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Chemical reactions

Chemical reactions. Page 12. 44. Balance these equations and Classify as being a synthesis, decomposition, single or double replacement, or combustion reaction:. 2 K + Cl 2  2 KCl Type: Synthesis 2 AlBr 3 + 3 Na 2 (CO 3 )  Al 2 (CO 3 ) 3 + 6 NaBr Type: Double replacement

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Chemical reactions

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  1. Chemical reactions Page 12

  2. 44. Balance these equations and Classify as being a synthesis, decomposition, single or double replacement, or combustion reaction: • 2K + Cl2 2KCl Type: Synthesis • 2AlBr3 + 3 Na2(CO3) Al2(CO3)3 +6 NaBr Type: Double replacement • 2C2H6 + 7O24 CO2 +6 H2O Type:Combustion • 3Cu2(CrO4) + 2 Fe  Fe2(CrO4)3 + 6Cu Type: Single replacement • Mg(CO3) MgO+ CO2 Type: decomposition • 3H2 + Fe2S33 H2S + 2 Fe Type: Single replacement

  3. 45. Identify the solubility of each compound in the double replacement reaction above • 2AlBr3(aq)+ 3 Na2(CO3) (aq) Al2(CO3)3(s) +6 NaBr(aq) Precipitate! (The insoluble product)

  4. 46. Which single replacement reaction above will not occur? • 3H2 + Fe2S33 H2S + 2 Fe will not occur. Iron is more reactive than hydrogen. (Refer to the activity series on the back of your periodic table). The more reactive element will be in the compound.

  5. 47a. Write the skeleton equation then balance • Copper(II) chloride reacts with iron to produce iron(III) chloride and copper metal. Skeleton: CuCl2 + Fe  FeCl3 +Cu Balanced: 3CuCl2 + 2Fe  2FeCl3 +3Cu Type of reaction : Single replacement

  6. 47b. Write the skeleton equation then balance • Hydrogen gas and bromine liquid react to yield hydrogen bromide. Skeleton: H2 + Br2 HBr Balanced: H2 + Br2 2HBr Type of reaction : Synthesis

  7. 47C. Write the skeleton equation then balance • Carbon tetrahydride reacts with oxygen to produce carbon dioxide and water vapor. Skeleton: CH4 + O2 CO2+ H2O Balanced: CH4 + 2O2 CO2 + 2H2O Type of reaction : Combustion

  8. Stoichiometry Pages 13 and 14

  9. 48. How many moles of calcium chloride can be theoretically yielded from 1.53 moles of hydrogen chloride? ___HCl + ___Ca(OH)2 ___CaCl2 + ___H2O Balance: 2HCl + 1Ca(OH) 21CaCl2 +2H2O Use the mole ratio from the balanced equation to convert from moles of HCl to moles of CaCl2 : 2HCl + 1Ca(OH) 21CaCl2 +2H2O 1.53 mol HCl1 mol CaCl2 = .765 mol CaCl2 2 mol HCl

  10. 49. How many grams of iron(III) bromide will be theoretically yielded from the reaction of 2.05 moles of iron? ___HBr + ___Fe  ___FeBr3 + ___H2 Balance: 6HBr + 2Fe 2FeBr3 + 3H2 • Use the mole ratio from the balanced equation to convert from moles of Fe to moles of FeBr3: • Use molar mass to convert between moles & grams. 6HBr + 2Fe 2FeBr3 + 3H2 2.05 mol Fe 2 mole FeBr3 295.557g = 606g FeBr3 2 mole Fe 1 mole FeBr3 Molar mass of FeBr3

  11. 50. In a reaction of sodium with chlorine, 2.30 grams of chlorine are consumed. How many grams of product can be theoretically yielded? ___Na + ___Cl2 ___NaCl Balance: 2Na + 1Cl22NaCl • Use the mole ratio from the balanced equation to convert from moles of Na to moles of Cl2: • Use molar mass to convert between moles & grams. 2Na + 1Cl22NaCl 2.3g Na 1 mol Na 2 mol NaCl 58.44g NaCl = 5.8g NaCl 22.990g Na 2 mol Na 1 mol NaCl Molar mass of Na Molar mass of NaCl

  12. 51. Nitrogen gas can be prepared by passing gaseous ammonia (NH3) over solid copper(II) oxide at hightemperatures. The other products of the reaction are solid copper and water vapor. A. If a sample containing 18.1 grams of NH3 reacted with 90.4 grams of copper(II) oxide, which is the limiting reactant? 2NH3 +3CuO 1N2 + 3Cu + 2H2O 18.1g NH3 1 mol NH31 mol N2 28.014 g N2 = 14.9 g N2 17.031g NH3 2 mol NH3 1 mol N2 90.4g CuO 1 mol CuO 1 mol N2 28.014 g N2 = 10.6 g N2 79.545g CuO 3 mol CuO 1 mol N2 A. CuO is the limiting reactant. B. How many grams of N2 will theoretically be formed? 10.6g N2

  13. 52. A student adds 200.0g of C7H6O3 to an excess of C4H6O3, this produces C2H4O2 and C2H4O2. Calculatethe percent yield if 231 g of aspirin (C9H8O4) is produced. 1C7H6O3 + 1C4H6O31C9H8O4 + 1C2H4O2 200.0g C7H6O3 1 mol C7H6O3 1 mol C9H8O4 180.069gC9H8O4 = 138.052g C7H6O3 1 mol C7H6O3 1 mol C9H8O4 Answer = 260.9g theoretical Actual x 100 = % yield Theoretical 231 x100 = 88.5% 260.9

  14. 53. Benzene (C6H6) reacts with bromine to form bromobenzene (C6H5Br ) in the reaction below. A. What is the theoretical yield of bromobenzene in this reaction when 30.0 grams of benzene reacts with 65.0 grams of bromine? 1C6H6 + 1Br2 1C6H5Br + 1HBr 30.0g C6H6 1 mol C6H6 1 molC6H5Br 156.95g C6H5Br = 60.32gC6H5Br 78.054g C6H6 1 mol C6H6 1 mol C6H5Br 65.0g Br2 1 mol Br2 1 mol C6H5Br 156.95g C6H5Br = 63.8 C6H5Br 159.808g Br2 1 mol Br2 1 mol C6H5Br C6H6 was the limiting reactant, so only 60.32 g of C6H5Br would be theoretically produced.

  15. 53.B If the actual yield of bromobenzene was 42.3 grams, what was the percent yield? C6H6 was the limiting reactant, so only 60.32 g of C6H5Br would be theoretically produced. Actual x 100 = % yield Theoretical 42.3 x100 = 65.82% 60.32

  16. States of matter Pages 15,16,17, 18

  17. Matter Pure Substances Mixtures Elements Compounds Homogeneous Mixtures Heterogeneous Mixtures 54. Classification of Matter

  18. 55. Define: a. matter – anything that has mass or takes up space b. physical property – property of matter that can be observed or measured without changing the substance • Examples: color, density, mass, volume

  19. 55. define: c. extensive physical property – depends on amount of substance present • Examples: mass, volume, length d. intensive physical property – does not depends on the amount of a substance • Examples: density, color, melting point, boiling point

  20. 55. define e. Chemical property- used to describe ability Examples: reactivity, flammability, separating mixtures f. Physical change- alters the appearance but does not change the composition of the substance. Examples: phase change, separating mixtures, dissolving, evaporating

  21. 56. Which physical seaparation technique? • a. Peas and carrots- chromatography to separate by color. • b. charcoal powder and iron powder-chromatography to separate by color. • c. salt water- salt dissolves in water so you would use crystallization • d. pigments in green food coloring- if the pigments are dissolved use crystallization, if not dissolved use filtration • e. rubbing alcohol and water- distillation to separate by boiling points

  22. 57 and 58 • 57. Define chemical change – change occurs when one or more substances undergoes a chemical reaction to form a new substance Examples: cooking, combustion, oxidation, fizzes • 58. List AND EXPLAIN the four indicators of a chemical change. a.Color change c. gas evolution b. formation of a precipitate d. odor

  23. 59. Indicate endothermic or exothermic • a. A student pours hydrochloric acid into a test tube containing a white, crystalline powder. The mixture begins to bubble and the test tube begins to feel cold. ENDOTHERMIC • b. A student pours HCl at 25.2 C into a test tube containing a small metal strip. The mixture begins to fizz and the temperature of the mixture rises to 38.6 C. EXOTHERMIC

  24. 60. States of matter

  25. 61.List the six phase changes, define them, and give at least one example of each: A. melting- solid to liquid B. Freezing- liquid to solid C. Vaporization- liquid to gas D. condensation- gas to liquid E. Sublimation- solid to gas F. deposition- gas to solid

  26. 62. critical point- anything above this point will be a gas Liquid Solid Gas Triple point- point where all 3 phases coexist

  27. 63. Define • a. Pure substance – uniform unchanging composition, ex: elements and compounds • b. Element- single type of atom ex: gold (Au), Hydrogen (H) • c. Compound- more than one element combined ex: NaCl

  28. 63. Define • d. Mixture- combination of two or more pure substances ex: salt water • e. Homogeneous mixture- constant composition throughout and are always in one phase • f. Solution- homogeneous mixture

  29. 63. Define • g. Heterogeneous mixture- mixtures do not blend together smoothly and the individual substances remain distinct ex: colloid, suspension • h. Colloid- one substance is suspended evenly throughout another substance ex: milk, fog, jello • i. Suspension-large substance particles are suspended in another substance. Ex: muddy water, paint

  30. Refer to graph for questions 64- 69

  31. 64-66 Use solubility graph 64. 85 grams 65. 130 grams 66. 50 oC

  32. Solutions Pages 19,20

  33. Refer to graph for questions 64- 69

  34. 67. unsaturated solution • 68. 30g will dissolve, 20 grams will remain at the bottom of the beaker • 69. 25 more grams will dissolve

  35. 70. Molarity formula • Molarity = moles of solute Liters of solution Molarity units = M (concentration) You may have to covert mL to L Given mL 1 L 1000 mL You may have to covert grams to moles in order to solve for Molarity Given grams 1 mole molar mass

  36. 71. What formula is used for diluting a concentrated solution to a given molarity? • M1V1= M2V2 • M= Molarity • V= Volume

  37. 72. Determine the molarity of 500 mL of a solution containing 7.20 g of sodium acetate • Molarity = moles of solute Liters of solution • Convert Grams  moles 7.20g 1 mole = .087 moles 83 g • Convert mL L 500mL 1L = .500L 1000 mL • Plug into equation: Molarity = .087 = .18M .500

  38. 73. How many moles of calcium chloride are in 1.35 L of a 2.5 M solution of calcium chloride? • Molarity = moles of solute Liters of solution Plug into equation: 2.5 = X = 3.375 mol 1.35

  39. 74. How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 100.0 mL of 0.25 MH2SO4?? • Formula: M1V1=M2V2 • M = Molarity, V = volume • Solving for V1 (5.0M)(V1)= (.25M)(100mL) (5.0M)(V1)= 25 (V1)= 5mL Multiply .25 x 100 Divide both sides by 5.0 to get (V1) by itself

  40. 75. If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of thedilute solution? • Formula: M1V1=M2V2 • M = Molarity, V = volume • Solving for M2 (3.5M)(20mL)= (M2)(100mL) 70 = (M2)(100mL) .7M= (M2) Multiply 3.5 x 20 Divide both sides by 100 to get (M2) by itself

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