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Chapter 4 Chemical Quantities and Aqueous Reactions

Chapter 4 Chemical Quantities and Aqueous Reactions. Part 1. Global Warming avg 0.6 °C rise in atmospheric temp since 1860 During same period atmospheric CO 2 levels have risen 25% Are the two trends causal ?. Tro: Chemistry: A Molecular Approach, 2/e. 2. Sources of Increased CO 2

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Chapter 4 Chemical Quantities and Aqueous Reactions

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  1. Chapter 4Chemical Quantities and Aqueous Reactions Part 1

  2. Global Warming • avg 0.6 °C rise in atmospheric temp since 1860 • During same period atmospheric CO2 levels have risen 25% • Are the two trends causal? Tro: Chemistry: A Molecular Approach, 2/e 2

  3. Sources of Increased CO2 One source is combustion rxtns of fossil fuels Another source of CO2 is volcanic action How can we judge whether global warming is natural or due to our use of fossil fuels? 3

  4. Quantities in Chemical Rxtns Law of Conservation of Mass Balancing equations by balancing atoms H2 + O2 H2O Stoichiometry: numerical relationship btwn chemquantities in a chemical rxtn C6H12O6 + O2 CO2 + H2O # atoms of elements # molecules reacting and being made # of moles needed and produced 4

  5. Reaction Stoichiometry Coefficients in balanced chem eqtn specify relative amounts (in moles) of each substance involved in rxtn 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O (g) Why are “moles” so important? Can be used to Tro: Chemistry: A Molecular Approach, 2/e 5

  6. Reaction Stoichiometry • 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O (g) to form (produce) • 2 molecules • of C8H18 • react w/ 25 molecules • of O2 16 molecules of CO2 & 18 molecules of H2O 6

  7. Reaction Stoichiometry • 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O (g) to form (produce) • 2 moles • of C8H18 • react w/ 25 moles • of O2 16 moles of CO2 & 18 moles of H2O Are # moles or # molecules balanced? Are # of atoms balanced? 7

  8. Making Pizza # of pizzas you can make depends on amount of the ingredients you use 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza 8

  9. Making Pizza Want to make more or less than one pizza? Use amount of cheese you have to determine # of pizzas you can make, assuming there’s enough crusts & tomato sauce Tro: Chemistry: A Molecular Approach, 2/e 9

  10. Predicting Amounts from Stoichiometry • amounts of any other substance in chemrxtn can be determined from amount of • just one substance • How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18? • 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) • Use Molar Ratios: Tro: Chemistry: A Molecular Approach, 2/e 10

  11. Predicting Amounts from Stoichiometry • How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18? • 2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g) Tro: Chemistry: A Molecular Approach, 2/e 11

  12. Practice Using the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C6H12O6 + 6 O2® 6 CO2 + 6 H2O Tro: Chemistry: A Molecular Approach, 2/e 12

  13. # moles of water made in combustion of 0.10 moles of glucose? Given: Find: 0.10 moles C6H12O6 moles H2O Conceptual Plan: Relationships: mol C6H12O6 mol H2O Solution: Check: 0.6 mol H2O = 0.60 mol H2O because 6x moles of H2O as C6H12O6, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 13

  14. Use Molar Mass & Molar Ratios to estimate mass of CO2produced in 2007 by combustion of 3.5 x 1015 g gasolne • Assuming gasoline is octane, C8H18, • eqtn for the rxtn is: • 2 C8H18(l) + 25 O2 (g) 16 CO2(g) + 18 H2O (g) • 2 moles : 25 moles  16 mole : 18 moles Tro: Chemistry: A Molecular Approach, 2/e 14

  15. g C8H18 mol C8H18 mol CO2 g CO2 The eqtn for the rxtn gives the mole ratio btwn amount of C8H18 and CO2, but we need to know mass relationship, so the conceptual plan will be: Tro: Chemistry: A Molecular Approach, 2/e 15

  16. g C8H18 mol C8H18 mol CO2 g CO2 estimate mass of CO2 produced in 2007 by combustion of 3.5 x 1015 g gasolne Given: Find: 3.4 x 1015 g C8H18 g CO2 Conceptual Plan: Relationships: Solution: Check: because 8x moles of CO2 as C8H18, but molar mass of C8H18 is 3x CO2, the # makes sense Tro: Chemistry: A Molecular Approach, 2/e 16

  17. Which Produces More CO2; Volcanoes or Fossil Fuel Combustion? World produced 1.1 x 1016 g of CO2 just from petroleum combustion in 2007 Tro: Chemistry: A Molecular Approach, 2/e 17

  18. Which Produces More CO2; Volcanoes or Fossil Fuel Combustion? • 1.1 x 1016 g of CO2 =1.1 x 1013kg CO2 • Estimates of volcanic CO2 production • are 2 x 1011 kg/year Tro: Chemistry: A Molecular Approach, 2/e 18

  19. g CO2 mol CO2 mol C6H12O6 g C6H12O6 1 mol 180.2 g 1 mol C H O 6 12 6 44.01 g 1 mol 6 mol CO 2 How many g of glucose can be synthesized from 37.8 g of CO2 in photosynthesis? Given: Find: 37.8 g CO2, 6 CO2 + 6 H2O  C6H12O6+ 6 O2 g C6H12O6 Conceptual Plan: Relationships: Solution: Check: because 6x moles of CO2 as C6H12O6, but the molar mass of C6H12O6 is 4x CO2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 19

  20. Clicker Question #1:

  21. Clicker Question #1: The overall eqtn involved in photosynthesis is: 6 CO2 + 6 H2O  C6H12O6 + 6 O2

  22. More Making Pizzas • We know that 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza • But what would happen if we had 4 crusts, • 15 oz. tomato sauce, and 10 cu cheese? Tro: Chemistry: A Molecular Approach, 2/e 22

  23. More Making Pizzas, Continued Each ingredient could potentially make a different # of pizzas But all ingredients have to work together! We have enough sauce  3 pizzas, after 3 pizzas, sauce runs out no matter how much other ingredients we have. Tro: Chemistry: A Molecular Approach, 2/e 23

  24. Tomato sauce limits amount of pizzas we can make. Maximum # of pizzas we can make depends on this one ingredient. In chemrxtns we call this the Tro: Chemistry: A Molecular Approach, 2/e 24

  25. Rxtns with multiple reactants, • one reactant will be used up • before the others: • the reaction stops no more product is made • in chemrxtns, this leads to • NOTE: it also determines • amounts of other ingredients we will use! Tro: Chemistry: A Molecular Approach, 2/e 25

  26. The Limiting Reactant • The reactant used up is called the • limiting reactant(or limiting reagent) • Reactants not completely consumed = • (Hint: in combustion, ) • Amount of product • made from limiting reactant • = theoretical yield Tro: Chemistry: A Molecular Approach, 2/e 26

  27. Limiting & Excess Reactants in the Combustion of Methane CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) 1 molecule of CH4 reacts with 2 molecules of O2 Tro: Chemistry: A Molecular Approach, 2/e 27

  28. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) Given: 5 molecules CH4 + 8 molecules O2, which is the limiting reactant? Tro: Chemistry: A Molecular Approach, 2/e 28

  29. How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 mole of N2 in the reaction 3 Si + 2 N2 Si3N4? Tro: Chemistry: A Molecular Approach, 2/e 29

  30. Pick least amount mol N2 mol Si mol Si3N4 mol Si3N4 Limiting reactant and theoretical yield Given: Find: 1.20 mol Si, 1.00 mol N2 mol Si3N4 Conceptual Plan: Relationships: Solution: Tro: Chemistry: A Molecular Approach, 2/e 30

  31. Assume that as we are still making pizzas: • we burn a pizza, • we drop one on the floor, etc. • We only make 2 pizzas. • Actual amount of product made = Tro: Chemistry: A Molecular Approach, 2/e 31

  32. Efficiency is determined by calculating % of max # of pizzas we actually make. In chemrxtns, this = Tro: Chemistry: A Molecular Approach, 2/e 32

  33. Theoretical and Actual Yield to determine theoretical yield, we use reaction stoichiometry. This determines amount of each product that our reactants could make Tro: Chemistry: A Molecular Approach, 2/e 33

  34. Theoretical and Actual Yield • Theoretical yield will always be • least possible amount of product. • always comes from the limiting reactant • Because of both controllable & uncontrollable factors: • - Actual yield of product • will always be < theoretical yield • - Could it ever be >? Tro: Chemistry: A Molecular Approach, 2/e 34

  35. e.g: • When 28.6 kg of C react with 88.2 kg of TiO2 in the reaction above, 42.8 kg of Ti are obtained. Find: • limiting reactant, • theoretical yield, • % yield. Tro: Chemistry: A Molecular Approach, 2/e 35

  36. Write down the given quantity & its units Given: 28.6 kg C 88.2 kg TiO2 42.8 kg Ti produced Tro: Chemistry: A Molecular Approach, 2/e 36

  37. Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Write down the quantity to find and/or its units Find: limiting reactant (LR) theoretical yield (TY) percent yield (PY) Tro: Chemistry: A Molecular Approach, 2/e 37

  38. } smallest amount is from limiting reactant kg C kg TiO2 smallest mol Ti • Write a conceptual plan Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. Tro: Chemistry: A Molecular Approach, 2/e 38

  39. Collect needed relationships 1000 g = 1 kg Molar Mass TiO2 = 79.87 g/mol Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol 1 mole TiO2: 1 mol Ti (from chem. equation) 2 mole C : 1 mol Ti (from chem. equation) Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct g rct molrct  mol Ti pick smallest # moles produced, then convert to kg, then get %Y # moles Ti  TY # kg Ti # kg Ti  %Y Ti Tro: Chemistry: A Molecular Approach, 2/e 39

  40. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) • Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e 40

  41. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) • Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e 41

  42. Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e 42

  43. limiting reactant = TiO2 theoretical yield = 52.9 kg percent yield = 80.9% Because Ti has lower molar mass than TiO2, the T.Y. makes sense and the % yield makes sense as it is < 100% • Check the solutions Tro: Chemistry: A Molecular Approach, 2/e 43

  44. Practice:How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO?2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 44

  45. g NH3 g CuO mol NH3 mol CuO mol N2 mol N2 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) Given: Find: 9.05 g NH3, 45.2 g CuO g N2 Conceptual Plan: Relationships: 1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g 2 mol NH3 : 1 mol N2, 3 mol CuO : 1 mol N2 Choose smallest g N2 Tro: Chemistry: A Molecular Approach, 2/e 45

  46. 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the % yield? Solution: because the percent yield is less than 100, the answer makes sense Check: Tro: Chemistry: A Molecular Approach, 2/e 46

  47. Clicker question #2: Ammonia is produced using the Haber process: 3 H2 + N2 2 NH3

  48. Given: 3 H2 + N2 2 NH3 15.0 kg each of H2 and N2, (= 1.50 x 104 g) 13.7 kg of product recovered (= 1.37 x 104 g)

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