When copper (II) reacts with silver nitrate, the number of grams of copper required to produce 432 grams of silver is:Warm-Up CuAgNO 3 Ag22+Cu(NO 3 ) g Ag1 mol Ag g Ag2 mol Ag 1 mol Cu63.55 g Cu 1 mol Cu = g Cu  +

Limiting Factors and Percent Yield

Hot Dogs in the News Takeru Kobayashi of Japan downed 44½ hot dogs in 12 minutes. Source: CNN.com WHAT IF… One hot dog = one hot dog + one bun. Mr. Kobayashi didn’t do his math correctly. He bought 5 packs of hot dogs (10 per package) and 5 packs of hot dog buns (8 per package). How many hot dogs (according to the official formula) could he have eaten?

Hot Dogs in the News Source: CNN.com WHAT IF… One hot dog = one hot dog + one bun. Mr. Kobayashi didn’t do his math correctly. He bought 5 packs of hot dogs (10 per package) and 5 packs of hot dog buns (8 per package). How many hot dogs (according to the official formula) could he have eaten? 5 hot dog pack10 hot dogs 1 hot dog pack 50 hot dogs = 5 bun pack8 buns 1 bun pack 40 buns = 40 possible hot dogs

Limiting Reactants Limiting ReactantLimiting Reactant –used up in a reaction –determines the amount of product Excess ReactantExcess Reactant –added to ensure that the other reactant is completely used up –cheaper & easier to recycle

Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: –limiting reactant –amount of product

Limiting Reactants in Chemistry 5.0 moles of chlorine gas react with 5.0 moles of sodium to produce sodium chloride. Which reagent is the limiting factor? How much of the excess reactant is left over? Cl 2 (g)+NaNaCl mol Cl 2 1 mol Cl 2 2 mol NaCl =10. mol NaCl 5.0 mol Na 2 mol NaCl 2 mol Na =5.0 mol NaCl EXCESS LIMITING 5.0 mol Na 2 mol Na 1 mol Cl 2 = 2.5 mol Cl mol Cl 2 given 2.5 mol Cl 2 used 2.5 mol Cl 2 left

Practice How many grams of copper are produced if 2.5 moles of copper sulfate and 5 moles of aluminum react? Identify the limiting and excess reactants. 3 CuSO 4 +2 AlAl 2 (SO 4 ) 3 3 Cu+ 2.5 mol CuSO 4 3 mol Cu 3 mol CuSO g Cu 1 mol Cu = g Cu 5 mol Al 3 mol Cu 2 mol Al 63.55g Cu 1 mol Cu = g Cu Excess = Al Limiting = CuSO 4

Percent Yield calculated on paper measured in lab

Johnny took a quiz yesterday. He missed 4 questions and earned 63 points out of 70. -Was he perfect? -What was his possible score? -What was his actual percent score? Get Real! (63/70)X100 =

Percent Yield When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g? g actual: 46.3 g

Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g ? g actual: 46.3 g Theoretical Yield:

Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g49.4 g actual: 46.3 g