Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Figure 4.21A: Gravimetric analysis for barium: solution is poured. Photo courtesy of James Scherer.Figure 4.21B: Gravimetric analysis for barium: solution filtered.Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Figure 4.22A: Titration of an unknown amount of HCl with NaOH (#1). Photo courtesy of American Color.Figure 4.22B: Titration of an unknown amount of HCl with NaOH (#2). Photo courtesy of American Color.Figure 4.22C: Titration of an unknown amount of HCl with NaOH (#3). Photo courtesy of American Color.Slide 21Slide 22Slide 23Slide 24Second Exam: Friday February 15Chapters 3 and 4. Please note that there is a class at 1 pm so you will need to finish by 12:55 pm.Electronic Homework due R by 11:30 pmOffice hours this week: T 2-3 pm R 1-2 pmSL 130Solving Limiting Reactant Problems in Solution - Precipitation Problem - IProblem: Lead has been used as a glaze for pottery for years, and can bea problem if not fired properly in an oven, and is leachable from the pottery. Vinegar is used in leaching tests, followed by Lead precipitated as a sulfide. If 257.8 ml of a 0.0468 M solution of Lead nitrate is added to 156.00 ml of a 0.095 M solution of Sodium sulfide, what mass of solid Lead Sulfide will be formed?Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass.Solution: Writing the balanced equation:Pb(NO3)2 (aq) + Na2S (aq) 2 NaNO3 (aq) + PbS (s)Volume (L) of Pb(NO3)2 solutionMass (g) of PbSAmount (mol) of Pb(NO3)2Volume (L) of Na2S solutionAmount (mol) of Na2SAmount (mol) of PbSAmount (mol) of PbSMultiply by M (mol/L)Multiply by M (mol/L)Molar Ratio Molar RatioChoose the lower numberof PbS and multiply by M (g/mol)Volume (L) of Pb(NO3)2 solutionVolume (L) of Na2S solutionAmount (mol) of Pb(NO3)2Amount (mol) of Na2SAmount (mol) of PbSMass (g) of PbSMultiply by M (mol/L)Multiply by M (mol/L)Molar RatioDivide byequationcoefficientDivide byequationcoefficientSmallestSolving Limiting Reactant Problems inSolution - Precipitation Problem - IIMoles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = =Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) =Solving Limiting Reactant Problems inSolution - Precipitation Problem - IIMoles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2Calculation of product yield:Therefore Lead Nitrate is the Limiting Reactant!Solving Limiting Reactant Problems inSolution - Precipitation Problem - IIMoles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2Therefore Lead Nitrate is the Limiting Reactant!Calculation of product yield:Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+21 mol PbS1 mol Pb(NO3)2Solving Limiting Reactant Problems inSolution - Precipitation Problem - IIMoles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2Therefore Lead Nitrate is the Limiting Reactant!Calculation of product yield:Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+21 mol PbS1 mol Pb(NO3)20.012065 Mol Pb+2 = 0.012065 Mol PbS0.012065 Mol PbS x = 2.89 g PbS239.3 g PbS1 Mol PbSFigure 4.21A: Gravimetric analysis for barium: solution is poured.Photo courtesy of James Scherer.Figure 4.21B: Gravimetric analysis for barium: solution filtered.Percent Yield / Limiting Reactant Problem - IProblem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction.N2 (g) + 3 H2 (g) 2 NH3 (g)Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield.Solution: Moles of Nitrogen and Hydrogen:moles N2 = =85.90 g N228.02 g N2 1 mole N2 moles H2 = =21.66 g H22.016 g H21 mole H2Percent Yield / Limiting Reactant Problem - IProblem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction.N2 (g) + 3 H2 (g) 2 NH3 (g)Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield.Solution: Moles of Nitrogen and Hydrogen:moles N2 = = 3.066 mol N285.90 g N228.02 g N2 1 mole N2 moles H2 = = 10.74 mol H221.66 g H22.016 g H21 mole H2Percent Yield / Limiting Reactant Problem - IProblem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction.N2 (g) + 3 H2 (g) 2 NH3 (g)Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield.Solution: Moles of Nitrogen and Hydrogen:3.066 mol N2 2 mol NH3 1 mol N2 10.74 mol H2 mol NH33 mol H2= 6.132 mol NH3= 7.16 mol NH3N2 is Limiting!Percent Yield of a reaction: Actual Yield x 100 Theortetical YieldPercent