II. Limiting Reactants Stoichiometry – 3.7

A. Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

A. Limiting Reactants b In a laboratory, usually one or more of the reactants are present in excess. There is more than the exact amount required to react b Once one of the reactants is used up, no more product can form

A. Limiting Reactants b Limiting Reactant used up in a reaction Limits the amount of reactant that can combine and determines amount of product determines the amount of product that can form b Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

B. 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

B. Limiting Reactants b 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

B. Limiting Reactants 79.1 g Zn 1 mol Zn g Zn = 27.1 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

B. Limiting Reactants 22.4 L H 2 1 mol H L 2.5 mol HCl 1 L = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M

B. Limiting Reactants Zn: 27.1 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc

LIMITING REACTANT b Try Example Problem #2 Method 1: Convert both reactants to product. See which is less. Method 2: Convert one reactant to another. See how much is needed.

LIMITING REACTANT b Problem #2: HF: limiting 4.0 mol excess SiO 2

C. Percent Yield 1. actual yield: measured amount of product obtained from a reaction; measured in actual lab; less than theoretical yield due to experimental errors 2. theoretical yield: maximum amt. of product that could ideally be obtained from a given amount of reactant

C. Percent Yield calculated w/stoich. measured in lab

C. Percent Yield b 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

C. 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:

C. 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