZnS is heated in the presence of 3.04 mol O2, which reactant will be used up? ZnS
  58. 58. Practice Problem 3 Use the following equation for the oxidation of aluminum in the following problems: 4Al + 3O2  2Al2O3
  59. 59. 4Al + 3O2  2Al2O3 Which reactant is limiting if 0.32 mol Al and 0.26 mol O2 are available? Al How many moles of Al2O3 are formed from the reaction of 6.38 x 10-3 mol of O2 and 9.15 x 10-3 mol of Al? 4.25 x 10-3 mol Al2O3 If 3.17 g Al and 2.55 g of O2 are available, which reactant is limiting? O2
  60. 60. Practice Problem 4ZrSiO4 + 2Cl2  ZrCl4 + SiO2 + O2 What mass of ZrCl4 can be produced if 862 g of ZrSiO4 and 950 g of Cl2 are available? 1.10 x 103 g
  61. 61. Percent Yield Amounts of products calculated in problems represent theoretical yields Theoretical yield  maximum amount of product that can be made from a given amount of reactant In lab, amount of product is usually less than theoretical yield
  62. 62. Why? Some reactant may be used competing in side reactions that reduce amount of product Once product is formed, usually collected in impure form Some product lost during purification Actual yield  measured amount of product gotten from reaction
  63. 63.  Chemists usually interested in efficiency of reaction Expressed by comparing actual and theoretical yields Percent yield  ratio of actual yield to theoretical yield multiplied by 100
  64. 64. Sample Problem
  65. 65. 1. Analyze Given: Mass C6H6 = 36.8 g Mass of Cl2 = excess Actual yield of C6H5Cl = 38.8 g Unknown: Percent yield of C6H5Cl
  66. 66. 2. Plan First do mass-mass calculation to find theoretical yield
  67. 67.  Then percent yield can be found
  68. 68. 3. Compute
  69. 69. Practice Problem Calculate the percent yield in each of the following cases T Yield = 50.0 g, A Yield = 41.9 g 83.8% T yield = 290 kg, A Yield = 270 kg 93% T Yield = 0.00192 g, A yield = 0.00089 g 46%

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