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Chapter 4

Chapter 4. Chemical Equations and Stoichiometry. CHEMICAL REACTIONS. Reactants: Zn + I 2. Product: Zn I 2. Chemical Equations. Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s)

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Chapter 4

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  1. Chapter 4 Chemical Equations and Stoichiometry Dr. S. M. Condren

  2. CHEMICAL REACTIONS Reactants: Zn + I2 Product: Zn I2 Dr. S. M. Condren

  3. Chemical Equations Depict the kind of reactantsand productsand their relative amounts in a reaction. 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds, (aq) refers to an aqueous or water solution. Dr. S. M. Condren

  4. The Mole and Chemical Reactions:The Nano-Macro Connection 2 H2 + O2 -----> 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules 2 H2 moles molecules 1 O2 mole molecules 2 H2O moles molecules 4 g H2 32 g O2 36 g H2O Dr. S. M. Condren

  5. Stoichiometry stoi·chi·om·e·trynoun 1. Calculation of the quantities of reactants and products in a chemical reaction. 2. The quantitative relationship between reactants and products in a chemical reaction. Dr. S. M. Condren

  6. Chemical Equations 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s) This equation means 4 Al atoms + 3 O2 molecules ---give---> 2 “molecules” of Al2O3 4 moles of Al + 3 moles of O2 ---give---> 2 moles of Al2O3 Dr. S. M. Condren

  7. Chemical Equations • Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. • The Law of the Conservation of Matter Demo of conservation of matter 2HgO(s) ---> 2 Hg(liq) + O2(g) Dr. S. M. Condren

  8. Combination Reaction PbNO3(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq) Colorless yellow yellow colorless Dr. S. M. Condren

  9. Lavoisier, 1788 Chemical Equations Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Dr. S. M. Condren

  10. Balancing Equations ___ Al(s) + ___ Br2(liq) ---> ___ Al2Br6(s) Dr. S. M. Condren

  11. Balancing Equations ____C3H8(g) + _____ O2(g) ----> _____CO2(g) + _____ H2O(g) ____B4H10(g) + _____ O2(g) ----> ___ B2O3(g) + _____ H2O(g) Dr. S. M. Condren

  12. EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2? 2 2 H2 + O2 -----> H2O #mol H2O = (3.3 mol O2) (2 mol H2O) (1 mol O2) Now we need the stoichiometric factor = 6.6 mol H2O Dr. S. M. Condren

  13. Stoichiometric Roadmap Dr. S. M. Condren

  14. EXAMPLE How much H2O, in grams results from burning an excess of H2 in 3.3 moles of O2? 2 2 H2 + O2 -----> H2O (18.0 g H2O) #g H2O = (3.3 mol O2) (2 mol H2O) (1 mol O2) (1 mol H2O) = 1.2x102 g H2O Dr. S. M. Condren

  15. EXAMPLE How much H2O, in grams results from burning an excess of H2 in 3.3 grams of O2? 2 2 H2 + O2 -----> H2O #g H2O = (3.3 g O2) (18.0 g H2O) (1 mol O2) (2 mol H2O) (1 mol O2) (32.g O2) (1 mol H2O) = 3.7 g H2O Dr. S. M. Condren

  16. Thermite Reaction Dr. S. M. Condren

  17. Thermite Reaction Fe2O3(s) + 2Al(s) Al2O3(s) + 2 Fe(l) Dr. S. M. Condren

  18. Thermite Reaction Dr. S. M. Condren

  19. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Dr. S. M. Condren

  20. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 103 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Dr. S. M. Condren

  21. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Balanced chemical equation: Fe2O3 + 2 Al ---> 2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? (1 mol Fe2O3) ------------------- (2 mol Fe) (159.7 g Fe2O3) ---------------------- (1 mol Fe2O3) (1 mol Fe) ---------------- (55.85 g Fe) (167 g Fe) #g Fe2O3 = = 238 g Fe2O3 Dr. S. M. Condren

  22. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? The mass of iron in a weld adding 10% mass: #g = (1.67 X 103 g) (0.10) = 167 g Balanced chemical equation: Fe2O3 + 2 Al ---> 2 Fe + Al2O3 What mass of Al is required for the thermite process? (2 mol Al) -------------- (2 mol Fe) (26.9815 g Al) ------------------- (1 mol Al) (1 mol Fe) ---------------- (55.85 g Fe) (167 g Fe) #g Al = = 80.6 g Al Dr. S. M. Condren

  23. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe #g Fe2O3 = 238 g Fe2O3 #g Al = 80.6 g Al Dr. S. M. Condren

  24. Reactions Involving aLIMITING REACTANT • In a given reaction, there is not enough of one reagent to use up the other reagent completely. • The reagent in short supply LIMITS the quantity of product that can be formed. Dr. S. M. Condren

  25. 2 NO(g) + O2 (g) 2 NO2(g) LIMITING REACTANTS Reactants Products Limiting reactant = ___________ Excess reactant = ____________ Dr. S. M. Condren

  26. EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) balanced equation relates: 2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g) have only: 1Fe2S3 (S) <=> 2H2O(l) <=> 3O2(g) not enough H2O to use all Fe2S3 plenty of O2 Dr. S. M. Condren

  27. EXAMPLEWhat is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) if use all Fe2S3: (1.0 mol Fe2S3) (4 mol Fe(OH)3) #mol Fe(OH)3 = ------------------------------------------ (2 mol Fe2S3) = 2.0 mol Fe(OH)3 Dr. S. M. Condren

  28. EXAMPLE:What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) if use all H2O: (2.0 mol H2O) (4 mol Fe(OH)3) #mol Fe(OH)3 = ----------------------------------------- (6 mol H2O) = 1.3 mol Fe(OH)3 Dr. S. M. Condren

  29. EXAMPLE:What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) if use all O2 (3.0 mol O2) (4 mol Fe(OH)3) #mol Fe(OH)3 = --------------------------------------- (3 mol O2) = 4.0 mol Fe(OH)3 Dr. S. M. Condren

  30. EXAMPLE:What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react? 2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S) 1.0 mol Fe2S3 => 2.0 mol Fe(OH)3 2.0 mol H2O => 1.3 mol Fe(OH)3 3.0 mol O2 => 4.0 mol Fe(OH)3 least amount Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3, then H2O is the limiting reactant. Thus the maximum number of moles of Fe(OH)3 that can be produced by this reaction is 1.3 moles. Dr. S. M. Condren

  31. Theoretical Yield the amount of product produced by a reaction based on the amount of the limiting reactant Dr. S. M. Condren

  32. Actual Yield amount of product actually produced in a reaction Dr. S. M. Condren

  33. Percent Yield actual yield % yield = --------------------- * 100 theoretical yield Dr. S. M. Condren

  34. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process Dr. S. M. Condren

  35. EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl2) #kg N2H4 = --------------------- Dr. S. M. Condren

  36. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl2) (1000 g Cl2) #kg N2H4 = ----------------------------------- (1 kg Cl2) metric conversion Dr. S. M. Condren

  37. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00) (1000 g Cl2) (1 mol Cl2) #kg N2H4 = ----------------------------------------- (1) (70.9 g Cl2) molar mass Dr. S. M. Condren

  38. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1 mol Cl2) #kg N2H4 = ----------------------------------------- (1)(70.9) Dr. S. M. Condren

  39. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4+ 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1 mol Cl2)(1 mol N2H4) #kg N2H4 = ------------------------------------------------- (1) (70.9)(1 mol Cl2) Dr. S. M. Condren

  40. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1 mol N2H4) #kg N2H4 = ------------------------------------------------- (1) (70.9)(1) Dr. S. M. Condren

  41. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1 mol N2H4) (32.0 g N2H4) #kg N2H4 = -------------------------------------------------------- (1)(70.9)(1)(1 mol N2H4) molar mass Dr. S. M. Condren

  42. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0 g N2H4)(1 kg N2H4) #kg N2H4 = ---------------------------------------------------------- (1)(70.9)(1)(1) (1000 g N2H4) metric conversion Dr. S. M. Condren

  43. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0)(1 kg N2H4) #kg N2H4 = ---------------------------------------------------------- (1)(70.9)(1)(1)(1000) Dr. S. M. Condren

  44. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0)(1 kg N2H4) #kg N2H4 = ---------------------------------------------------------- (1)(70.9)(1)(1)(1000) = 0.451 kg N2H4 Dr. S. M. Condren

  45. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) theoretical yield #kg N2H4 = 0.451 kg N2H4 (b) actual yield (0.299 kg product) # kg N2H4 = -------------------------- Dr. S. M. Condren

  46. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) theoretical yield #kg N2H4 = 0.451 kg N2H4 (b) actual yield (0.299 kg product) (98.0 kg N2H4) # kg N2H4 = -------------------------------------------- (100 kg product) purity factor Dr. S. M. Condren

  47. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) theoretical yield #kg N2H4 = 0.451 kg N2H4 (b) actual yield (0.299 kg product) (98.0 kg N2H4) # kg N2H4 = -------------------------------------------- (100 kg product) = 0.293 kg N2H4 Dr. S. M. Condren

  48. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) theoretical yield #kg N2H4 = 0.451 kg N2H4 (b) actual yield # kg N2H4 = 0.293 kg N2H4 Dr. S. M. Condren

  49. EXAMPLEA chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4? 2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O (a) theoretical yield #kg N2H4 = 0.451 kg N2H4 (b) actual yield # kg N2H4 = 0.293 kg N2H4 (c) percent yield 0.293 kg % yield = -------------- X 100 = 65.0 % yield 0.451kg Dr. S. M. Condren

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