Chem 161-2011 Lecture 14 2PLAN FOR TODAYCOMPLETE CHAPTER 8 Reaction yield Periodic trends in reactivity of the main group elements General trends in reactivity Hydrogen (1s1) Reactions of the active metals Group 1A Elements (ns1, n ≥2) Group 2A Elements (ns2, n ≥2) Reactions of other main group elements Group 3A Elements (ns2np1, n ≥2) Group 4A Elements (ns2np2, n ≥2) Group 5A Elements (ns2np3, n ≥2) Group 6A Elements (ns2np4, n ≥2) Group 7A Elements (ns2np5, n ≥2) Group 8A Elements (ns2np6, n ≥2) Comparison of Group 1A and Group 1B elements CHAPTER 9, CHEMICAL REACTIONS IN AQUEOUS SOLUTIONSGeneral properties of aqueous solutions Electrolytes and nonelectrolytes Strong electrolytes and weak electrolytes Precipitation reactions Solubility guidelines for ionic compounds in water Molecular equations Ionic equations Net ionic equations Acid-base reactions Strong acids and bases Brønsted acids and bases Acid-base neutralization Oxidation-reduction reactions Oxidation numbers Oxidation of metals in aqueous solutions Balancing simple redox equations Other types of redox reactions Combination reactions Decomposition Concentration of solutions Molarity Dilutuon Serial dilution Solution stoichiometry Aqueous reactions and chemical analysis Gravimetric analysis Acid-Base titrations
Chem 161-2011 Lecture 14 3PERCENT YIELDTERMS: YIELD = ACTUAL YIELD, i.e., ACTUAL QUANTITY OF PRODUCT OBTAINED THEORETICAL YIELD = CALC. YIELD FROM EQUATION PERCENT YIELD = PERCENT OF THEORETICAL YIELD ET: e.g., 6 sandwiches + 14 cookies →6 box lunches, but only 4 box lunches made. Yield = 4 box lunches; theoretical yield = 6 box lunches; percent yield = 67%. Hill & Petrucci Aluminum burns in bromine, producing aluminum bromide: 2Al(s)+ 3Br2(l)→2AlBr3(s)In a certain experiment, 6.0 g aluminum was reacted with an excess of bromine to yield 50.3 g aluminum bromide. Calculate the theoretical and percent yields for this experiment. 2Al(s)+ 3Br2(l)→2AlBr3(s)6.0 g XS 50.3 g actual ? g theoretically Theoretical Yield (= Calculated yield)Plan: g Al →mol Al →mol AlBr3→g AlBr3AWAl= 26.98 g/mol MW Br2= 79.90 x 2 = 159.8 g/mol MWAlBr3= 26.98 + (3 x 79.90) = 266.68 g/mol 6.0 g Al/(26.98 g Al/mol Al) x (2 mol AlBr3/2 mol Al) x (266.68 g AlBr3/mol AlBr3) = 59.31g AlBr3= theoretical yield of AlBr3Percent Yield (= Percent of theoretical yield)(50.3 g actual yield/59.31g theoretical yield) x 100 = 84.8 percent of theoretical yield = 84.8% yield
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