2. 1 1)Write correct chemical formulas, know types of reactions and write a balanced equation!!! 2)a) Convert g, L or # or particles to mole and b) use mole ratios to switch to new substance 3)Determine the limiting and excess reagent 4)Determine theoretical yield 5)From actual yield determine % yield 6)Determine amount of reagents remaining Chapter 11 Stoichiometry skills required

3. 2 Volume of gas (STP) 1.00 mol 22.4L 1.00 mol molar mass Particles 1.00 mol 6.02 x10 23 p. Mass Mole SubstanceA Substance A Volume of gas (STP) 1.00 mol 22.4L 1.00 mol molar mass Particles 1.00 mol 6.02 x10 23 p. Mass Mole Substance B Mole ratio X moleA X mole A Y mole B

4. 3 Limiting Reagent Any reactant that is used up first in a chemical reaction; it determines the amount of product that can be formed in the reaction. The reactant(s) not completely used up is/are the Excess Reagent(s) Two possible method to determine the limiting reagent: 1) calculate each amount separately. The result with less product is the most you can produce from the limiting reagent. Or 2) convert each reactant to mole, then compare mole ratios to determine which is the limiting reagent.

5. 4 Theoretical Yield Theoretical Yield: Maximum amount of product that will form from given amount of reactants Actual Yield Actual Yield: Amount of product actually formed when the reaction is carried out. Percent Yield Percent Yield: The ratio of the actual yield to the theoretical yield expressed as a percent Percent Yield = Actual Yield x 100 % Theoretical Yield Theoretical Yield Percent Yield

6. 5 Problem: 5.000g of Al was used to recover silver from an aqueous solution containing 65.35g of silver nitrate a)Write a balanced equation b)If any, which is the limiting reagent and the excess reagent? c)What is the theoretical yield of elemental silver that can be recovered? d)What is the % yield if 39.26g of Ag was recovered. e)How much Al, aqueous silver nitrate and silver ions remained at the end?

7. 6 5.000g of Al was used to recover silver from an aqueous solution containing 65.35g of silver nitrate a)Al + 3 AgNO 3  3 Ag + Al(NO 3 ) 3 b)5.000g Al  59.97 g Agmore 65.35g AgNO 3  42.44g Ag less = limiting AgNO 3 is the limiting, Al the excess reactant the theoretical yield is 42.44g Ag c) % yield is 94.60% d) remaining: 1.727g Al, 5.4% aqueous silver nitrate = 3.529g AgNO 3 and 3.529x107.89/169.88 = 2.24g silver ions Ag+.