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ADVANCED CHEM Chapter 12.1 What is Stoichiometry?

Stoichiometric Calculations. ADVANCED CHEM Chapter 12.1 What is Stoichiometry?. Date. Stoichiometry. In Greek means, “ to measure the elements ” Study of Quantitative relationships in a chemical reaction between: Amounts of reactants used Amounts of products formed

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ADVANCED CHEM Chapter 12.1 What is Stoichiometry?

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  1. Stoichiometric Calculations ADVANCED CHEMChapter 12.1What is Stoichiometry? Date

  2. Stoichiometry • In Greek means, “to measure the elements” • Study of Quantitative relationships in a chemical reaction between: • Amounts of reactants used • Amounts of products formed • Based on the “Law of Conservation of Mass”

  3. Mole-Mass Relationships in Chemical Reactions • 4Fe(s) + 3O2 (g) -->2Fe2O3 (s) • Interpret in terms of • Particles • Moles! (Use for RATIOS!) • Mass • If given a balanced equation, you can figure # of grams of each reactant & product!

  4. Assignment #1 • Book # 53 pg. 378

  5. Mole Ratios • A ratio between the numbers of moles of any two substances in a balanced equation. • 2Al(s) + 3Br2 (l) -->2AlBr3 (s) • Make a Ratio for each. • 2 mol Al2 mol Al 3 mol Br2 2 mol AlBr3 • 3 mol Br23 mol Br2 2 mol Al2 mol AlBr3 • 2 mol AlBr32 mol AlBr3 2 mol Al3 mol Br2

  6. Assignment #2 • Mole Ratio Worksheet • Book # 54

  7. Stoichiometric Calculations Chapter 12.2Stoichiometric Calculations Date

  8. Mole to Mole Conversion • Add Potassium to Water and produce Potassium Hydroxide and Hydrogen gas • K(s) + H2O (l) -->KOH (aq) + H2 (g) • 2K(s) + 2H2O (l) -->2KOH (aq) + H2 (g) • GIVEN: .0400 mole of potassium • FIND: Moles of H2 gas produced • MOLE RATIO! • 2 mol K 1 mole H2 1 mole H2 2 mole K • .0400 mol Kx 1 mol H2 = .0200 mol H2 2 mol K Given #

  9. Assignment #3 • Book # 63

  10. Mole to Mass Conversion • Sodium plus Chlorine equals… :) • Na(s) + Cl2(g) --> NaCl (s) • 2Na(s) + Cl2(g) --> 2NaCl (s) • GIVEN: 1.25 mole of Chlorine • FIND: How many g of NaCl are produced? • MOLE RATIO! • 1 mol Cl22 moles NaCl 2 moles NaCl1 mole Cl2 • 1.25 mol Cl2 x 2 mole NaCl = 2.50 mol NaCl 1 mole Cl2 • 2.5 mol NaClx 58.44 g = 146 g NaCl 1 mole NaCl

  11. Assignment #4 • Book # 66, 68

  12. Do#70 maybe? Mass to Mass Conversion • Ammonium Nitrate produces N2O gas and water when it decomposes. • NH4NO3(s) -->N2O (g) + H2O • NH4NO3(s) -->N2O (g) + 2H2O • GIVEN: 25 g of NH4NO3 • FIND: Mass of water produced? • 25 g NH4NO3 x 1 mol NH4NO3 = .312 mol 80.04 gNH4NO3 • .312 mol NH4NO3 x 2 moles H2O =.624 mol 1 mole NH4NO3 • .624 mol H2Ox 18.02 g H2O = 11.2 grams1 mole H2O

  13. HOW MUCH PRODUCT? • Theoretical Yield (in theory using math) • Max amt of product that can be produced from a given amount of reactant • Actual Yield (in lab) • Amt of product actually produced by a chemical reaction carried out. • Percent Yield • RATIO of actual yield to theoretical yield (expressed as a percent) • Percent Yield = Actual Yield (from Experiment) x 100 = Theoretical Yield (from stoichcalcs)

  14. Calculating Percent Yield • Potassium Chromate is added to a solution containing .500 g Silver Nitrate, solid silver chromate is formed. • DETERMINE: Theoretical Yield of silver chromate precipitate • ALSO: If .455 g of silver chromate is obtained, calc % yield. • AgNO3(aq) + K2CrO4 (aq) --> Ag2CrO4 (s) + KNO3 (aq) • 2AgNO3(aq) + K2CrO4 (aq) --> Ag2CrO4 (s) +2KNO3 (aq) • .500 g AgNO3 x 1 mol AgNO3 =2.94 x 10-3mol AgNO3 169.9gAgNO3 • 2.94 x 10-3mol AgNO3x 1 mol Ag2CrO4 = 1.47 x10-3mol Ag2CrO42 mol AgNO3 • 1.47 x 10-3mol Ag2CrO4x 331.7 g Ag2CrO4= .488 g Ag2CrO41 mol Ag2CrO4 • .455 g x 100 = 93.2% Ag2CrO4 .488 g

  15. Assignment #5 • Book # 73, 74

  16. Steps in Stoichiometric Calculations • Write a BALANCED equation. • Determine the MOLES of the GIVENsubstance using a mass (grams) to mole conversion. (Use Molar mass) • Determine the MOLES of the UNKNOWN substance from the moles of the GIVEN substance. (Use Mole Ratio) • From the MOLES of the UNKNOWN substance, determine the MASS (grams) of the unknown substance using a mole to mass (grams) conversion (Use Molar mass)

  17. Assignment • Book Problems!

  18. Chapter 12.3Limiting Reactants Date

  19. Why Do Reactions Stop? • Limiting Reactants • Reactant that limits or CONTROLS the extent of the reaction • Excess Reactants • Left over reactants • Sometimes use excess on purpose

  20. Calculating a PRODUCTWhen a Reactant is Limited • Disulfur dichloride is used to make rubber harder, stronger and less brittle. • Molton sulfur reacts with chlorine gas to produce this! • S8(l) + Cl2 (g) --> S2Cl2 (l) • 1S8(l) + 4Cl2 (g) -->4S2Cl2 (l) • GIVEN: 200.0 g of sulfur reacts with 100.0 g of chlorine • FIND: Mass of the product disulfur dichloride produced? • 200.00 g S8 x 1 mol S8 = .7797 mol S8 256.5gS8 • 100.0 g Cl2 x 1 mol CL2 = 1.410 mol Cl2 70.91gCl2 • FIGURE A MOLE RATIO YOU CAN UNDERSTAND!! • 1.410 mol Cl2 = 1.808 mol Cl2 Available .7797 mol S81 mol S8 Limiting Reactant !

  21. Calculating a PRODUCTWhen a Reactant is Limited • S8(l) + 4Cl2 (g) --> 4S2Cl2 (l) • GIVEN: 200.0 g of sulfur reacts with 100.0 g of chlorine • FIND: Mass of the product disulfur dichloride produced? • 1.410 mol Cl2 X 4 mol S2Cl2 = 1.410 mol S2Cl2 4 mol Cl2 • 1.410 mol S2Cl2 X 135.0 g S2Cl2 = 190.4 g S2Cl2 1 mol S2Cl2 • FIND: How much Sulfur actually reacted? • 1.410 mol Cl2 X 1mol S8 = .3525 mol S8 4 mol Cl2 • .3525 mol S8 X 256.5. g S8 = 90.42 g S8 used 1 mol S8 • We have 200.0 g of Sulfur. So 109.6 g S8is in excess

  22. Assignment • Do #76 together • Book # 81 • 79, 82

  23. Chapter 12.4Percent Yield Date

  24. Assignment • Do #84 & 88 in class • Book #89

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