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CHAPTER THREE

CHAPTER THREE. CHEMICAL EQUATIONS & REACTION STOICHIOMETRY. Chapter Three Goals. Chemical Equations Calculations Based on Chemical Equations Percent Yields from Chemical Reactions The Limiting Reactant Concept Concentrations of Solutions Dilution of solutions

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CHAPTER THREE

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  1. CHAPTER THREE • CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

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

  3. Chemical Equations • A chemical process is represented by a chemical equation • Reaction of methane with O2: CH4 + 2O2 CO2 + 2H2O reactantsproducts • reactants on left side of reaction • products on right side of equation • relative amounts of each using stoichiometric coefficients

  4. Chemical Equations

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

  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,C3H8, burns in oxygen to give carbon dioxide and water.

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

  8. Law of Conservation of Matter • C7H16 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. Chemical Equations • Look at the information an equation provides: reactantsyieldsproducts 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. 1Fe2O3 needs 3 CO 25Fe2O3 needs ? CO Calculations Based on Chemical Equations • How many CO molecules are required to react with 25 formula units of Fe2O3? 25 Fe2O3 + ? CO  Product

  11. 1Fe2O3 gives 2Fe 2.5 X 105 Fe2O3 gives ? Fe Calculations Based on Chemical Equations • 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? Fe2O3 + excess CO  2Fe + 3CO2

  12. Fe2O3 + 3CO  Product MW(Fe2O3) needs 3MW(CO) 146 g needs ?g CO Calculations Based on Chemical Equations • What mass of CO is required to react with 146 g of iron (III) oxide?

  13. 1mol Fe2O3 gives 3 mol CO2 0.540 molFe2O3 gives ? mol CO2 ? mol CO2 = ? g CO2/MW(g/mol) CO2 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? Fe2O3 + excess CO  2Fe + 3CO2

  14. Calculations Based on Chemical Equations • 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! Fe2O3 + excess CO  2Fe + 3CO2 ? g Fe2O3 = 9.57 g Fe2O3

  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. Percent Yields from Reactions • 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? CH3COOH + C2H5OH  CH3COOC2H5 + H2O MW  MW 10.0 g X (Theoretical Yield)

  17. Percent Yields from Reactions

  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 C7H6O3 + C4H6O3 2 C9H8O4 + H2O salicylic acid acetic anhydride aspirin MW = 180 g/mol MW = 138 g/mol 29 billion tablets are consumed by Americans each year

  19. actual Yield (150 g) 78.5 = x 100 Theoretical Yield (g) Percent Yields from Reactions 2 C7H6O3 + C4H6O32 C9H8O4 + H2O salicylic acid acetic anhydride aspirin 2MW  2MW X  191.08(Theoretical Yield) Answer:X = 146.5 g

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

  21. Limiting Reactant Concept

  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 + Br2 HgBr2 Thus the limiting reagent is

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

  25. Limiting Reactant Concept 3PbO2 + Cr2(SO4)3 + K2SO4 + H2O  3PbSO4 + K2Cr2O7 + H2SO4 If 25.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? (a) PbO2 (b) H2O (c) K2SO4 (d) PbSO4 (e) Cr2(SO4)3 2MnO2 + 4KOH + O2 + Cl2 2KMnO4 + 2KCl + 2H2O If 20.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? (a) MnO2 (b) KOH (c) O2 (d) Cl2 (e) KMnO4

  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, H2O: NaCl: solute, H2O: solvent

  27. Amount of solute Mass or Volume of solution Concentration of Solutions Concentration = • 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. 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, H2O.

  29. g KNO3 x 100 (a) % by mass = g solution g KNO3 x 100 20.0 % = 250.0 g 20.0 % x 250.0 g g KNO3 = = 50.0 g 100 Concentration of Solutions • Calculate the mass of potassium nitrate, KNO3 required to prepare 250.0 g of solution that is 20.0 % KNO3 by mass. • What is the mass of water in the solution?

  30. Concentration of Solutions (b) mass of solution = mass of KNO3 + mass H2O mass H2O = mass of solution - mass of KNO3 mass H2O = 250.0 g - 50.0 g mass H2O = 200.0 g

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

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

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

  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. Concentrations of Solutions • Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . You do it!

  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. 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. 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 M1V1 = M2V2 is appropriate for dilutions, but not for chemical reactions.

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

  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. 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. 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. 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. Using Solutions in Chemical Reactions • What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4?

  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(NO3)3?

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

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

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