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Second Exam: Friday February 15  Chapters 3 and 4. Please note that there is a

Second Exam: Friday February 15  Chapters 3 and 4. Please note that there is a class at 1 pm so you will need to finish by 12:55 pm. Electronic Homework due R by 11:30 pm Office hours this week: T 2-3 pm R 1-2 pm SL 130 .

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Second Exam: Friday February 15  Chapters 3 and 4. Please note that there is a

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  1. Second Exam: Friday February 15 Chapters 3 and 4. Please note that there is a class at 1 pm so you will need to finish by 12:55 pm. Electronic Homework due R by 11:30 pm Office hours this week: T 2-3 pm R 1-2 pm SL 130

  2. Figure 4.9: Reaction of a carbonate with an acid.Photo courtesy of American Color.

  3. Molarity (Concentration of Solutions)= M Moles of Solute Moles Liters of Solution L M = = solute = material dissolved into the solvent In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes. In sea water , Water is the solvent, and salt, magnesium chloride, etc. are the solutes. In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

  4. Fig. 3.11

  5. Preparing a Solution - I • Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 l ! • What is the Molarity of the salt and each of the ions? • Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)

  6. Preparing a Solution - II • Mol wt of Na3PO4 = 163.94 g / mol • 3.95 g / 163.94 g/mol = 0.0241 mol Na3PO4 • dissolve and dilute to 300.0 ml • M = 0.0241 mol Na3PO4 / 0.300 l = 0.0803 M Na3PO4 • for PO4-3 ions = 0.0803 M • for Na+ ions = 3 x 0.0803 M = 0.241 M

  7. Dilution of Solutions • Take 25.00 ml of the 0.0400 M KMnO4 • Dilute the 25.00 ml to 1.000 l - What is the resulting Molarity of the diluted solution? • # moles = Vol x M • 0.0250 l x 0.0400 M = 0.00100 Moles • 0.00100 Mol / 1.00 l = 0.00100 M

  8. Converting a Concentrated Solution to a Dilute Solution

  9. The Dilution Dogma: NEVER FORGET IT! M1V1=M2V2

  10. How could you make 5.0 L of 0.025 M sucrose from a solution which is 0.100 M sucrose? Mix 1.250 L of 0.100 M sucrose with 3.75 L water.

  11. Chemical Equation Calculation - III Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles Molarity moles / liter Solutions

  12. Calculating Mass of Solute from a Given Volume of Solution Volume (L) of Solution Molarity M = (mol solute / Liters of solution) = M/L Moles of Solute Molar Mass (M) = ( mass / mole) = g/mol Mass (g) of Solute

  13. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol Al(OH)3 Moles of Al(OH)3 molar ratio Moles of HCl M ( mol/L) Volume (L) of HCl

  14. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol = 0.128 mol Al(OH)3 Moles of Al(OH)3 0.128 mol Al(OH)3 x = molar ratio Moles HCl Moles of HCl M ( mol/L) Volume (L) of HCl

  15. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol = 0.128 mol Al(OH)3 Moles of Al(OH)3 3 moles HCl moles Al(OH)3 0.128 mol Al(OH)3 x = molar ratio 0.385 Moles HCl Moles of HCl Vol HCl M ( mol/L) Volume (L) of HCl

  16. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq) Given 10.0 g Al(OH)3, what volume of 1.50 M HCl is required to neutralize the base? Mass (g) of Al(OH)3 M (g/mol) 10.0 g Al(OH)3 78.00 g/mol = 0.128 mol Al(OH)3 Moles of Al(OH)3 3 moles HCl moles Al(OH)3 0.128 mol Al(OH)3 x = molar ratio 0.385 Moles HCl Moles of HCl 1.00 L HCl 1.50 Moles HCl Vol HCl = x 0.385 Moles HCl Vol HCl = 0.256 L = 256 ml M ( mol/L) Volume (L) of HCl

  17. Solving Limiting Reactant Problems in Solution - Precipitation Problem - I Problem: Lead has been used as a glaze for pottery for years, and can be a problem if not fired properly in an oven, and is leachable from the pottery. Vinegar is used in leaching tests, followed by Lead precipitated as a sulfide. If 257.8 ml of a 0.0468 M solution of Lead nitrate is added to 156.00 ml of a 0.095 M solution of Sodium sulfide, what mass of solid Lead Sulfide will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Writing the balanced equation: Pb(NO3)2 (aq) + Na2S (aq) 2 NaNO3 (aq) + PbS (s)

  18. Volume (L) of Pb(NO3)2 solution Volume (L) of Na2S solution Multiply by M (mol/L) Multiply by M (mol/L) Amount (mol) of Pb(NO3)2 Amount (mol) of Na2S Molar Ratio Molar Ratio Amount (mol) of PbS Amount (mol) of PbS Choose the lower number of PbS and multiply by M (g/mol) Mass (g) of PbS

  19. Volume (L) of Pb(NO3)2 solution Volume (L) of Na2S solution Multiply by M (mol/L) Multiply by M (mol/L) Amount (mol) of Pb(NO3)2 Divide by equation coefficient Amount (mol) of Na2S Divide by equation coefficient Smallest Molar Ratio Amount (mol) of PbS Mass (g) of PbS

  20. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) =

  21. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2 Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield:

  22. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2 Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: 1 mol PbS 1 mol Pb(NO3)2 Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+2

  23. Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2 Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: 1 mol PbS 1 mol Pb(NO3)2 Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+2 0.012065 Mol Pb+2 = 0.012065 Mol PbS 0.012065 Mol PbS x = 2.89 g PbS 239.3 g PbS 1 Mol PbS

  24. Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: 85.90 g N2 28.02 g N2 1 mole N2 moles N2 = = 21.66 g H2 2.016 g H2 1 mole H2 moles H2 = =

  25. Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: 85.90 g N2 28.02 g N2 1 mole N2 moles N2 = = 3.066 mol N2 21.66 g H2 2.016 g H2 1 mole H2 moles H2 = = 10.74 mol H2

  26. Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: 2 mol NH3 1 mol N2 = 6.132 mol NH3 3.066 mol N2 = 7.16 mol NH3 2 mol NH3 3 mol H2 10.74 mol H N2 is Limiting!

  27. Percent Yield of a reaction: Actual Yield x 100 Theortetical Yield

  28. Percent Yield/Limiting Reactant Problem - II N2 (g) + 3 H2 (g) 2 NH3 (g) Solution Cont. Since Nitrogen is limiting, the theoretical yield of ammonia is: 6.132 mol NH3 x = 104.427 g NH3 (Theoretical Yield) 17.03 g NH3 1 mol NH3 Actual Yield Theoretical Yield Percent Yield = x 100% 98.67 g NH3 104.427 g NH3 Percent Yield = x 100% = 94.49 %

  29. CaCO3(s) + 2 HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) 2 g 10 mL 0.75 M Which is limiting? 2 g CaCO3 x 1 mol CaCl2 = 0.01 mol CaCl2 100 g CaCO3 1 mol CaCO3 0.01 L HCl x 0.75 mol HCl x 1 mol CaCl2= L HCl 2 mol HCl 0.004 mol CaCl2 What is the [Cl-] after the reaction? How many g of CaCO3 remain?

  30. Fig. 3.14

  31. Figure 4.22A: Titration of an unknown amount of HCl with NaOH (#1). Photo courtesy of American Color.

  32. Figure 4.22B: Titration ofan unknown amount of HCl with NaOH (#2). Photo courtesy of American Color.

  33. Figure 4.22C: Titration of an unknown amount of HCl with NaOH (#3). Photo courtesy of American Color.

  34. Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO4 and has a molecular mass of 158.04 g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO4 into sufficient water to make 250.00 ml of solution. What is the molarity of this diluted solution? 1 mole KMnO4 158.04 g KMnO4 1.58 g KMnO4 x =

  35. Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO4 and has a molecular mass of 158.04 g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO4 into sufficient water to make 250.00 ml of solution. What is the molarity of this diluted solution? 1 mole KMnO4 158.04 g KMnO4 1.58 g KMnO4 x = 0.0100 moles KMnO4

  36. Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO4 and has a molecular mass of 158.04 g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO4 into sufficient water to make 250.00 ml of solution. What is the molarity of this diluted solution? 1 mole KMnO4 158.04 g KMnO4 1.58 g KMnO4 x = 0.0100 moles KMnO4 0.0100 moles KMnO4 0.250 liters Molarity = = 0.0400 M Molarity of K+ ion = [K+] ion = [MnO4-] ion = 0.0400 M

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