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Reading: All of Chapter 9 HW 9: DUE 7/9/14 ( only assignment from this chapter)

Reading: All of Chapter 9 HW 9: DUE 7/9/14 ( only assignment from this chapter) Chap. 9, #'s 13, 15, 21-29 odd, 37, 43-45, 47, 51, 57, 61, 62, 63, 65, 69, 71 Lab Tomorrow (WET LAB!) Experiment 7B (IN LAB BOOK!) Objective (leave rest of page blank) Procedure Data Tables

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Reading: All of Chapter 9 HW 9: DUE 7/9/14 ( only assignment from this chapter)

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  1. Reading: All of Chapter 9 • HW 9: DUE 7/9/14 (only assignment from this chapter) • Chap. 9, #'s 13, 15, 21-29 odd, 37, 43-45, 47, 51, 57, 61, 62, 63, 65, 69, 71 • Lab Tomorrow (WET LAB!) • Experiment 7B (IN LAB BOOK!) • Objective (leave rest of page blank) • Procedure • Data Tables • Wednesday – DA/Math Proficiency Quiz 2

  2. The mole 81.534 grams of water is equivalent to how many moles of water? game plan: game plan: g game plan: g mol

  3. The mole Only 2 options of what to do with the mole:

  4. The mole Only 2 options of what to do with the mole: mole-to-mole ratio mole “X” mole “Y”

  5. The mole Only 2 options of what to do with the mole: 2 places to find a mole-to-mole ratio: a chemical formula H2O mole-to-mole ratio comes from the subscripts of the formula

  6. The mole Only 2 options of what to do with the mole: 2 places to find a mole-to-mole ratio: a chemical equation 3 H2(g) + 1 N2(g) 2NH3(l) mole-to-mole ratio comes from the coefficients of the balanced chemical equation.

  7. Avogadro’s # The mole Only 2 options of what to do with the mole: mole particles molar mass mole mass mole-to-mole ratio mole “X” mole “Y”

  8. The mole How many ATOMS of nitrogen are in 6.119 moles of ammonium sulfide?

  9. ANY TIME you are going from an amount of one substance to an amount of another substance, you MUST use a mole to mole ratio.

  10. ANY TIME you are going from an amount of one thing to an amount of another thing, work in 3 steps: Step 1: convert the given information to moles Step 2: use a mole-to-mole ratio Step 3: convert the moles to whatever is asked for

  11. The mole How many ATOMS of nitrogen are in 6.119 moles of ammonium sulfide? (NH4)2S

  12. A sample of aluminum bicarbonate contains 1.75 grams of carbon, how many molecules of aluminum bicarbonate are present? grams of carbon  molecules of aluminum bicarbonate amount of another thing amount of one thing MOLE TO MOLE RATIO Al(HCO3)3

  13. A sample of aluminum bicarbonate contains 1.75 grams of carbon, how many molecules of aluminum bicarbonate are present? Al(HCO3)3

  14. How many atoms of C are in an aspirin (C9H8O4) pill that contains 0.2000 grams of aspirin?

  15. How many atoms of C are in an aspirin (C9H8O4) pill that contains 0.2000 grams of aspirin? (180.17 g/mol) game plan: g C9H8O4 mol C9H8O4  mol C game plan: g C9H8O4 mol C9H8O4 game plan: g C9H8O4 game plan: game plan: g C9H8O4 mol C9H8O4  mol Catoms C molar mass mole-to-mole ratio Avogadro’s#

  16. mole-to-mole ratios Silver nitrate is reacted with barium chloride: How many grams of barium chloride are needed to make 2.24 grams of solid? 2 AgNO3(aq) + BaCl2(aq) 2 AgCl(s) + Ba(NO3)2(aq) AgCl = BaCl2 = game plan: game plan: g AgCl game plan: g AgCl mol AgCl game plan: g AgCl mol AgCl  mol BaCl2  g BaCl2 game plan: g AgCl mol AgCl  mol BaCl2 ANY TIME you are going from an amount of one substance to an amount of another substance, you MUST use a mole to mole ratio.

  17. mole-to-mole ratios Chromium (III) chlorate is reacted with sodium oxide: a) What is the balanced chemical equation? b) How many grams of sodium oxide are needed to react with 15.67 grams of chromium (III) chlorate

  18. mole-to-mole ratios Chromium (III) chlorate is reacted with sodium oxide: 2 Cr(ClO3)3(aq) + 3 Na2O(aq) Cr2O3(s) + 6 NaClO3(aq) Cr(ClO3)3 = Na2O = 61.98 g/mol game plan: g Cr(ClO3)3 mol Cr(ClO3)3  mol Na2O game plan: g Cr(ClO3)3 mol Cr(ClO3)3 game plan: g Cr(ClO3)3 game plan: game plan: g Cr(ClO3)3 mol Cr(ClO3)3  mol Na2Og Na2O

  19. Mole – Mole revisited What mass of liquid will be produced from the reaction of bismuth (V) hydroxide with 3.28 moles of sulfuric acid? 2 Bi(OH)5(s) + 5 H2SO4(aq) Bi2(SO4)5(aq) + 10 H2O(l) game plan: mol H2SO4 mol H2O  g H2O game plan: mol H2SO4 mol H2O game plan: mol H2SO4 game plan:

  20. mole-to-mole ratios What mass of liquid will be produced from the reaction of 203 g of bismuth (V) hydroxide with a sulfuric acid solution? 2 Bi(OH)5(s) + 5 H2SO4(aq) Bi2(SO4)5(aq) + 10 H2O(l) Bi(OH)5 = 294.03 g/mol game plan: g Bi(OH)5 mol Bi(OH)5  mol H2Og H2O game plan: g Bi(OH)5 mol Bi(OH)5  mol H2O game plan: g Bi(OH)5 mol Bi(OH)5 game plan: g Bi(OH)5 game plan:

  21. mole-to-mole ratios 2 Bi(OH)5(s) + 5 H2SO4(aq) Bi2(SO4)5(aq) + 10 H2O(l) Start with 3.28 mol H2SO4(aq) Can make 118 grams of water Start with 203 grams of Bi(OH)5 Can make 62.2 grams of water What if we start with 3.28 mol H2SO4(aq) AND 203 grams of Bi(OH)5?

  22. ACME Tricycle Company Tricycles are made of 1 frame, 2 pedals, and 3 wheels A shipment comes in consisting of 1387 f, 2744 p, 4188 w. How many tricycles can be built?

  23. ACME Tricycle Company The smallest amount of product is how much can actually be made! (only 1372 tricycles can be made in this case) The starting material that you run out of is the limiting reactant (pedals in this case) The starting material that you have left over is the excess reactant (frames and wheels in this case) You KNOW you are doing a limiting reactant problem when you have amounts of more than one reactant (starting material)

  24. Limiting Reactants Hydrogen gas and nitrogen gas combine to form ammonia gas (NH3). If you start with 18 moles of hydrogen and 11 moles of nitrogen, how many moles of ammonia can be made? Hydrogen gas and nitrogen gas combine to form ammonia gas (NH3). If you start with 18 moles of hydrogen and 11 moles of nitrogen, how many moles of ammonia can be made? 3 H2(g) + N2(g)  2 NH3(g) H2(g) + N2(g)  NH3(g) H2(g) + N2(g)  2 NH3(g) H2 is the limiting reactant (you run out of it first) N2 is the excess reactant (you have more than you need) Only 12 moles of NH3 can be made

  25. 3 H2 + N2 2 NH3 +

  26. Rules for Limiting Reactant Problems • Is it even a L.R.? • Step 1: How much product can be made from each given reactant? • Step 2: Identify the limiting reactant.

  27. 3 H2 + N2 2 NH3 E.R. (left over) L.R. (ran out) +

  28. Limiting Reactants Started with 18 moles of hydrogen and 11 moles of nitrogen. 3 H2(g) + N2(g)  2 NH3(g) Theoretical Yield

  29. H2 + N2 NH3 E.R. (left over) L.R. (ran out) 0 mol H2 left 5 mol N2 left 12 mol NH3 made +

  30. CrCl3 = Pb(NO3)2 = PbCl2 = 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl3(aq) + 3 Pb(NO3)2(aq)  2 Cr(NO3)3(aq) + 3 PbCl2(s)

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