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3 Chemical Equations & Reaction Stoichiometry

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2 Chapter Three Goals 1.Chemical Equations 2.Calculations Based on Chemical Equations 3.The Limiting Reactant Concept 4.Percent Yields from Chemical Reactions 5.Sequential Reactions 6.Concentrations of Solutions 7.Dilution of solutions 8.Using Solutions in Chemical Reactions 9.Synthesis Question

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3 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

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

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5 Chemical Equations Look at the information an equation provides:

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

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

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8 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

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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 132g

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

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

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

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

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

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15 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

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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).

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17 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 ?

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18 Calculations Based on Chemical Equations Example 3-2: 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?

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19 Calculations Based on Chemical Equations Example 3-2: 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?

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20 Calculations Based on Chemical Equations Example 3-2: 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?

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

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

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23 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!!!

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24 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?

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25 Limiting Reactant Concept Mix PacketsEggsMilk 11 dozen1 gallon limiting reactant is the muffin mix 21 dozen1 gallon 31 dozen1 gallon 41 dozen1 gallon 51 dozen1 gallon 61 dozen1 gallon 71 dozen1 gallon limiting reactant is the dozen eggs

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

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27 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?

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28 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?

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29 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?

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30 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?

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

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

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33 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.

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34 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?

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35 Percent Yields from Reactions

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36 Percent Yields from Reactions

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37 Percent Yields from Reactions

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38 Percent Yields from Reactions

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39 Sequential Reactions Example 3-10: Starting with 10.0 g of benzene (C 6 H 6 ), calculate the theoretical yield of nitrobenzene (C 6 H 5 NO 2 ) and of aniline (C 6 H 5 NH 2 ).

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40 Sequential Reactions

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

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42 Sequential Reactions

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43 Sequential Reactions

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44 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:

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45 Concentration of Solutions Example 3-11: What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% w/w NaOH?

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46 Concentration of Solutions Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.

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47 Concentration of Solutions Example 3-13: 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!

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48 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/mL. You do it!

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49 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/mL.

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50 Concentrations of Solutions Second common unit of concentration:

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51 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!

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52 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.

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53 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.

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54 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!

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55 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.

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56 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 M V = mol

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57 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?

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58 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?

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59 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?

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60 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.

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

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62 Dilution of Solutions Example 3-18: 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?

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63 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!

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64 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?

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65 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.

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66 Using Solutions in Chemical Reactions Example 3-20: What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

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67 Using Solutions in Chemical Reactions Example 3-20: What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

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68 Using Solutions in Chemical Reactions Example 3-20: What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

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69 Using Solutions in Chemical Reactions Example 3-21: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO 3 ) 3 ?

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70 Using Solutions in Chemical Reactions Example 3-20: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate?

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

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

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73 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

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74 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

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75 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

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76 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

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77 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

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3 Chemical Equations & Reaction Stoichimoetry

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