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Plan for Wed, 1 Oct 08

Plan for Wed, 1 Oct 08. Lecture Limiting Reactants and Percent Yield (3.10) Molecular vs. Ionic compounds. What happens when they dissolve in water? (4.1-2) …undergo a phase change? (1.9) Composition of solutions (4.3) Also read 4.5-10. Stoichiometry Concept Quiz.

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Plan for Wed, 1 Oct 08

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  1. Plan for Wed, 1 Oct 08 • Lecture • Limiting Reactants and Percent Yield (3.10) • Molecular vs. Ionic compounds. What happens when they dissolve in water? (4.1-2) • …undergo a phase change? (1.9) • Composition of solutions (4.3) • Also read • 4.5-10

  2. Stoichiometry Concept Quiz 2 Al(s) + 3 CuCl2(aq) 3 Cu(s) + 2 AlCl3(aq) • In which case will all the Al AND all the CuCl2 be used up? • 1:1 mass ratio (1 gram Al and 1 gram CuCl2) • 1:1 mole ratio (1 mol Al and 1 mol CuCl2) • 2:3 mass ratio (2 grams Al and 3 grams CuCl2) • 2:3 mole ratio (2 mol Al and 3 mol CuCl2)

  3. Limiting Reactants • Given a balanced chemical equation, the amount of reactants available to react determines the amount of products we can produce. 1 ¾ cup cake flour 3 tsp baking powder 2 oz. baking chocolate 1 ½ cup sugar ½ cup butter 4 eggs ½ cup milk 1 ¾ cup cake flour 3 tsp baking powder 2 oz. baking chocolate 1 ½ cup sugar ½ cup butter 2 eggs ½ cup milk one whole chocolate cake Having only 2 eggs available limits the amount of cake we can make, even though we have plenty of the other ingredients. one half of a chocolate cake

  4. Limiting Reactants (cont) • The same kind of thing happens in chemical reactions… • …because we have to count particles by weighing, it is rare to measure the exact number of particles of each reactant required in a chemical equation. • More often than not one reactant is completely used up before the other reactant(s) are.

  5. Limiting Reactants (cont) • Consider the combination of solid carbon with oxygen: C(s) + O2(g)  CO2(g)

  6. LRs in Action CO(g) + 2H2(g)  CH3OH(l) Both reactants are completely used up…neither CO or H2 limit the production of CH3OH.

  7. LRs in Action (cont) CO(g) + 2H2(g)  CH3OH(l) CO is used up before H2 … CO limits the production of CH3OH.

  8. LRs in Action (cont) CO(g) + 2H2(g)  CH3OH(l) H2 is used up before CO… H2limits the production of CH3OH.

  9. LRs in Action (cont) • Chemistry happens at the particulate level… • …so the limiting reactant in a chemical equation must be determined by comparing numbers of moles. • In the last example, we were given moles of reactants, so we could determine the LR by inspection. • In the lab, you will measure reactant amounts in terms of g, mg, kg, etc. • These measurements will need to be converted to moles.

  10. Example • Suppose 25.0 kg of nitrogen gas and 5.00 kg of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion. N2(g) + H2(g)  NH3(g) 2 3 25.0 kg ? kg 5.00 kg

  11. MM MM MM mol NH3 mol LR Example (cont) N2(g) + 3H2(g)  2NH3(g) 25.0 kg 5.00 kg ? kg Use mole ratios to determine LR mol N2 mol H2 mol LR mol NH3

  12. 1000 g N2 1000 g H2 1 mol H2 1 mol N2 1 kg N2 1 kg H2 2.016 g H2 28.01 g N2 25.0 kg ? kg 5.00 kg N2(g) + 3H2(g)  2NH3(g) ? mol N2 = 25.0 kg N2 = 892.54 mol N2 ? mol H2 = 5.00 kg H2 = 2480.16 mol H2 To determine which of this reactants is limiting, ask the following question: “How many moles of N2 would we need to react completely with 2480.16 mol of H2??”

  13. 1 mol N2 3 mol H2 25.0 kg 892.5 mol 5.00 kg 2480.2 mol ? kg N2(g) + 3H2(g)  2NH3(g) “How many moles of N2 would we need to react completely with 2480.16 mol of H2??” ? mol N2 = 2480.2 mol H2 = 826.73 mol N2 We have 892 mol N2, but we only need 827 mol N2 to react completely with 2480 mol of H2. H2 IS THE LIMITING REAGENT.

  14. 3 mol H2 1 mol N2 25.0 kg 892.5 mol 5.00 kg 2480.2 mol ? kg N2(g) + 3H2(g)  2NH3(g) What if instead we asked:“How many moles of H2 would we need to react completely with 892.5 mol of N2??” ? mol H2 = 892.5 mol N2 = 2677.5 mol H2 We need 2680 mol H2 to react completely with 892 mol of N2, but we only have 2480 mol H2. H2 IS THE LIMITING REAGENT.

  15. 2 mol NH3 17.03 g NH3 1 kg NH3 3 mol H2 1 mol NH3 1000 g NH3 25.0 kg 892.5 mol 5.00 kg 2480.2 mol ? kg N2(g) + 3H2(g)  2NH3(g) LIMITING REAGENT ? kg NH3 = 2480.2 mol H2 = 28.1585 kg NH3 = 28.2 kg NH3

  16. Making the LR work for YOU. • To determine which reactant is the LR, we must compare the following quantities… • what we have(based on the problem set up) • what we need (based on the balanced chemical equation) • The mole ratios in a chemical equation are crucialfor determining the LR… • The reactant for which we have fewer moles is not necessarily the LR!!

  17. Theoretical vs. Actual Yield • The theoretical yield of a reaction is the amount of product that would be formed under ideal reaction conditions in which starting materials are completely consumed (up to the LR). This is a calculated number. • The actual yield is the amount of product that is actually produced in real life (in the lab). • The actual yield is always less than the theoretical yield because… • starting materials may not be completely consumed • side reactions may occur • the reverse reaction may occur • there may be loss of product during purification steps

  18. Percent Yield • A comparison of… how much product we actually produced to how much product we could theoretically produce Gives us the percent yield of a chemical reaction. actual yield % yield = 100 theoretical yield

  19. 25.0 kg 892.5 mol 5.00 kg 2480.2 mol 28.2 kg N2(g) + 3H2(g)  2NH3(g) Theoretical Yield LIMITING REAGENT In a certain experiment, the actual yield of this reaction was only 26.7 kg NH3. What was the percent yield? 26.7 kg NH3 % yield = = 94.7% 100 = 94.68% 28.2 kg NH3 Percent yield should never be greater than 100%. If it is, there is something wrong with your theoretical yield calculation, or with the actual yield you obtained experimentally.

  20. Tips for Stoichiometric Success • Make sure your chemical equation is balanced. • Always compare moles, not mass. • Use mole ratios from your balanced chemical equation to determine the limiting reactant. • Remember… • theoretical yield is a calculatedquantity. • actual yield is ameasured quantity. • If your percent yield is greater than 100%... • something is wrong with the theoretical yield you calculated, or… • something is wrong with the actual yield you measured in your experiment.

  21. QUESTION If two compounds are about to react, which statement about the reaction states an accurate observation? 1. If temperature conditions can be kept optimal, both compounds will fully react with no excess. 2. The reactant present with the fewest grams will be the limiting reactant. 3. The reactant present with the fewest moles will be the limiting reactant. 4. None of the statements are accurate. A limiting reactant can be determined only when both the moles of the two compounds AND the mole ratio by which the compounds react are known.

  22. Ch 4 – Chemical Rxns and Solution Stoichiometry • Molecular vs. Ionic compounds. What happens when they dissolve in water? (4.1-2) • Composition of solutions (4.3) Also read… • 4.5-10

  23. Ionic vs. Molecular Compounds Solid sodium chloride Solid water Let’s compare sodium chloride and water. What are the microscopic particles in the samples of these two compounds?

  24. Molecular Solid: H2O Molecular solids consist of discrete molecules or atoms held together by intermolecular forces. Intermolecular attraction (need ~10-40 kJ/mol to break) Covalent O-H bond (need ~100-1000 kJ/mol to break)

  25. Ionic Solid: NaCl Ionic solids consist of arrays of anions and cations held together in a net of mutualelectrostatic attraction. Ionic bond: ~100-1000 kJ/mol needed to break bond Formula Unit: NaCl

  26. Copper sulfate, CuSO4 O O S – O O S – O O O O What about ionic compounds with polyatomic ions? Cu2+

  27. Ionic compounds in solution Polar bond

  28. Molecular compounds in solution Some molecular compounds have polar bonds like water does. These compounds will be somewhat soluble in water.

  29. Anti-freeze is soluble in water. Motor oil is not soluble in water. No polar bonds to attract water Fig. 16-8, p. 464

  30. Charged species in solution conduct electricity…these are called electrolytes. Electrolytes are either strong or weak Strong: soluble salts (NaCl, KNO3, etc), strong acids (HCl, H2SO4, etc) Weak: insoluble salts (BaSO4, Mg(OH)2, etc), weak acids (CH3COOH, HF, etc) Solutions of non-electrolytes (like sugar in water) will not conduct electricity Electrical Conductivity of Aqueous Solutions Strong Weak Non

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