1. Chemical Reactions

2. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen.

3. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l ) 

4. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g)

5. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g) The equation is not yet balanced. Hydrogens come in twos on the left, and three hydrogens are on the right side of the equation. The equation is not yet balanced. Hydrogens come in twos on the left, and three hydrogens are on the right side of the equation.

6. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g) Try a “2” in front of the water. Try a “2” in front of the water.

7. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + 2 H 2 O(l )  NaOH(aq) + H 2 (g) We now have two O atoms on the left, so we need to put a 2 before NaOH. We now have two O atoms on the left, so we need to put a 2 before NaOH.

8. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The two sodium atoms on the right require that we put a 2 in front of Na on the left. The two sodium atoms on the right require that we put a 2 in front of Na on the left.

9. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The two sodium atoms on the right require that we put a 2 in front of Na on the left. The equation is now balanced. The two sodium atoms on the right require that we put a 2 in front of Na on the left. The equation is now balanced.

10. Balancing Chemical Equations- Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) Left SideRight Side Na- 2Na- 2 H- 4H- 4 O- 2O- 2

11. Chemical Equations 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 2 atoms of sodium react with 2 molecules of water to produce 2 molecules of sodium hydroxide and a molecule of hydrogen.

12. Chemical Equations 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 200 atoms of sodium react with 200 molecules of water to produce 200 molecules of sodium hydroxide and 100 molecules of hydrogen.

13. Chemical Equations 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) The balanced chemical equation can be interpreted in a variety of ways. It is usually interpreted as 2 moles of sodium will react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen. It is usually interpreted as 2 moles of sodium will react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen. The balanced equation tells us nothing about the masses of reactants or products. The balanced equation tells us nothing about the masses of reactants or products.

14. Types of Chemical Reactions A combination reaction is a reaction with two or more reactants, and a single product. 2 CO (g) + O 2 (g)  2 CO 2 (g)

15. Types of Chemical Reactions A decomposition reaction is a reaction has one reactant, and two or more products. CaCO 3 (s)  CaO(s) + CO 2 (g)

16. Types of Chemical Reactions In a combustion reaction, a substance burns in the presence of oxygen. If the reactant is a compound containing on carbon, hydrogen and oxygen, the products are water and carbon dioxide. CH 4 (g) +2 O 2 (g)  2 H 2 O (l) + CO 2 (g)

17. Chemical Composition Usually, the compound is combusted in the presence of oxygen. Any carbon in the compound is collected as carbon dioxide (CO 2 ), and any hydrogen is collected as water (H 2 O).

18. Determining Empirical Formulas If given combustion data: The ultimate goal is to get the simplest whole number ratio of the elements in the compound. Usually the compound contains carbon, hydrogen and perhaps oxygen or nitrogen. 1. Use the information about CO 2 to determine the moles and mass of carbon in the compound.

19. Determining Empirical Formulas If given combustion data: 2. Use the information about H 2 O to determine the moles and mass of hydrogen in the compound. 3. The mass and moles of oxygen (or a third element) can be obtained by difference. 4. Once moles of each element is obtained, find the relative number of moles and empirical formula (as with % composition).

20. Formulas from Combustion Data This method assumes the compound contains only C and H.

21. Empirical Formula using Combustion Data- Problem A compound, which contains C, H and O, is analyzed by combustion. If 10.68 mg of the compound produces 16.01 mg of carbon dioxide and 4.37 mg of water, determine the empirical formula of the compound. A compound, which contains C, H and O, is analyzed by combustion. If 10.68 mg of the compound produces 16.01 mg of carbon dioxide and 4.37 mg of water, determine the empirical formula of the compound. If the compound has a molar mass of 176.1 g/mol, determine the molecular formula of the compound. If the compound has a molar mass of 176.1 g/mol, determine the molecular formula of the compound.

22. Stoichiometry Stoichiometry is a Greek word that means using chemical reactions to calculate the amount of reactants needed and the amount of products formed. Stoichiometry is a Greek word that means using chemical reactions to calculate the amount of reactants needed and the amount of products formed. Amounts are typically calculated in grams (or kg), but there are other ways to specify the quantities of matter involved in a reaction. Amounts are typically calculated in grams (or kg), but there are other ways to specify the quantities of matter involved in a reaction.

23. Stoichiometry A balanced chemical equation or reaction is needed before any calculations can be made. The formulas of all reactants and products are written before attempting to balance the equation.

24. Stoichiometry Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. How many grams of sodium are needed to produce 50.0g of hydrogen? Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. How many grams of sodium are needed to produce 50.0g of hydrogen? A balanced chemical equation is needed before any calculations can be made.

25. Stoichiometry Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. How many grams of sodium are needed to produce 50.0g of hydrogen? Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. How many grams of sodium are needed to produce 50.0g of hydrogen? Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g) Na(s) + H 2 O(l )  NaOH(aq) + H 2 (g)

26. Stoichiometry Problem Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. How many grams of sodium are needed to produce 50.0g of hydrogen? Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen. How many grams of sodium are needed to produce 50.0g of hydrogen? 2 Na(s) +2 H 2 O(l )  2 NaOH(aq) + H 2 (g) 2 Na(s) +2 H 2 O(l )  2 NaOH(aq) + H 2 (g)

27. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) How many grams of sodium are needed to produce 50.0g of hydrogen? How many grams of sodium are needed to produce 50.0g of hydrogen? 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g

28. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g Although the question doesn’t state it, you can assume enough water is present for complete reaction. Although the question doesn’t state it, you can assume enough water is present for complete reaction. We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na

29. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na We use the molar mass of H 2 to go from grams of H 2 to moles of H 2.

30. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 molar mass of H 2

31. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams 50.0 g ? grams 50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 molar mass of H 2 We use the coefficients from the balanced equation to go from moles of H 2 to moles of Na. We use the coefficients from the balanced equation to go from moles of H 2 to moles of Na.

32. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of H 2 coefficients

33. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of H 2 coefficients We use the molar mass of Na to go from moles of Na to grams of Na. We use the molar mass of Na to go from moles of Na to grams of Na.

34. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g We can map out the problem: We can map out the problem: g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of Na molar mass of H 2 coefficients molar mass of Na

35. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) ? grams50.0 g ? grams50.0 g g H 2  moles H 2  moles Na  grams Na molar mass of H 2 coefficients molar mass of Na molar mass of H 2 coefficients molar mass of Na (50.0 g H 2 ) (1 mol H 2 ) (2 moles Na) ( 22.99 g Na) (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na) (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na)

36. Stoichiometry Problem 2 Na(s) + 2 H 2 O(l )  2NaOH(aq) + H 2 (g) (50.0 g H 2 ) (1 mol H 2 ) (2 moles Na) ( 22.99 g Na) = (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na) (2.02 g H 2 ) (1 mol H 2 ) (1 mol Na) = 1,138 grams Na = 1.14 x 10 3 g Na = 1.14 kg Na

37. Limiting Reagent Problems Sometimes you are given quantities of more than one reactant, and asked to calculate the amount of product formed. The quantities of reactants might be such that both react completely, or one might react completely, and the other(s) might be in excess. These are called limiting reagent problems, since the quantity of one of the reacts will limit the amount of product that can be formed. Sometimes you are given quantities of more than one reactant, and asked to calculate the amount of product formed. The quantities of reactants might be such that both react completely, or one might react completely, and the other(s) might be in excess. These are called limiting reagent problems, since the quantity of one of the reacts will limit the amount of product that can be formed.

38. Limiting Reagent - Problem Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 1. First write the formulas for reactants and products. Al + Br 2  AlBr 3

39. Limiting Reagent - Problem Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 2. Now balance the equation by adding coefficients. 2 Al +3 Br 2  2 AlBr 3

40. Limiting Reagent - Problem Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? Aluminum and bromine react to form aluminum bromide. If 500. g of bromine are reacted with 50.0 g of aluminum, what is the theoretical yield of aluminum bromide? 2 Al +3 Br 2  2 AlBr 3 The theoretical yield is the maximum amount of product that can be formed, given the amount of reactants. It is usually expressed in grams.

41. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g500.g ? grams There are several ways to solve this problem. One method is to solve the problem twice. Once, assuming that all of the aluminum reacts, the other assuming that all of the bromine reacts. The correct answer is whichever assumption provides the smallest amount of product. There are several ways to solve this problem. One method is to solve the problem twice. Once, assuming that all of the aluminum reacts, the other assuming that all of the bromine reacts. The correct answer is whichever assumption provides the smallest amount of product.

42. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g500.g ? grams The problem can be mapped: The problem can be mapped: Al: grams Al  moles Al  moles AlBr 3  g AlBr 3 molar mass Al coefficients molar mass AlBr 3 molar mass Al coefficients molar mass AlBr 3

43. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given:50.0g 500.g ? grams The problem can be mapped: The problem can be mapped: Al: grams Al  moles Al  moles AlBr 3  g AlBr 3 molar mass Al coefficients molar mass AlBr 3 (50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr 3 ) (266.7 g AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 )

44. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g ? grams Al: (50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr 3 ) (266.7 g AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 ) (26.98 g Al) ( 2 mol Al) (mol AlBr 3 ) = 494. g AlBr 3 (if all of the Al reacts)

45. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g ? grams The calculation is repeated for Br 2. g Br 2  moles Br 2  moles AlBr 3  g AlBr 3 molar mass Br 2 coefficients molar mass AlBr 3

46. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g ? grams The calculation is repeated for Br 2. g Br 2  moles Br 2  moles AlBr 3  g AlBr 3 molar mass Br 2 coefficients molar mass AlBr 3 (500. g Br 2 ) (1 mol Br 2 ) (2 moles AlBr 3 ) (266.7 g AlBr 3 ) (159.8 g Br 2 ) (3 moles Br 2 ) (1 mol AlBr 3 ) (159.8 g Br 2 ) (3 moles Br 2 ) (1 mol AlBr 3 ) = 556. grams of AlBr 3

47. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g500.g ? grams Summary: Summary: We have enough Al to produce 494. g AlBr 3 We have enough Br 2 to produce 556. grams of AlBr 3 The theoretical yield is 494. grams of AlBr 3.

48. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g 494. g Summary: All of the Al reacts, so Al is limiting. All of the Al reacts, so Al is limiting. Bromine is in excess. Bromine is in excess. Additional questions: Additional questions: 1. How much bromine is left over? 2. If 418 grams of AlBr 3 is obtained, what is the % yield?

49. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g 494. g 1. How much bromine is left over? Since all 50.0 g of the Al reacts, the product must contain 494.g -50.g = 444. g of bromine. Therefore, 500.g- 444.g = 56. g of Br 2 are left over.

50. Limiting Reagent - Problem 2 Al +3 Br 2  2 AlBr 3 Given: 50.0g 500.g 494. g 2.If 418 grams of AlBr 3 is obtained, what is the % yield? The percent yield is (actual yield) (100%) (theoretical yield) % yield = (418 g) (100%) = 84.6 % (494g) (494g)

51. Trends in Reactivity of Main Group Elements In using an element’s position on the periodic table to predict reactivity, there are two underlying factors to consider.  The smallest member of a family or group often exhibits different behavior than the rest of the family due to its small size.  Elements on a diagonal often exhibit similar behavior even though they are in different groups.

52. Trends in Reactivity of Main Group Elements For example, both Li and Mg react with nitrogen to form nitrides (Li 3 N and Mg 3 N 2 ), although nitrogen is inert with most elements.

53. Metallic Character

54. Hydrogen Hydrogen is a non metal. It bonds covalently, with the only ionic form being the hydride ion, H –. In this way it is similar to the halogens. And is sometimes placed in both group IA and group VIIA on periodic tables.

55. Metallic Character Across a period, metallic behavior decreases. Non-metals are often crumbly solids, liquids or gases at room temperature.

56. Metallic Character Metallic behavior increases going down a group.

57. Group IA – the Alkali Metals In discussing the chemistry, preparation and properties of the group IA elements, it is important to remember that hydrogen is not a group IA metal. It’s properties and reactivity would place it within group 7A (diatomic non- metals), rather than group IA.

58. Group 1A Metals The group 1A metals are soft shiny metals with fairly low densities (Li, Na and K are less dense than water) and low melting points. Sodium melts at 98 o C, and cesium melts at 29 o C. The softness, low density and low melting points are the result of weaker metallic bonding due to only one valence electron in this group.

59. Group 1A Metals - Production Due to the high reactivity with oxygen and water, all of the metals are found in nature in ionic form (M 1+ ). The pure metal must be produced in an oxygen and water-free environment. Typically, an electrical current is passed through the melted chloride salt. The metal and the chlorine gas are collected separately.

60. Reactivity Trends The chemical behavior of the group IA metals illustrates periodic trends. As the valence electron occupies a higher quantum level, it experiences less nuclear attraction, and is more easily removed.

61. Group 1A Metals + Water The reaction with water forms hydrogen gas and the aqueous metal hydroxide. The reaction is so vigorous, that the hydrogen may ignite. 2 M(s) + 2 H 2 O(l)  H 2 (g) + 2 MOH(aq)

62. Metallic Character The group IA metals react with water to produce hydrogen and the metal hydroxide. Metallic behavior increases going down a group.