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. 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.

17. 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.

18. 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.

19. 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)

20. 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)

21. 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

22. 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

23. 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.

24. 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

25. 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.

26. 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

27. 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.

28. 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

29. 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)

30. 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

31. 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.

32. 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

33. 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

34. 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.

35. 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.

36. 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

37. 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 )

38. 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)

39. 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

40. 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

41. 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.

42. 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?

43. 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.

44. 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)

45. 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)

46. Alkali Metal Reactions The group IA metals will donate electrons to non-metals. Reactions with the halogens (group VIIA) produces ionic salts. 2 Na(s) + Cl 2 (g)  2 NaCl(s) Reactions with water produces hydrogen gas and the metal hydroxide. 2 Na(s) + 2 H 2 O(l)  2 NaOH(aq) + H 2 (g)

47. Reactions of the Halogens The elements of group VIIA readily pick up electrons from metals to form ionic compounds (salts). In addition, the halogens react with hydrogen gas to form covalently bonded hydrogen halides. H 2 (g) + Cl 2 (g)  2 HCl(g)

48. Reactions of the Halogens The halogens can also react with each other to form interhalogen compounds. The products are covalently bonded. Br 2 (g) + F 2 (g)  2 BrF(g)