1. 1 Chapter 5 chemical reaction

2. 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 6.022x10 23 A mole is a quantity that contains 6.02 x 10 23 items.

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

4. 4 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 10 23 atoms or 6.02 x 10 23 atoms 1 mol 6.02 x 10 23 molecules or 6.02 x 10 23 molecules 1 mol

5. 5 Coefficients are used to form mole ratios, which can serve as conversion factors. N 2 (g) + O 2 (g)2 NO(g) Mole ratios: 1 mol N 2 1 mol O 2 1 mol N 2 2 mol NO 1 mol O 2 2 mol NO Mole Calculations in Chemical Equations

6. 6 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product Moles of reactant Moles of reactant Grams of product Grams of product mole–mole conversion factor mole–mole conversion factor molar mass conversion factor molar mass conversion factor Moles of product Moles of product Grams of reactant Grams of reactant molar mass conversion factor molar mass conversion factor [1] [2] [3]

7. 7 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. The actual yield is the amount of product isolated from a reaction. Usually, however, the amount of product formed is less than the maximum amount of product predicted.

8. 8 5.7 Percent Yield Percent yield = actual yield (g) theoretical yield (g) x 100% 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? = 15 g 23 g x 100% =65%

9. 9 5.8 Limiting Reactants The limiting reactant is the reactant that is completely used up in a reaction.

10. 10 Limiting Reactant A limiting reactant in a chemical reaction is the substance that Is used up first. Stops the reaction. Limits the amount of product that can form.

11. 11 Determining the Limiting Reactant Sample Problem 5.18 [1]: Determine how much of one reactant is needed to react with a second reactant. 2 H 2 (g) + O 2 (g) 2 H 2 O(l) chosen to be “Original Quantity” chosen to be “Unknown Quantity” There are 4 molecules of H 2 in the picture.

12. 12 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 H 2 (g) + O 2 (g) 2 H 2 O(l) 2 molecules H 2 1 molecule O 2 2 molecules H 2 Choose this conversion factor to cancel molecules H 2

13. 13 Determining the Limiting Reactant Sample Problem 5.18 [3]: Calculate the number of moles (molecules) of the second reactant needed for complete reaction. 2 H 2 (g) + O 2 (g) 2 H 2 O(l) 1 molecule O 2 2 molecules H 2 4 molecules H 2 x = 2 molecules O 2

14. 14 Determining the Limiting Reactant Sample Problem 5.18 [4]: Analyze the two possible outcomes: If the amount present of the second reactant is less than what is needed, the second reactant is the limiting reagent. If the amount present of the second reactant is greater than what is needed, the second reactant is in excess.

15. 15 Determining the Limiting Reactant Sample Problem 5.18

16. 16 Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 Using the balanced equation, determine the limiting reactant when 10.0 g of N 2 (MM = 28.02 g/mol) react with 10.0 g of O 2 (MM = 32.00 g/mol). N 2 (g) + O 2 (g) 2 NO(g)

17. 17 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.

18. 18 Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 [2] Determine the limiting reactant by choosing N 2 as the original quantity and converting to mol O 2. 1 mol O 2 1 mol N 2 0.357 mol N 2 x = 0.357 mol O 2 mole–mole Conversion factor mole–mole Conversion factor The amount of O 2 we started with (0.313 mol) is less than the amount we would need (0.357 mol) so O 2 is the limiting reagent.

19. End of chapter 5 question 87 Question 87 The local anesthetic ethyl chloride (C 2 H 5 Cl, molar mass is 64.51g/mol) can be prepared by reaction of ethylene (C 2 H 4 molar mass 28.05g/mol) according to the balanced equation A. if 8.00g of ethylene and 12.0g of HCl are used, how many moles of each reactant are used? What is the limiting reactant How many grams of the product formed? If a 10.6g of product are formed, what is the percent yield of the reaction? C 2 H 4 + HCl C 2 H 5 Cl

20. Problem 5.87 The local anesthetic ethyl chloride (C 2 H 5 Cl, molar mass 64.51g/mole) can be prepared by reaction of ethylene (C 2 H 4, 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? C 2 H 4 + HCl C 2 H 5 Cl 1 mol C 2 H 4 28.05g C 2 H 4 8.00g C 2 H 4 x = 0.285 mol C 2 H 4 1 mol HCl 36.46g HCl 12.0g HCl x = 0.329 mol HCl

21. b. What is the limiting reactant 0.285 mol C 2 H 4 is completely used up in the reaction so it is the limiting reactant c. how many moles of product are formed Problem 5.87 C 2 H 4 + HCl C 2 H 5 Cl 1 mol C 2 H 5 Cl 1mol C 2 H 4 0.285mol C 2 H 4 x = 0.285 mol C 2 H 5 Cl

22. d. How many grams of the product formed e. if 10.6g of product are formed, what is the percent yield of the reaction? Problem 5.87 C 2 H 4 + HCl C 2 H 5 Cl 64.51gC 2 H 5 Cl 1mol C 2 H 5 Cl 0.285mol C 2 H 5 Cl x = 18.4 g C 2 H 5 Cl Percent yield = actual yield (g) theoretical yield (g) x 100% Percent yield = 10.6 g C 2 H 5 Cl 18.4 g C 2 H 5 Cl x 100% Percent yield = 57.6%

23. Alka seltzer Calculations 1. NaHCO 3 (s)  Na + (aq) + HCO 3 - (aq) 2. HCO 3 - (aq) + H 3 O + (aq)  2H 2 O(l) + CO 2 (g) For example if you determined the mass of CO 2 lost is 0.512g determine moles of CO 2 (g) lost Use mole ratio to calculate moles of NaHCO 3 (s) 1 mol CO 2 44.01g CO 2 0.50g CO 2 x = 0.0163 mol CO 2 1 mole NaHCO 3 (s) 1 mole CO 2 0.0163moles CO 2 x = 0.0163 mol NaHCO 3 (s)

24. Mass of NaHCO 3 (s) Calculated Mass% of NaHCO 3 reacted in tablet (printed on label is 1.916g of NaHCO 3 83.00g NaHCO 3 (s) 1 mole NaHCO 3 (s) 0.0163moles NaHCO 3 x = 1.35g NaHCO 3 (s) 1.35g NaHCO 3 (s) 1.916g in tablet x 100% = 70 % NaHCO 3 (s) reacted Alka seltzer Calculations

25. 25 5.9 Oxidation and Reduction 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. A. General Features

26. 26 Zn + Cu 2+ Zn 2+ + Cu Zn loses 2 e – Cu 2+ gains 2 e − Zn loses 2 e − to form Zn 2+, so Zn is oxidized. Cu 2+ gains 2 e − to form Cu, so Cu 2+ is reduced. 5.9 Oxidation and Reduction

27. 27 Oxidation half reaction:ZnZn 2+ + 2 e − Each of these processes can be written as an individual half reaction: Zn + Cu 2+ Zn 2+ + Cu Zn loses 2 e – Cu 2+ gains 2 e − loss of e − Reduction half reaction:Cu 2+ + 2e − Cu gain of e − 5.9 Oxidation and Reduction

28. 28 Zn + Cu 2+ Zn 2+ + Cu Zn acts as a reducing agent because it causes Cu 2+ to gain electrons and become reduced. A compound that is oxidized while causing another compound to be reduced is called a reducing agent. oxidizedreduced 5.9 Oxidation and Reduction

29. 29 Zn + Cu 2+ Zn 2+ + Cu A compound that is reduced while causing another compound to be oxidized is called an oxidizing agent. Cu 2+ acts as an oxidizing agent because it causes Zn to lose electrons and become oxidized. oxidizedreduced 5.9 Oxidation and Reduction

30. 30 5.9 Oxidation and Reduction

31. 31 Iron Rusting 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) Fe 3+ O 2– neutral Feneutral O Fe loses e – and is oxidized. O gains e – and is reduced. Examples of Oxidation–Reduction Reactions

32. 32 Inside an Alkaline Battery Zn + 2 MnO 2 ZnO + Mn 2 O 3 neutral ZnMn 4+ Zn 2+ Mn 3+ Zn loses e − and is oxidized. Mn 4+ gains e − and is reduced. Examples of Oxidation–Reduction Reactions

33. 33 Zn + 2 MnO 2 ZnO + Mn 2 O 3 Examples of Oxidation–Reduction Reactions

34. 34 Oxidation results in the:Reduction results in the: Gain of oxygen atoms Loss of hydrogen atoms Loss of oxygen atoms Gain of hydrogen atoms Examples of Oxidation–Reduction Reactions

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

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

37. 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. Mg + Fe 2+ Mg 2+ + Fe Sn + Cu 2+ Sn 2+ + Cu 4Na + O 2 2Na 2 O

38. 38 Oxidation half reaction:MgMg 2+ + 2 e − Each of these processes can be written as an individual half reaction: Mg + Fe 2+ Mg 2+ + Fe Mg loses 2 e – Fe 2+ gains 2 e − loss of e − Reduction half reaction:Fe 2+ + 2e − Fe gain of e − Mg + Fe 2+ Mg 2+ + Fe

39. 39 Oxidation half reaction:SnSn 2+ + 2 e − Each of these processes can be written as an individual half reaction: Sn + Cu 2+ Sn 2+ + Cu Sn loses 2 e – Cu 2+ gains 2 e − loss of e − Reduction half reaction:Cu 2+ + 2e − Cu gain of e − Sn + Cu 2+ Sn 2+ + Cu

40. Chapter 5 question 91 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. Fe + Cu 2+ Fe 2+ + Cu Cl 2 + 2I - I2 + 2Cl - 2Na + Cl 2 2NaCl

41. 41 Oxidation half reaction:4Na4Na 1+ + 4 e − Each of these processes can be written as an individual half reaction: 4Na + O 2 2Na 2 O gain of oxygen Loss of oxygen loss of e − Reduction half reaction:O 2 + 4e − 2O -2 gain of e − 4Na + O 2 2Na 2 O

42. 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. Mg is oxidized becomes the reducing agent Fe 2+ is reduced becomes the oxidizing agent Zn is oxidized becomes the reducing agent Cu 2+ is reduced becomes the oxidizing agent Na is oxidized becomes the reducing agent O is reduced becomes the oxidizing agent Mg + Fe 2+ Mg 2+ + Fe Sn + Cu 2+ Sn 2+ + Cu 4Na + O 2 2Na 2 O