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THIS. IS. JEOPARDY. Your. With. Host. DR. MILLER. Jeopardy. Mole-Mole Mass-Mass. Limiting & Excess. Yield. Equations. Rxn Types. Predicting Products. 100. 100. 100. 100. 100. 100. 200. 200. 200. 200. 200. 200. 300. 300. 300. 300. 300. 300. 400. 400. 400. 400.

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  1. THIS IS JEOPARDY

  2. Your With Host... DR. MILLER

  3. Jeopardy Mole-Mole Mass-Mass Limiting & Excess Yield Equations Rxn Types Predicting Products 100 100 100 100 100 100 200 200 200 200 200 200 300 300 300 300 300 300 400 400 400 400 400 400 500 500 500 500 500 500

  4. 2 NH3 N2 + 3 H2 How many moles of nitrogen are produced when 1.5 moles of ammonia decomposes? A 100

  5. 1.5 = X0.75 mol 2 1 A 100

  6. 4 Al + 3 O2 2 Al2O3 How many moles of aluminum are needed to completely react with 2 moles of oxygen? A 200

  7. X = 22.67 mol 4 3 A 200

  8. C3H8 + 5 O2 3 CO2 + 4 H2O How many grams of water are produced when 45.0 grams of C3H8 combusts completely? A 300

  9. 45.0 g C3H8 x 1 mol = 1.023 mol 44 g 1.023 = X 1 4 4.09 mol H2O x 18 g = 73.6 g 1 mol A 300

  10. C3H8 + 5 O2 3 CO2 + 4 H2O When 150 grams of oxygen react, how many grams of carbon dioxide will be produced? A 400

  11. 150 g O2 x 1 mol = 4.7 mol 32 g 4.7 = X 5 3 2.813 mol CO2 x 44 g = 123.75 g 1 mol A 400

  12. C3H8 + 5 O2 3 CO2 + 4 H2O If 36 grams of water are produced, how much carbon dioxide was also produced? A 500

  13. 36 g H2O x 1 mol = 2 mol 18 g 2 = X 4 3 1.5 mol CO2 x 44 g = 66 g 1 mol A 500

  14. Define Limiting Reagent and Excess Reagent B 100

  15. Limiting Reagent – substance that gets completely used up in a reaction (limiting the amount of product formed) Excess Reagent – substance that is not completely used up in a reaction (some is left over) B 100

  16. How do you determine which reactant is the limiting reagent? B 200

  17. Find the amount of product that can be formed for each of the reactants. The one that can produce the least amount of product is the limiting reagent B 200

  18. 4 Al + 3 O2 2 Al2O3 Which is the limiting reagent if 7 moles of aluminum reacts with 5 moles of oxygen? B 300

  19. Al 7 = X 3.5 mol 4 2 O25 = X 3.3 mol 3 2 B 300

  20. 4 Al + 3 O2 2 Al2O3 What is the maximum amount of aluminum oxide that can be produced if 7 moles of aluminum reacts with 5 moles of oxygen? B 400

  21. Al 7 = X 3.5 mol 4 2 O25 = X3.3 mol 3 2 B 400

  22. 4 Fe + 3 O2 Fe2O3 What is the limiting reagent and what quantity of Fe2O3 can be formed if 74.55 g of iron reacts with 27.65 g of oxygen? B 500

  23. Fe 74.55 g x 1 mol = 1.33 mol 1.33 = X 0.3325 mol x 160 g = 54.25 g 56 g 4 1 1 mol O2 27.65 g x 1 mol = 0.864 mol 0.864 = X 0.288 mol x 160 g = 46.08 g 32 g 3 1 1 mol O2 / 46.08 g B 500

  24. When you use a balanced equation to predict the number of grams of product in a reaction, you get the ______________. C 100

  25. Theoretical Yield C 100

  26. In an experiment, students determined the theoretical yield to be 20 grams. Their actual yield was 15 grams. What was their percent yield? C 200

  27. 15 x 100 = 75% 20 C 200

  28. In an experiment, students determined the theoretical yield to be 12 grams. Their percent yield was 90 percent. What was their actual yield? C 300

  29. X x 100 = 90% 12 10.8 g C 300

  30. Is it possible to have a percent yield greater than 100% C 400

  31. No, the maximum amount of product that can actually be produced can never be greater than the calculated theoretical yield C 400

  32. 4 NH3 + 5 O2 4 NO + 6 H2O The reaction of 0.68 g of NH3 with excess O2 according to the following reaction yields 0.98 g of NO. What is the percent yield? C 500

  33. 0.68 g NH3 x 1 mol = 0.04 0.04 = X 0.04 mol NO x 30 g = 1.2 g 17 g 4 4 1 mol 0.98 x 100 = 81.7% 1.2 C 500

  34. Name the reactants MnO2 + Al → Al2O3 + Mn D 100

  35. Manganese (IV) oxide & Aluminum D 100

  36. Balance this equation MnO2 + Al → Al2O3 + Mn D 200

  37. 3MnO2 + 4Al → 2Al2O3 + 3Mn D 200

  38. What is meant by a balanced equation D 300

  39. Obeying the law of conservation of matter: same number of atoms of each element in reactants and products D 300

  40. Write a complete chemical equation for aluminum sulfate + sodium hydroxide → aluminum hydroxide + sodium sulfate D 400

  41. Al2(SO4)3 +6NaOH  2Al(OH)3 + 3Na2SO4 D 400

  42. Write the combustion reaction for C2H4ç D 500

  43. C2H4 + 3O2 2H2O + 2CO2 D 500

  44. Reaction Type? C10H8 + 12O2 10CO2 + 4H2O E 100

  45. Combustion E 100

  46. Reaction Type? 2H2O  2H2 + O2 E 200

  47. Decomposition E 200

  48. Reaction Type? Mg + 2H2O  Mg(OH)2 + H2 E 300

  49. Single Replacement E 300

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