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Warm-Up

When copper (II) reacts with silver nitrate, the number of grams of copper required to produce 432 grams of silver is:. 2. 2. Warm-Up. Cu. +. AgNO 3. . Ag. +. Cu(NO 3 ) 2. 432 g Ag. 1 mol Ag. 1 mol Cu. 63.55 g Cu. 107.87 g Ag. 2 mol Ag. 1 mol Cu. = 127.25 g Cu.

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Warm-Up

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  1. When copper (II) reacts with silver nitrate, the number of grams of copper required to produce 432 grams of silver is: 2 2 Warm-Up Cu + AgNO3  Ag + Cu(NO3)2 432 g Ag 1 mol Ag 1 mol Cu 63.55 g Cu 107.87 g Ag 2 mol Ag 1 mol Cu = 127.25 g Cu

  2. Limiting Factors and Percent Yield

  3. Hot Dogs in the News Takeru Kobayashi of Japan downed 44½ hot dogs in 12 minutes. One hot dog = one hot dog + one bun. WHAT IF… 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? Source: CNN.com

  4. Hot Dogs in the News One hot dog = one hot dog + one bun. WHAT IF… 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? Excess 5 hot dog pack 10 hot dogs 50 hot dogs = 1 hot dog pack Limiting Factor 40 possible hot dogs 5 bun pack 8 buns 40 buns = 1 bun pack Source: CNN.com

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

  6. 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

  7. 2 2 LIMITING 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? Cl2 (g) + Na NaCl EXCESS 5.0 mol Cl2 2 mol NaCl = 10. mol NaCl 1 mol Cl2 2 mol NaCl 5.0 mol Na = 5.0 mol NaCl 5.0 mol Cl2 given 2 mol Na 2.5 mol Cl2 used 5.0 mol Na 1 mol Cl2 2.5 mol Cl2 = 2.5 mol Cl2 left 2 mol Na

  8. Practice 3 CuSO4 + 2 Al Al2(SO4)3 + 3 Cu 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. 2.5 mol CuSO4 3 mol Cu 63.55g Cu Limiting = CuSO4 = 158.87g Cu 3 mol CuSO4 1 mol Cu Excess = Al 5 mol Al 3 mol Cu 63.55g Cu = 476.63g Cu 1 mol Cu 2 mol Al

  9. Percent Yield measured in lab calculated on paper

  10. Get Real! Johnny took a quiz yesterday. He missed 4 questions and earned 63 points out of 70. (63/70)X100= • Was he perfect? • What was his possible score? • What was his actual percent score? 70 points = 100 % 90 %

  11. Percent Yield When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  12. Percent Yield Theoretical Yield: K2CO3 + 2HCl  2KCl + H2O + CO2 ? g 45.8 g actual: 46.3 g 45.8 g K2CO3 1 mol K2CO3 138.21 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

  13. Percent Yield Theoretical Yield = 49.4 g KCl 46.3 g 49.4 g K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g actual: 46.3 g  100 = 93.7% % Yield =

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