CHAPTER THREE

# CHAPTER THREE

## CHAPTER THREE

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##### Presentation Transcript

1. CHAPTER THREE • CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

2. Chapter Three Goals • Chemical Equations • Calculations Based on Chemical Equations • The Limiting Reactant Concept • Percent Yields from Chemical Reactions • Sequential Reactions • Concentrations of Solutions • Dilution of solutions • Using Solutions in Chemical Reactions • Synthesis Question

3. Chemical Equations • Symbolic representation of a chemical reaction that shows: • reactants on left side of reaction • products on right side of equation • relative amounts of each using stoichiometric coefficients

4. Chemical Equations • Attempt to show on paper what is happening at the laboratory and molecular levels.

5. Chemical Equations • Look at the information an equation provides:

6. Chemical Equations • Look at the information an equation provides: reactants yields products

7. Chemical Equations • Look at the information an equation provides: reactants yields products 1formula unit 3 molecules 2 atoms 3 molecules

8. Chemical Equations • Look at the information an equation provides: reactants yields products 1formula unit 3 molecules 2 atoms 3 molecules 1 mole 3 moles 2 moles 3 moles

9. Chemical Equations • Look at the information an equation provides: reactants yields products 1formula unit 3 molecules 2 atoms 3 molecules 1 mole 3 moles 2 moles 3 moles 159.7 g 84.0 g 111.7 g 132 g

10. Chemical Equations • Law of Conservation of Matter • There is no detectable change in quantity of matter in an ordinary chemical reaction. • Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. • This law was determined by Antoine Lavoisier. • Propane,C3H8, burns in oxygen to give carbon dioxide and water.

11. Law of Conservation of Matter • NH3 burns in oxygen to form NO & water You do it!

12. Law of Conservation of Matter • NH3 burns in oxygen to form NO & water

13. Law of Conservation of Matter • C7H16 burns in oxygen to form carbon dioxide and water. You do it!

14. Law of Conservation of Matter • C7H16 burns in oxygen to form carbon dioxide and water.

15. Law of Conservation of Matter • C7H16 burns in oxygen to form carbon dioxide and water. • Balancing equations is a skill acquired only with lots of practice • work many problems

16. Calculations Based on Chemical Equations • Can work in moles, formula units, etc. • Frequently, we work in mass or weight (grams or kg or pounds or tons).

17. Calculations Based on Chemical Equations • Example 3-1: How many CO molecules are required to react with 25 formula units of Fe2O3?

18. Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

19. Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

20. Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

21. Calculations Based on Chemical Equations • Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

22. Calculations Based on Chemical Equations • Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

23. Calculations Based on Chemical Equations • Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

24. Calculations Based on Chemical Equations • Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

25. Calculations Based on Chemical Equations • Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

26. Calculations Based on Chemical Equations • Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

27. Calculations Based on Chemical Equations • Example 3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? You do it!

28. Calculations Based on Chemical Equations • Example 3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

29. Calculations Based on Chemical Equations • Example 3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide? You do it!

30. Calculations Based on Chemical Equations YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!!

31. Limiting Reactant Concept • Kitchen example of limiting reactant concept. 1 packet of muffin mix + 2 eggs + 1 cup of milk • 12 muffins • How many muffins can we make with the following amounts of mix, eggs, and milk?

32. Limiting Reactant Concept • Mix Packets Eggs Milk 1 1 dozen 1 gallon limiting reactant is the muffin mix 2 1 dozen 1 gallon 3 1 dozen 1 gallon 4 1 dozen 1 gallon 5 1 dozen 1 gallon 6 1 dozen 1 gallon 7 1 dozen 1 gallon limiting reactant is the dozen eggs

33. Limiting Reactant Concept • Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box?

34. Limiting Reactant Concept • Look at a chemical limiting reactant situation. Zn + 2 HCl ZnCl2 + H2

35. Limiting Reactant Concept • Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

36. Limiting Reactant Concept • Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

37. Limiting Reactant Concept • Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

38. Limiting Reactant Concept • Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

39. Limiting Reactant Concept What do we do next? You do it!

40. Limiting Reactant Concept • Which is limiting reactant? • Limiting reactant is O2. • What is maximum mass of sulfur dioxide? • Maximum mass is 147 g.

41. Percent Yields from Reactions • Theoretical yield is calculated by assuming that the reaction goes to completion. • Determined from the limiting reactant calculation. • Actual yield is the amount of a specified pure product made in a given reaction. • In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried. • Percent yield indicates how much of the product is obtained from a reaction.

42. Percent Yields from Reactions • Example 3-9: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?

43. Percent Yields from Reactions

44. Percent Yields from Reactions

45. Percent Yields from Reactions

46. Percent Yields from Reactions

47. Sequential Reactions • Example 3-10: Starting with 10.0 g of benzene (C6H6), calculate the theoretical yield of nitrobenzene (C6H5NO2) and of aniline (C6H5NH2).

48. Sequential Reactions

49. Sequential Reactions • Next calculate the mass of aniline produced. You do it!

50. Sequential Reactions