3 Chemical Equations Reaction Stoichiometry Chapter Three Goals

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3 Chemical Equations & Reaction Stoichiometry

3 Chemical Equations & Reaction Stoichiometry

Chapter Three Goals 1. 2. 3. 4. 5. 6. 7. 8. 9. Chemical Equations

Chapter Three Goals 1. 2. 3. 4. 5. 6. 7. 8. 9. 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 2

Chemical Equations • Symbolic representation of a chemical reaction that shows: 1. reactants on

Chemical Equations • Symbolic representation of a chemical reaction that shows: 1. reactants on left side of reaction 2. products on right side of equation 3. relative amounts of each using stoichiometric coefficients 3

Chemical Equations • Attempt to show on paper what is happening at the laboratory

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

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

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

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

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

Chemical Equations • Look at the information an equation provides: reactants 1 formula unit

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

Chemical Equations • Look at the information an equation provides: reactants 1 formula unit

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

Chemical Equations • Look at the information an equation provides: reactants yields 1 formula

Chemical Equations • Look at the information an equation provides: reactants yields 1 formula unit 3 molecules 1 mole 3 moles 159. 7 g 84. 0 g products 2 atoms 2 moles 111. 7 g 132 g 9

Chemical Equations • Law of Conservation of Matter – There is no detectable change

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, C 3 H 8, burns in oxygen to give carbon dioxide and water. 10

Law of Conservation of Matter • NH 3 burns in oxygen to form NO

Law of Conservation of Matter • NH 3 burns in oxygen to form NO & water You do it! 11

Law of Conservation of Matter • NH 3 burns in oxygen to form NO

Law of Conservation of Matter • NH 3 burns in oxygen to form NO & water 12

Law of Conservation of Matter • C 7 H 16 burns in oxygen to

Law of Conservation of Matter • C 7 H 16 burns in oxygen to form carbon dioxide and water. You do it! 13

Law of Conservation of Matter • C 7 H 16 burns in oxygen to

Law of Conservation of Matter • C 7 H 16 burns in oxygen to form carbon dioxide and water. 14

Law of Conservation of Matter • C 7 H 16 burns in oxygen to

Law of Conservation of Matter • C 7 H 16 burns in oxygen to form carbon dioxide and water. • Balancing equations is a skill acquired only with lots of practice – work many problems 15

Calculations Based on Chemical Equations • Can work in moles, formula units, etc. •

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). 16

Calculations Based on Chemical Equations • Example 3 -1: How many CO molecules are

Calculations Based on Chemical Equations • Example 3 -1: How many CO molecules are required to react with 25 formula units of Fe 2 O 3? 17

Calculations Based on Chemical Equations • Example 3 -2: How many iron atoms can

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? 18

Calculations Based on Chemical Equations • Example 3 -2: How many iron atoms can

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

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 -3: What mass of CO is

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

Calculations Based on Chemical Equations • Example 3 -3: What mass of CO is

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

Calculations Based on Chemical Equations • Example 3 -3: What mass of CO is required to react with 146 g of iron (III) oxide? YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!! 23

Limiting Reactant Concept • Kitchen example of limiting reactant concept. 1 packet of muffin

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? 24

Limiting Reactant Concept • Mix Packets Eggs Milk 1 1 dozen 1 gallon limiting

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 25

Limiting Reactant Concept • Look at a chemical limiting reactant situation. Zn + 2

Limiting Reactant Concept • Look at a chemical limiting reactant situation. Zn + 2 HCl Zn. Cl 2 + H 2 26

Limiting Reactant Concept • Example 3 -8: What is the maximum mass of sulfur

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? 27

Limiting Reactant Concept • Example 3 -8: What is the maximum mass of sulfur

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? 28

Limiting Reactant Concept • Example 3 -8: What is the maximum mass of sulfur

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? 29

Limiting Reactant Concept • Example 3 -8: What is the maximum mass of sulfur

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? 30

Limiting Reactant Concept What do we do next? You do it! 31

Limiting Reactant Concept What do we do next? You do it! 31

Limiting Reactant Concept • • Which is limiting reactant? Limiting reactant is O 2.

Limiting Reactant Concept • • Which is limiting reactant? Limiting reactant is O 2. What is maximum mass of sulfur dioxide? Maximum mass is 147 g. 32

Percent Yields from Reactions • Theoretical yield is calculated by assuming that the reaction

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. 33

Percent Yields from Reactions • Example 3 -9: A 10. 0 g sample of

Percent Yields from Reactions • Example 3 -9: 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? 34

Percent Yields from Reactions 35

Percent Yields from Reactions 35

Percent Yields from Reactions 36

Percent Yields from Reactions 36

Percent Yields from Reactions 37

Percent Yields from Reactions 37

Percent Yields from Reactions 38

Percent Yields from Reactions 38

Sequential Reactions • Example 3 -10: Starting with 10. 0 g of benzene (C

Sequential Reactions • Example 3 -10: Starting with 10. 0 g of benzene (C 6 H 6), calculate theoretical yield of nitrobenzene (C 6 H 5 NO 2) and of aniline (C 6 H 5 NH 2). 39

Sequential Reactions 40

Sequential Reactions 40

Sequential Reactions • Next calculate the mass of aniline produced. You do it! 41

Sequential Reactions • Next calculate the mass of aniline produced. You do it! 41

Sequential Reactions 42

Sequential Reactions 42

Sequential Reactions 43

Sequential Reactions 43

Concentration of Solutions • Solution is a mixture of two or more substances dissolved

Concentration of Solutions • Solution is a mixture of two or more substances dissolved in another. – Solute is the substance present in the smaller amount. – Solvent is the substance present in the larger amount. – In aqueous solutions, the solvent is water. • The concentration of a solution defines the amount of solute dissolved in the solvent. – The amount of sugar in sweet tea can be defined by its concentration. • One common unit of concentration is: 44

Concentration of Solutions • Example 3 -11: What mass of Na. OH is required

Concentration of Solutions • Example 3 -11: What mass of Na. OH is required to prepare 250. 0 g of solution that is 8. 00% w/w Na. OH? 45

Concentration of Solutions • Example 3 -12: Calculate the mass of 8. 00% w/w

Concentration of Solutions • Example 3 -12: Calculate the mass of 8. 00% w/w Na. OH solution that contains 32. 0 g of Na. OH. 46

Concentration of Solutions • Example 3 -13: Calculate the mass of Na. OH in

Concentration of Solutions • Example 3 -13: Calculate the mass of Na. OH in 300. 0 m. L of an 8. 00% w/w Na. OH solution. Density is 1. 09 g/m. L. You do it! 47

Concentrations of Solutions • Example 3 -14: What volume of 12. 0% KOH contains

Concentrations of Solutions • Example 3 -14: What volume of 12. 0% KOH contains 40. 0 g of KOH? The density of the solution is 1. 11 g/m. L. You do it! 48

Concentrations of Solutions • Example 3 -14: What volume of 12. 0% KOH contains

Concentrations of Solutions • Example 3 -14: What volume of 12. 0% KOH contains 40. 0 g of KOH? The density of the solution is 1. 11 g/m. L. 49

Concentrations of Solutions • Second common unit of concentration: 50

Concentrations of Solutions • Second common unit of concentration: 50

Concentrations of Solutions • Example 3 -15: Calculate the molarity of a solution that

Concentrations of Solutions • Example 3 -15: Calculate the molarity of a solution that contains 12. 5 g of sulfuric acid in 1. 75 L of solution. You do it! 51

Concentrations of Solutions • Example 3 -15: Calculate the molarity of a solution that

Concentrations of Solutions • Example 3 -15: Calculate the molarity of a solution that contains 12. 5 g of sulfuric acid in 1. 75 L of solution. 52

Concentrations of Solutions • Example 3 -15: Calculate the molarity of a solution that

Concentrations of Solutions • Example 3 -15: Calculate the molarity of a solution that contains 12. 5 g of sulfuric acid in 1. 75 L of solution. 53

Concentrations of Solutions • Example 3 -16: Determine the mass of calcium nitrate required

Concentrations of Solutions • Example 3 -16: Determine the mass of calcium nitrate required to prepare 3. 50 L of 0. 800 M Ca(NO 3)2. You do it! 54

Concentrations of Solutions • Example 3 -16: Determine the mass of calcium nitrate required

Concentrations of Solutions • Example 3 -16: Determine the mass of calcium nitrate required to prepare 3. 50 L of 0. 800 M Ca(NO 3)2. 55

Concentrations of Solutions • One of the reasons that molarity is commonly used is

Concentrations of Solutions • One of the reasons that molarity is commonly used is because: M x L = moles solute and M x m. L = mmol solute M V = mol 56

Concentrations of Solutions • Example 3 -17: The specific gravity of concentrated HCl is

Concentrations of Solutions • Example 3 -17: The specific gravity of concentrated HCl is 1. 185 and it is 36. 31% w/w HCl. What is its molarity? 57

Concentrations of Solutions • Example 3 -17: The specific gravity of concentrated HCl is

Concentrations of Solutions • Example 3 -17: The specific gravity of concentrated HCl is 1. 185 and it is 36. 31% w/w HCl. What is its molarity? 58

Concentrations of Solutions • Example 3 -17: The specific gravity of concentrated HCl is

Concentrations of Solutions • Example 3 -17: The specific gravity of concentrated HCl is 1. 185 and it is 36. 31% w/w HCl. What is its molarity? 59

Dilution of Solutions • To dilute a solution, add solvent to a concentrated solution.

Dilution of Solutions • To dilute a solution, add solvent to a concentrated solution. – One method to make tea “less sweet. ” – How fountain drinks are made from syrup. • The number of moles of solute 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. 60

Dilution of Solutions • Common method to dilute a solution involves the use of

Dilution of Solutions • Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb. 61

Dilution of Solutions • Example 3 -18: If 10. 0 m. L of 12.

Dilution of Solutions • Example 3 -18: If 10. 0 m. L of 12. 0 M HCl is added to enough water to give 100. m. L of solution, what is the concentration of the solution? 62

Dilution of Solutions • Example 3 -19: What volume of 18. 0 M sulfuric

Dilution of Solutions • Example 3 -19: 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! 63

Dilution of Solutions • Example 3 -19: What volume of 18. 0 M sulfuric

Dilution of Solutions • Example 3 -19: What volume of 18. 0 M sulfuric acid is required to make 2. 50 L of a 2. 40 M sulfuric acid solution? 64

Using Solutions in Chemical Reactions • Combine the concepts of molarity and stoichiometry to

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. 65

Using Solutions in Chemical Reactions • Example 3 -20: What volume of 0. 500

Using Solutions in Chemical Reactions • Example 3 -20: What volume of 0. 500 M Ba. Cl 2 is required to completely react with 4. 32 g of Na 2 SO 4? 66

Using Solutions in Chemical Reactions • Example 3 -20: What volume of 0. 500

Using Solutions in Chemical Reactions • Example 3 -20: What volume of 0. 500 M Ba. Cl 2 is required to completely react with 4. 32 g of Na 2 SO 4? 67

Using Solutions in Chemical Reactions • Example 3 -20: What volume of 0. 500

Using Solutions in Chemical Reactions • Example 3 -20: What volume of 0. 500 M Ba. Cl 2 is required to completely react with 4. 32 g of Na 2 SO 4? 68

Using Solutions in Chemical Reactions • Example 3 -21: (a)What volume of 0. 200

Using Solutions in Chemical Reactions • Example 3 -21: (a)What volume of 0. 200 M Na. OH will react with 50. 0 m. L 0 f 0. 200 M aluminum nitrate, Al(NO 3)3? 69

Using Solutions in Chemical Reactions • Example 3 -20: (a)What volume of 0. 200

Using Solutions in Chemical Reactions • Example 3 -20: (a)What volume of 0. 200 M Na. OH will react with 50. 0 m. L 0 f 0. 200 M aluminum nitrate? 70

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

Using Solutions in Chemical Reactions • (b)What mass of Al(OH)3 precipitates in (a)? You do it! 71

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

Using Solutions in Chemical Reactions • (b) What mass of Al(OH)3 precipitates in (a)? 72

Using Solutions in Chemical Reactions • Titrations are a method of determining the concentration

Using Solutions in Chemical Reactions • Titrations are a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. – Requires special lab glassware • Buret, pipet, and flasks – Must have an indicator also 73

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of a KOH solution if 38. 7 m. L of the KOH solution is required to react with 43. 2 m. L of 0. 223 M HCl? 74

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of a KOH solution if 38. 7 m. L of the KOH solution is required to react with 43. 2 m. L of 0. 223 M HCl? 75

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of a KOH solution if 38. 7 m. L of the KOH solution is required to react with 43. 2 m. L of 0. 223 M HCl? 76

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of

Using Solutions in Chemical Reactions • Example 3 -22: What is the molarity of a KOH solution if 38. 7 m. L of the KOH solution is required to react with 43. 2 m. L of 0. 223 M HCl? 77

3 Chemical Equations & Reaction Stoichimoetry

3 Chemical Equations & Reaction Stoichimoetry