CHEM 1211 Lecture 10Outline of Last Lecture I. Example ReactionsA. Clicker QuestionsII. Chemical CalculationsA. Mom’s Pound CakeIII. Conceptual UnderstandingOutline of Current Lecture I. StoichiometryII. Clicker QuestionsIII. Concept BuildingIV. Aqueous ReactionsA. MolarityCurrent LectureI. StoichiometrygA molA molB gB = gB gA molA molB1. How many CO molecules are required to react with 25 formula units of Fe2O3?Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)25 f.u. Fe2O3 x 3 molecule CO = 75 molecules CO 1 f.u. Fe2O32. 146g of iron(III) oxide reacts and produces 99.8 g of iron. What is the percent yield for the reaction?(molar masses: Fe2O3 = 159.1, Fe = 55.84, CO = 28.00)Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)146g Fe2O31 mol Fe2O32 mol Fe 55.84g Fe = 102.48g Fe159.1g Fe2O3mol Fe2O3 mol Fe Theoretical Yield%Yield: 99.8g Fe x 100% = 97.4%102.48g Fe3. Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)How many pounds of iron (III) oxide is required to produce 125 pounds of CO2?(molar masses: Fe2O3 = 159.1, Fe = 55.84, CO2 = 44.00)125 lbs CO2454.59g CO21 mol CO21 mol Fe2O3159.1g Fe2O3 1 lb CO244.00g CO2 3 mol CO2 1 mol Fe2O3 1lb Fe2O3454.59g Fe2O3= 151 lb. Fe2O3II. Clicker Questions1. 3N2 + 9H2 6NH3: N2 + 3H2 2NH32. Limiting Reagent: a. NH3 – Product… has to be reactantsb. H2 – all used upc. N2 – all used upd. None3. Under Concept Building: 1b4. Under Concept Building: 2a5. Under Concept Building: 2bIII. Concept Building1. What is the maximum mass of sulfur dioxide that can be produced by a reaction of 95.6g of carbon disulfide with 110.g of oxygen?CS2 + 3O2 CO2 + 2SO2a. 95.6g of CS21 mol CS22 mol SO264.19g SO2=161g SO276.20g CS2 1mol CS2 1 mol SO2b. 110.g of O21 mol O22 mol SO264.10g SO2=147g SO2 32g O23 mol O2 1 mol SO22. An aqueous solution of iron (III) chloride and sodium sulfide react to form iron (III) sulfide and sodium chloride. a. What is the sum of the coefficients?1FeCl2 + 1Na2S 1FeS + 2NaCl = 5b. If you combine 40.g each of Sodium sulfide and iron (III) chloride who is the limiting reagent? FeCl2IV. Aqueous ReactionsA. Molarity = # mol solute M = mol = mmol # L of solution L mL1. Calculate molarity of a solution that contains 12.5g of sulfuric acid in 1.75L of solution.12.5g H2SO41mol H2SO4= 7.28 x 10-2M H2SO4 1.75L 98.1g H2SO42. Mass of calcium nitrate required to prepare 3.50L of 0.800M calcium nitrate?M x V = mol!0.800M Ca(NO3)2 x 3.50L = 2.8 mol Ca(NO3)20.800M Ca(NO3)2 x 3.50L x 164.09g Ca(NO3)2 = 459g Ca(NO3)2L 1 mol
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