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ADVANCED CHEM Chapter 12.1 What is Stoichiometry?

This chapter delves into the fundamental concepts of stoichiometry, the study of quantitative relationships in chemical reactions. We explore mole-mass relationships, mole ratios, and various conversion methods, including mole-to-mole and mass-to-mass conversions. Specific chemical reactions such as the formation of potassium hydroxide and sodium chloride are analyzed in terms of reactants and products. Additionally, we cover limiting reactants and percent yield calculations, providing a comprehensive overview for students striving to master stoichiometric calculations in advanced chemistry.

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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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