Yields, Limiting Reactants, and Solutions Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 1/1

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/2 Yields of Chemical Reactions The amount actually formed in a reaction. The amount predicted by calculations. Actual Yield: Theoretical Yield: actual yield theoretical yield x 100%Percent Yield =

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/10 Reactions with Limiting Amounts of Reactants Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant. Excess Reactant: Any of the other reactants still present after determination of the limiting reactant.

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/11 Reactions with Limiting Amounts of Reactants Because water is so cheap and abundant, it is used in excess when compared to ethylene oxide. This ensures that all of the relatively expensive ethylene oxide is entirely consumed. At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l)

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/12 Reactions with Limiting Amounts of Reactants C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l) If 3 moles of ethylene oxide react with 5 moles of water, which reactant is limiting and which reactant is present in excess? At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers:

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/13 Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l)

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/14 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: If 80.0 g of water are to be removed and 65.0 g of Li 2 O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced?

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/15 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) Which reactant is limiting? 65.0 g Li 2 O 1 mol Li 2 O 1 mol H 2 O 29.9 g Li 2 O 1 mol Li 2 O = 4.44 moles H 2 O 80.0 g H 2 O 18.0 g H 2 O 1 mol H 2 O = 2.17 moles H 2 O Amount of H 2 O given: Amount of H 2 O that will react with 65.0 g Li 2 O: Li 2 O is limiting x x x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/16 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) 80.0 g H 2 O g H 2 O = 40.9 g H 2 O 2.17 mol H 2 O 1 mol H 2 O 18.0 g H 2 O = 39.1 g H 2 O (consumed) How many grams of excess H 2 O remain? remaininginitialconsumed x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/17 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) 2.17 mol H 2 O 1 mol LiOH 23.9 g LiOH = 104 g LiOH How many grams of LiOH are produced? 1 mol H 2 O 2 mol LiOH xx

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/30 Concentrations of Reactants in Solution: Molarity Molarity: The number of moles of a substance dissolved in each liter of solution. In practice, a solution of known molarity is prepared by weighing an appropriate amount of solute, placing it in a container called a volumetric flask, and adding enough solvent until an accurately calibrated final volume is reached. Solution: A homogeneous mixture. Solute: The dissolved substance in a solution. Solvent: The major component in a solution.

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/34 Concentrations of Reactants in Solution: Molarity Molarity converts between mole of solute and liters of solution: molarity = moles of solute liters of solution L mol or 1.00 M 1.00 L 1.00 mol = mol of sodium chloride placed in enough water to make 1.00 L of solution would have a concentration equal to:

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/35 Concentrations of Reactants in Solution: Molarity Molar mass C 6 H 12 O 6 = g/mol How many grams of solute would you use to prepare 1.50 L of M glucose, C 6 H 12 O 6 ? 1 mol mol180.0 g = 49.5 g 1 L 1.50 L0.250 mol = mol x x

H11 – C3 3.10, 3.11, 3.13, 3.17, 3.73a*, 3.75, 3.77*, 3.80b, 3.81b, 3.85*, 3.121*