Chapter 4§4.1 Stoichiometry§4.2 Stoichiometric Calculations§4.2 Stoichiometry§4.3 Cupcakes§4.3 Limiting ReactantSlide 7Slide 8§4.3 Percent YieldSlide 10Slide 11§4.4 Some Definitions§4.3 Solution Concentration§4.4 Molarity Calculations§4.5 ElectrolytesPowerPoint Presentation§4.5 Strong Electrolytes§4.5 Weak Electrolytes§4.4 Dilution§4.6 Three Types of Reactions§4.5 Solubility Rules§4.6 Precipitation Reactions§4.6 Precipitation ReactionsSlide 24§4.6 PrecipitationSlide 26Slide 27Slide 28§4.7 Writing Chemical Equations for SolutionsWrite the net ionic equation (if any) for these reactions.§4.8 Acid-Base Reactions§4.8 Acid-Base TitrationsSlide 33§4.9 Oxidation-Reduction ReactionsSlide 35Slide 36Slide 37§4.9 Oxidation State§4.9 Oxidation State RulesSlide 40Slide 41§4.9 Oxidation StatesSlide 43§4.6 Precipitation StoichiometrySlide 45Slide 46STOICHIOMETRYChapter 4§4.1 StoichiometryStoichiometry – calculating quantities of reactants and products in a chemical reactionStoichiometry is based on the Law of Conservation of Mass.§4.2 Stoichiometric CalculationsThe amounts of any other substance in a chemical reaction can be determined from the amount of just one substance:(where A and B are in the same reaction)§4.2 StoichiometryWhat mass of sulfuric acid (98.09 g/mol) would exactly react with 15.2 g aluminum (26.98 g/mol) ?3 H2SO4 (aq) + 2 Al(s) → Al2(SO4)3 (aq) + 3 H2(g) 15.2 g Al1 mol Al26.98 g Al3 mol H2SO4 2 mol Al98.09 g H2SO4 1 mol H2SO4= 82.9 g H2SO4§4.3 CupcakesIf you have 10 cupcakes, 2 cherries and enough whipped cream for 25 cupcakes, how many of these desserts can you make?§4.3 Limiting ReactantSometimes all reactants are completely consumed; sometimes one or more is in excess.A limiting reactant is present in lesser molar amounts and is consumed first, limiting the amount of product that can form.Always determine which reactant is limiting when performing stoichiometry calculations!§4.3 Limiting ReactantN2(g) + 3H2(g) → 2NH3(g) in excess limiting reagent§4.3 Limiting ReactantReaction of ammonia and oxygen gives nitrogen monoxide and water. Determine the mass of nitrogen monoxide produced from 10.0 g ammonia and 10.0 g oxygen. NH3(g) + O2(g)  NO(g) + H2O(l)4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)1. Balance the equationmol NH317.034 g NH3mol O232.00 g O22. Limiting reagent?10.0 g NH3 10.0 g O24 mol NO4 mol NH34 mol NO5 mol O2= 0.587 mol NO= 0.250 mol NO4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)Limiting Reagent30.01 g NO1 mol NO30.01 g NO1 mol NO= 17.7 g NO= 7.50 g NO§4.3 Percent YieldTheoretical Yield (calculated) – the maximum amount of product that can form during a reaction Actual Yield (measured) – the amount of pure product collected from a reaction(actual yield) ≤ (theoretical yield)Percent Yield = Actual Yield (100%) Theoretical Yield§4.3 Percent YieldIf 18.20 g of NH3 (mm 17.03 g/mol) are produced by a reaction mixture that initially contains 6.00 g H2 and an excess of N2, what is the percent yield of the reaction?N2(g) + 3H2(g) → 2NH3(g)Calculate theoretical yield of NH3 :1 mol H22.016 g H26.00 g H2 2 mol NH33 mol H217.03 g NH31 mol NH3Percent yield:18.20 g NH333.8 g NH3(100%)= 53.8%N2(g) + 3H2(g) → 2NH3(g)= 33.8 g NH3= 33.8 g NH3§4.3 Percent YieldReaction of titanium (IV) oxide with graphite produces titanium metal and carbon monoxide. When 28.6 kg of C is allowed to react with 88.2 kg of titanium (IV) oxide, 42.8 kg of Ti is produced. Find the percent yield of Ti from this reaction.TiO2(s) + C(s)  Ti(s) + CO(g)TiO2(s) + 2C(s)  Ti(s) + 2CO(g)1. Balance the equationmol TiO279.88 g TiO2mol C12.01 g C2. Limiting reagent?88.2 x 103 g TiO228.6 x 103 g C1 mol Ti1 mol TiO21 mol Ti2 mol C= 1104 mol Ti= 1191 mol TiTiO2(s) + 2C(s)  Ti(s) + 2CO(g)Limiting Reagent47.88 g Ti mol Ti42.8 kg Ti (100%) 42.8 kg Ti (100%)= 52.9 kg Ti= 80.9%= 5.29 x 104 g Ti§4.4 Some DefinitionsSolution – a homogenous mixtureSolvent – the majority solution component, retains its stateSolute – the minority solution component, becomes the state of the solventSalt – any ionic compound§4.3 Solution ConcentrationMolarity (M) = moles of solute liters of solution0.5 M CuSO40.3 M CuSO40.1 M CuSO4§4.4 Molarity CalculationsDetermine the molarity of a solution where 45.5 mg of NaCl is dissolved to give 154.4 mL of solution.How many moles of Cl- ion are contained in 114 mL of a 1.85 M CaCl2 solution?45.5 mg NaCl154.4 mL soln1000 mL soln1 L soln1 g NaCl1000 mg NaCl1 mol NaCl58.44 g NaCl= 5.04 × 10-3 M NaClCaCl2(s) -> Ca2+(aq) + 2Cl-(aq)114 mL soln1.85 mol CaCl21 L soln1 L soln1000 mL soln2 mol Cl-1 mol CaCl2= 0.422 mol Cl-§4.5 ElectrolytesAn electrolyte is a substance that when dissolved in water produces a solution that conducts electricity.Nonelectrolytes dissolve in water but don't ionize.§4.5 ElectrolytesElectrical conductivity is a qualitative measure of the degree to which a solute ionizes.weakelectrolytenon-electrolytestrongelectrolyte§4.5 Strong ElectrolytesStrong electrolytes completely ionize in water. There are three types:1) Soluble salts – NaCl, KBr, LiNO3, etc.2) Strong acids – HCl, HNO3, H2SO4, HClO43) Strong bases – NaOH, KOHExample: NaOH(s) → Na+(aq) + OH-(aq)100% ionized H2O§4.5 Weak ElectrolytesWeak electrolytes slightly ionize in water.Examples include weak acids such as HC2H3O2 and weak bases such as NH3.For instance, acetic acid is ~1% ionized in water:HC2H3O2(l) H+(aq) + C2H3O2-(aq)In contrast, strong electrolytes are 100% ionized.→←99% 1%§4.4 DilutionIt is common to make dilute solutions from more concentrated ones (stock solutions).For dilutions: more solvent is added but the amount of solute remains constant. Hence:M1V1 = M2V2Molarity = mol solute L solution(Molarity) (L solution) = mol soluteM1 = molarity of initial solutionV1 = volume of initial solutionM2 = molarity of final solutionV2 = volume of final solutionmol solute before dilution = mol solute after dilution§4.6 Three Types of Reactions1. Precipitation reactions2. Acid-base reactions3. Oxidation-reduction reactionsSkip gas-evolution reactions in this section.§4.5 Solubility RulesLearn these.§4.6 Precipitation ReactionsHow