Percent Yield and Limiting Reactants

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Presentation transcript:

Percent Yield and Limiting Reactants Unit 9

Percent Yield The percent yield is the percentage of a certain product actually produced in a chemical reaction. The theoretical yield is predicted by a stoichiometry problem.

A. Percent Yield measured in lab calculated on paper

A. 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

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

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

Percent Yield The reaction between SO2 and oxygen yields SO3.. Calculate the percent yield of SO3 if 40.0 grams of SO3is formed, when 32 grams of SO2 react with an excess of oxygen. SO2 + O2  SO3 2SO2 + O2  2SO3

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

B. Limiting Reactants Limiting Reactant Excess 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

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

Limiting Reactant Steps (version 2) Step one Write and balance the equation for the reaction. Step two Convert known masses to grams of product. Step three Determine limiting reactant and amount that can be made. Step four Determine the grams of excess from limiting reactant.

B. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 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  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 22.4 L H2 1 mol = 27.1 L H2

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 2.5 mol HCl 1 L 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2

A. Limiting Reactants Limiting reactant: HCl Excess reactant: Zn Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2 left over zinc

Limiting Reagent or Reactant 2 (problem #2) How many grams of ammonia will be produced when 20.0 grams of potassium hydroxide react with 15.0 grams of ammonium sulfate? Step 1 2KOH + (NH4)2SO4  K2SO4 +2NH3 +2H2O

(NH4)2SO4 is the limiting reagent because it will make the least amount of product. It is the reactant that will run out first.

Limiting Reagent or Reactant 2