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Reminders

Reminders. VSEPR lab is due on Wednesday Question 4: Although PH 3 and BH 3 both contain bonds of similar polarity, one molecule is polar while the other is nonpolar. Explain. Change it to

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Reminders

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  1. Reminders • VSEPR lab is due on Wednesday • Question 4: Although PH3 and BH3 both contain bonds of similar polarity, one molecule is polar while the other is nonpolar. Explain. • Change it to • Although PH3 and NH3 both have similar shape, one molecule is polar while the other is nonpolar. Explain.

  2. Determination of % NaHCO3 in Alka Seltzer Tablets Objectives: • To determine the amount of NaHCO3 in Alka Seltzer tablets by observing the amount of CO2 produced from the acid base reaction of HCO3- with acetic acid (in vinegar). • To practice stoichiometry. • To study the concept of limiting reactant.

  3. Determination of % NaHCO3 in Alka Seltzer Tablets • Alka Seltzer is an effervescent tablet that contains aspirin (acetylsalicylic acid), citric acid, and sodium bicarbonate (NaHCO3). As soon as the tablet dissolves in water, the NaHCO3 dissociates to form a bicarbonate ion (HCO3-) and a sodium ion (Na+) 1. NaHCO3(s)  Na+(aq) + HCO3-(aq) Acetic acid, which will ‘donate’ H+ to water to create H3O+(aq) is then added to the mixture. With the addition of acetic acid the following acid-base reaction then takes place: 2. HCO3-(aq) + H3O+(aq)  2H2O(l) + CO2 (g)

  4. Determination of % NaHCO3 in Alka Seltzer Tablets Weight of tablet is not • Alka Seltzer is an effervescent tablet that contains aspirin (acetylsalicylic acid), citric acid, and sodium bicarbonate (NaHCO3). • 1. NaHCO3(s)  Na+(aq) + HCO3-(aq) • 2. HCO3-(aq) + H3O+(aq)  2H2O(l) + CO2 (g) • Mass of CO2 (g) lost Use stoichiometry to go back and calculate %NaHCO3(s) in Alka Seltzer tablet

  5. Prelab • Due on Wednesday • Objective • Pre-lab Question: (To be submitted at the beginning of lab) • Summary of the procedure in your own words. • What are the bubbles that form when Alka Seltzer is dissolved in water? • What is the ratio of moles of CO2 produced to moles of NaHCO3 reacted?

  6. Chapter 5 chemical reaction

  7. Mole and Avogadro's number • Just as a grocer sells rice by weight rather than by counting grains; a chemist uses weight to count for atoms • As a dozen of anything contains 12 a mole of anything contains 6.022x1023 A mole is a quantity that contains 6.02 x 1023 items.

  8. Use of the mole? • 1mole = Avogadro's number • This graph will help you with most of chapter 5 calculations

  9. The Mole and Avogadro’s Number It can be used as a conversion factor to relate the number of moles of a substance to the number of atoms or molecules: 1 mol 6.02 x 1023 atoms 6.02 x 1023 atoms 1 mol or 1 mol 6.02 x 1023 molecules 6.02 x 1023 molecules 1 mol or

  10. Mole Calculations in Chemical Equations Coefficients are used to form mole ratios, which can serve as conversion factors. N2(g) O2(g) 2 NO(g) + Mole ratios: 1 mol N2 2 mol NO 1 mol O2 2 mol NO 1 mol N2 1 mol O2

  11. Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product mole–mole conversion factor [2] Moles of reactant Moles of product molar mass conversion factor molar mass conversion factor [3] [1] Grams of reactant Grams of product

  12. 5.7 Percent Yield • The theoretical yield is the amount of product • expected from a given amount of reactant based • on the coefficients in the balanced chemical • equation. • Usually, however, the amount of product formed • is less than the maximum amount of product • predicted. • The actual yield is the amount of product isolated • from a reaction.

  13. 5.7 Percent Yield Sample Problem 5.14 If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed? actual yield (g) theoretical yield (g) x 100% = Percent yield 15 g 23 g = x 100% = 65%

  14. 5.8 Limiting Reactants • The limiting reactant is the reactant that is completely used up in a reaction.

  15. Determining the Limiting Reactant Sample Problem 5.18 [1]: Determine how much of one reactant is needed to react with a second reactant. 2 H2(g) + O2(g) 2 H2O(l) chosen to be“Original Quantity” chosen to be“Unknown Quantity” There are 4molecules of H2in the picture.

  16. Determining the Limiting Reactant Sample Problem 5.18 [2]: Write out the conversion factors that relate the numbers of moles (or molecules) of reactants 2 H2(g) + O2(g) 2 H2O(l) 2 molecules H2 1 molecule O2 1 molecule O2 2 molecules H2 Choose this conversion factor to cancel molecules H2

  17. Determining the Limiting Reactant Sample Problem 5.18 [3]: Calculate the number of moles (molecules) of the second reactant needed for complete reaction. 2 H2(g) + O2(g) 2 H2O(l) 1 molecule O2 2 molecules H2 = 2 molecules O2 4 molecules H2 x

  18. Determining the Limiting Reactant Sample Problem 5.18 [4]: Analyze the two possible outcomes: • If the amount present of the second reactantis less than what is needed, the secondreactant is the limiting reagent. • If the amount present of the second reactant isgreater than what is needed, the secondreactant is in excess.

  19. Determining the Limiting Reactant Sample Problem 5.18

  20. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 Using the balanced equation, determine the limitingreactant when 10.0 g of N2 (MM = 28.02 g/mol) reactwith 10.0 g of O2 (MM = 32.00 g/mol). N2(g) + O2(g) 2 NO(g)

  21. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 [1] Convert the number of grams of each reactant into moles using the molar masses.

  22. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 [2] Determine the limiting reactant by choosing N2 as the original quantity and converting to mol O2. mole–mole Conversion factor 1 mol O2 1 mol N2 = 0.357 mol O2 0.357 mol N2 x The amount of O2 we started with (0.313 mol) isless than the amount we would need (0.357 mol) so O2 is the limiting reagent.

  23. Alka seltzer Calculations • 1. NaHCO3(s)  Na+(aq) + HCO3-(aq)  • 2. HCO3-(aq) + H3O+(aq)  2H2O(l) + CO2 (g) • For example if you determined the mass of CO2 lost is 0.512g determine moles of CO2 (g) lost Use mole ratio to calculate moles of NaHCO3(s) 1 mole NaHCO3(s) 1 mole CO2 1 mol CO2 44.01g CO2 = 0.0163 mol CO2 = 0.0163 mol NaHCO3(s) 0.50g CO2 x 0.0163moles CO2 x

  24. Alka seltzer Calculations • Mass of NaHCO3(s) • Calculated Mass% of NaHCO3 reacted in tablet (printed on label is 1.916g of NaHCO3 83.00g NaHCO3(s) 1 mole NaHCO3(s) 1.35g NaHCO3(s) 1.916g in tablet = 1.35g NaHCO3(s) 0.0163moles NaHCO3 x x 100% = 70 % NaHCO3(s) reacted

  25. Problem 5.87 • The local anesthetic ethyl chloride (C2H5Cl, molar mass 64.51g/mole) can be prepared by reaction of ethylene (C2H4, molar mass 28.05g/mole) with HCl (molar mass 36.46g/mole), according to the balanced equation, • a. if 8.00g of ethylene and 12.0g of HCl are used, how many moles of each reacted are used? C2H5Cl C2H4 + HCl

  26. Problem 5.87 • b. What is the limiting reactant • c. how many moles of product are formed C2H5Cl C2H4 + HCl

  27. Problem 5.87 • d. How many grams of the product formed • e. if 10.6g of product are formed, what is the percent yield of the reaction? C2H5Cl C2H4 + HCl

  28. 5.9 Oxidation and Reduction A. General Features • Oxidation is the loss of electrons from an atom. • Reduction is the gain of electrons by an atom. • Both processes occur together in a single reaction • called an oxidation−reduction or redox reaction. • Thus, a redox reaction always has two components, • one that is oxidized and one that is reduced. • A redox reaction involves the transfer of electrons • from one element to another.

  29. 5.9 Oxidation and Reduction Cu2+ gains 2 e− Zn2+ + Cu Zn + Cu2+ Zn loses 2 e– • Zn loses 2 e− to form Zn2+, so Zn is oxidized. • Cu2+gains 2 e−to form Cu, so Cu2+ is reduced.

  30. 5.9 Oxidation and Reduction Cu2+ gains 2 e− Zn2+ + Cu Zn + Cu2+ Zn loses 2 e– Each of these processes can be written as an individual half reaction: Oxidation half reaction: Zn Zn2+ + 2 e− loss of e− Reduction half reaction: Cu2+ + 2e− Cu gain of e−

  31. 5.9 Oxidation and Reduction Zn2+ + Cu Zn + Cu2+ oxidized reduced A compound that is oxidized while causing another compound to be reduced is called a reducing agent. • Zn acts as a reducing agent because it causes • Cu2+ to gain electrons and become reduced.

  32. 5.9 Oxidation and Reduction Zn2+ + Cu Zn + Cu2+ oxidized reduced A compound that is reduced while causing another compound to be oxidized is called an oxidizing agent. • Cu2+ acts as an oxidizing agent because it causes • Zn to lose electrons and become oxidized.

  33. 5.9 Oxidation and Reduction

  34. Examples of Oxidation–Reduction Reactions Iron Rusting O gains e– and is reduced. 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) neutral Fe neutral O Fe3+ O2– Fe loses e– and is oxidized.

  35. Examples of Oxidation–Reduction Reactions Inside an Alkaline Battery Mn4+ gains e− and is reduced. Zn + 2 MnO2 ZnO + Mn2O3 neutral Zn Mn4+ Zn2+ Mn3+ Zn loses e− and is oxidized.

  36. Examples of Oxidation–Reduction Reactions Zn + 2 MnO2 ZnO + Mn2O3

  37. Examples of Oxidation–Reduction Reactions Oxidation results in the: Reduction results in the: • Gain of oxygen atoms • Loss of hydrogen atoms • Loss of oxygen atoms • Gain of hydrogen atoms

  38. redox chemistry Oxidation occurs when a molecule does any of the following: • Loses electrons • Gains oxygen If a molecule undergoes oxidation, it is the reducing agent.

  39. redox chemistry Reduction occurs when a molecule does any of the following: Gains electrons Loses oxygen If a molecule undergoes reduction, it is the oxidizing agent.

  40. Question 5.92 • Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species. Mg2+ + Fe Mg + Fe2+ Sn2+ + Cu Sn + Cu2+ 2Na2O 4Na + O2

  41. Mg2+ + Fe Mg + Fe2+

  42. Sn2+ + Cu Sn + Cu2+

  43. 2Na2O 4Na + O2

  44. Question 5.92 • Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species. Mg2+ + Fe Mg + Fe2+ Sn2+ + Cu Sn + Cu2+ 2Na2O 4Na + O2

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