1. 1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

2. 2 Chapter Three Goals 1. Chemical Equations 2. Calculations Based on Chemical Equations 3. Percent Yields from Chemical Reactions 4. The Limiting Reactant Concept 5. Concentrations of Solutions 6. Dilution of solutions 7. Using Solutions in Chemical Reactions

3. 3 Chemical Equations A chemical process is represented by a chemical equation 1. Reaction of methane with O 2 : CH 4 + 2O 2 CO 2 + 2H 2 O reactantsproducts reactantsproducts 2. reactants on left side of reaction 3. products on right side of equation 4. relative amounts of each using stoichiometric coefficients

4. 4 Chemical Equations

5. 5 Look at the information an equation provides: reactants yields products 1 formula 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

6. 6 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. Propane,C 3 H 8, burns in oxygen to give carbon dioxide and water.

7. 7 Law of Conservation of Matter NH 3 burns in oxygen to form NO & water

8. 8 Law of Conservation of Matter You do it! C 7 H 16 burns in oxygen to form carbon dioxide and water. You do it! Balancing equations is a skill acquired only with lots of practice – work many problems

9. 9 Chemical Equations Look at the information an equation provides: reactants yields products 1 formula 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. 10 Calculations Based on Chemical Equations How many CO molecules are required to react with 25 formula units of Fe 2 O 3 ? 25 Fe 2 O 3 + ? CO  Product 1Fe 2 O 3 needs 3 CO 25Fe 2 O 3 needs ? CO

11. 11 Calculations Based on Chemical Equations How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide? Fe 2 O 3 + excess CO  2Fe + 3CO 2 1Fe 2 O 3 gives 2Fe 2.5 X 10 5 Fe 2 O 3 gives ? Fe

12. 12 Calculations Based on Chemical Equations What mass of CO is required to react with 146 g of iron (III) oxide? Fe 2 O 3 + 3CO  Product MW(Fe 2 O 3 ) needs 3MW(CO) 146 g needs ?g CO

13. 13 Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? Fe 2 O 3 + excess CO  2Fe + 3CO 2 1mol Fe 2 O 3 gives 3 mol CO 2 0.540 mol Fe 2 O 3 gives ? mol CO 2 ? mol CO 2 = ? g CO 2 /MW(g/mol) CO 2

14. 14 Calculations Based on Chemical Equations You do it! 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! Fe 2 O 3 + excess CO  2Fe + 3CO 2 ? g Fe 2 O 3 = 9.57 g Fe 2 O 3

15. 15 Percent Yields from Reactions Theoretical yield is calculated by assuming that the reaction goes to completion. 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.

16. 16 Percent Yields from Reactions A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What is the percent yield? CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5 + H 2 O MW  MW 10.0 g  X (Theoretical Yield)

17. 17 Percent Yields from Reactions

18. 18 Percent Yields from Reactions Salicylic acid reacts with acetic anhydride to form aspirin, acetylsalicylic acid. If the percent yield in this reaction is 78.5%, what mass of salicylic acid is required to produce 150. g aspirin? 2 C 7 H 6 O 3 + C 4 H 6 O 3  2 C 9 H 8 O 4 + H 2 O salicylic acid acetic anhydride aspirin 29 billion tablets are consumed by Americans each year MW = 138 g/mol MW = 180 g/mol

19. 19 Percent Yields from Reactions 2 C 7 H 6 O 3 + C 4 H 6 O 3  2 C 9 H 8 O 4 + H 2 O salicylic acid acetic anhydride aspirin 2MW  2MW X  191.08 (Theoretical Yield) Answer: X = 146.5 g actual Yield (150 g) Theoretical Yield (g) 78.5 =x 100

20. 20 Limiting Reactant Concept 1. In a given reaction, there is not enough of one reagent to use up the other reagents completely. 2. The reagent in short supply LIMITS the quantity of the product that can be formed. 3. How many bikes can be made from 10 frames and 16 wheels? 1 frame + 2 wheels 1 bike excess limiting

21. 21 Limiting Reactant Concept

22. 22 Limiting Reactant Concept When 100.0 g mercury is reacted with 100.0 g bromine to form mercuric bromide, which is the limiting reagent? Hg + Br 2 HgBr 2 Thus the limiting reagent is

23. 23 Limiting Reactant Concept 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?

24. 24 Limiting Reactant Concept Which is limiting reactant? Limiting reactant is O 2. What is maximum mass of sulfur dioxide? Maximum mass is 147 g.

25. 25 Limiting Reactant Concept If 25.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? (a) PbO 2 (b) H 2 O (c) K 2 SO 4 (d) PbSO 4 (e) Cr 2 (SO 4 ) 3 3PbO 2 + Cr 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O  3PbSO 4 + K 2 Cr 2 O 7 + H 2 SO 4 If 20.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? (a) MnO 2 (b) KOH (c) O 2 (d) Cl 2 (e) KMnO 4 2MnO 2 + 4KOH + O 2 + Cl 2  2KMnO 4 + 2KCl + 2H 2 O

26. 26 Concentration of Solutions Definition of Solution: a homogeneous mixture of two or more substances dissolved in another. A solution is composed of two parts: (1) Solute: dissolved substance (or substance in the lesser amount). (2) Solvent: dissolving substance (or substance in the greater amount). – In aqueous solutions, the solvent is water. Example: Solution of NaCl in water, H 2 O: NaCl: solute, H 2 O: solvent

27. 27 Concentration of Solutions Concentration = Amount of solute Mass or Volume of solution Relative terms: Dilute solution: small amount of solute in large amount of solvent. Concentrated solution: large amount of solute in smaller amount of solvent (e.g. the amount of sugar in sweet tea can be defined by its concentration). We will discuss 2 concentration units: 1- Percent by mass (do not confuse with % by mass of element in compound) 2- Molarity

28. 28 Concentration of Solutions 1- Percent by mass : Note: if the question says the solution is aqueous oe does not Specify the solvent, the solvent is water, H 2 O.

29. 29 Concentration of Solutions Calculate the mass of potassium nitrate, KNO 3 required to prepare 250.0 g of solution that is 20.0 % KNO 3 by mass. What is the mass of water in the solution? (a)% by mass = g KNO 3 g solution x 100 20.0 % = g KNO 3 250.0 g x 100 g KNO 3 = 100 20.0 % x 250.0 g = 50.0 g

30. 30 Concentration of Solutions (b) mass of solution = mass of KNO 3 + mass H 2 O mass H 2 O = mass of solution - mass of KNO 3 mass H 2 O = 250.0 g - 50.0 g mass H 2 O = 200.0 g

31. 31 Concentration of Solutions Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.

32. 32 Concentration of Solutions Calculate the mass of NaOH in 300.0 mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL. You do it!

33. 33 Concentrations of Solutions What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL. You do it!

34. 34 Concentrations of Solutions 2- Second common unit of concentration: Molarity

35. 35 Concentrations of Solutions 2- Second common unit of concentration: Molarity

36. 36 Concentrations of Solutions Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. You do it!

37. 37 Concentrations of Solutions Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO 3 ) 2. You do it!

38. 38 Concentrations of Solutions One of the reasons that molarity is commonly used is because: M x L = moles solute and M x mL = mmol solute

39. 39 Concentrations of Solutions The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?

40. 40 Dilution of Solutions To dilute a solution, add solvent to a concentrated solution. – One method to make tea “less sweet.” The number of moles of solute (before and after dilution) in the two solutions remains constant. The relationship M 1 V 1 = M 2 V 2 is appropriate for dilutions, but not for chemical reactions.

41. 41 Dilution of Solutions Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.

42. 42 Dilution of Solutions Take-Home Calculations (M x V) A = (M x V) B M x V = moles of solute M x V = W/MW (M x V) A = (W/MW) A OR (M x V) A = (W/MW) B W = M x V x MW

43. 43 Dilution of Solutions If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution?

44. 44 Dilution of Solutions What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? You do it!

45. 45 Using Solutions in Chemical Reactions Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.

46. 46 Using Solutions in Chemical Reactions What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

47. 47 Using Solutions in Chemical Reactions (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO 3 ) 3 ?

48. 48 Using Solutions in Chemical Reactions (a) What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate?

49. 49 Using Solutions in Chemical Reactions (b) What mass of Al(OH) 3 precipitates in (a)?

50. 50 Homework Assignment One-line Web Learning (OWL): Chapter 3 Exercises and Tutors – Optional

51. 51 End of Chapter 3