II. Stoichiometry in the Real World (p. 368 – 375) Stoichiometry – Ch. 12 II. Stoichiometry in the Real World (p. 368 – 375)

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

A. Limiting Reactants Available Ingredients 24 graham cracker squares 1 bag of marshmallows 12 pieces of chocolate Limiting Reactant chocolate Excess Reactants Marshmallows and graham crackers

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

A. Limiting Reactants Write the balanced equation. Identify and label known and unknown. For each reactant, calculate the amount of product formed. Smaller answer indicates: limiting reactant maximum amount of product actually possible

A. Limiting Reactants ? LR ? ER Zn + 2HCl  ZnCl2 + H2 79.1 g ? g 79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed? ? LR ? ER Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 2.02 g H2 1 mol = 2.44 g H2

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 68.1 g HCl 1 mol HCl 36.46 g HCl 1 mol H2 2 mol HCl 2.02 g H2 1 mol H2 = 1.89 g H2

A. Limiting Reactants Zn: 2.44 g H2 HCl: 1.89 g H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 1.89 g H2 left over zinc

A. Limiting Reactants #2 ? LR ? ER 2Mg + O2  2MgO 5.42 g 4.00 g ? g 5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed? ? LR ? ER 2Mg + O2  2MgO 5.42 g 4.00 g ? g

A. Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 5.42 g Mg 1 mol Mg 24.31 g Mg 2 mol MgO Mg 40.31 g MgO 1 mol = 8.99 g MgO

A. Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 4.00 g O2 1 mol O2 32.00 g O2 2 mol MgO 1 mol O2 40.31 g MgO 1 mol MgO = 10.1 g MgO

A. Limiting Reactants #2 Limiting reactant: Mg Excess reactant: O2 Mg: 8.99 g MgO O2: 10.1 g MgO Limiting reactant: Mg Excess reactant: O2 Product Formed: 8.99 g MgO Excess oxygen

A. Limiting Reactants What other information could you find in these problems? How much of each reactant is used – in grams, liters, moles How much of excess reactant is left over – in grams, liters, moles

A. Limiting Reactants – Shortcut 5.00 mol of magnesium ribbon react with 4.00 mol of oxygen gas. Identify the limiting and excess reactants. How many moles of excess reactant are left over? 2Mg + O2  2MgO 5.00 mol 4.00 mol ? mol ER left over

A. Limiting Reactants – Shortcut LR ER 2Mg + O2 divide divide M 2 1 A U L Mole Ratio multiply multiply 5.00 4.00 Subtract Subtract Available bring down bring down 8.00 2.50 Answer Answer Used 1.50 Left Over 1.50 mol of O2 left over

B. Percent Yield Percent of theoretical yield that is actually produced in the lab Theoretical yield = amount of product calculated on paper using stoichiometry Actual yield = amount of product actually formed in the lab

B. Percent Yield measured in lab calculated on paper

B. 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

B. 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

B. 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%