CHEM 101 1st Edition Lecture 10 Outline of Last Lecture I. Anhydrides of Acids and BasesII. Gas- Forming Reactions of Ionic CompoundsIII. Oxidation-Reduction ReactionsIV. Determining Oxidation NumbersOutline of Current Lecture I. StoichiometryII. Limiting ReactantsCurrent LectureChapter 4: Stoichiometry- Stoichiometry- the study of mass relationships in chemical reactionso Provides quantitative information about chemical reactionso Mass must be conserved in a chemical reactiono Cannot equal the mass of reactants and products must be CONVERTED to moles.1 mole of reactant A 1 mole of product B L> x(y mol product B/x mol reactant A) stoichiometric factor Convert mol of product B to mass to get expected outcomeo Ex: 454g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water; how many grams of dinitrogen monoxide and water are formed?Step 1: write balanced equation: NH4NO3(s) N2O + 2 H2O(l)Step 2: convert to moles 454 NH4NO3 x 1 mol = 5.67 mol N4NO3 1 mol NH4NO3Step 3: Convert moles of reactants to moles of products 5.67 mol NH4NO3 x 1 mol N 2 O = 5.67 mol N2O 1 mol NH4NO3 5.67 mol NH4NO3 x 2 mol H 2 O = 11.3 mol H2O 1 mol NH4NO3These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Step 4: Convert moles of product to masses 5.67 mol NH4NO3 x 80.04 g = 454 g NH4NO3 1 mol NH4NO3 5.67 mol NH4NO3 x 44.01 g N 2 O = 250 g N2O 1 mol NH4NO3 5.67 mol NH4NO3 x 18.02 g H 2 O = 204 g H2O 1 mol NH4NO3- Limiting Reactantso Some reactions are carried out with an excess of one reactant over that regulated by stoichiometry.CH4(g) + 2O2(g) CO3(g) + 2H2O(l)o Methane is the limiting reactanto The amount of product is determined by the amount of CH4 (limiting) not O2 (excess)o Ex: Which is the limiting reactant if we use 50.0 g Al and 50.0 O2 for the reaction between these two elements? What is the mass of the product?Step 1: Write balanced equation 4Al(s) + 3 O2(g) 2Al2O3(s)Step 2: Convert masses to moles 50.0 Al x 1 mol = 1.85 mol Al 1 mol Al 50.0 O2 x 1 mol = 1.56 mol O2 1 mol O2Step 3: Which is the limiting reactant? Use ratios: Actual Theoretical 1.85 mol Al = 1.19 mol Al 4 mol Al = 1.37 mol Al 1.56 mol O2 1.00 mole O2 3 mol O2 1.00 mol O2 1.19 < 1.33 * Al= limiting reactant because we have less than what we should have * O2 is excessStep 4: Mass of products in grams: *Always convert to grams from the limiting reactant 1.85 mol Al x 2 mol Al 2 O 3 x 101.96 g Al 2 O 3 = 94.3 g Al2O3 =theoretical yield 4 mol Al 1 mol Al2O3 How many dioxygen remain? (using law of conservation)masses of AL + O2 = masses of 2Al2O3 + O2(excess) 50 + 50 = 94.3 + 5.7g O2 excess If 48.5g of Al2O3 was collected at the completion of the reaction (actual yield), what is the percent yield of this reaction?% yield = actual yield x 100 theoretical yield = 48.5 g x 100 = 51.4%
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