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Stoichiometry in the Real World Stoichiometry – Ch. 11.

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Presentation on theme: "Stoichiometry in the Real World Stoichiometry – Ch. 11."— Presentation transcript:

1 Stoichiometry in the Real World Stoichiometry – Ch. 11

2 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

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

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

5 Limiting Reactant Steps 1. Write the balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product actually possible

6 Limiting Reactants b 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? Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 68.1 g

7 Limiting Reactants 79.1 g Zn 1 mol Zn 65.39 g Zn = 2.44 g H 2 1 mol H 2 1 mol Zn 2.02 g H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 68.1 g

8 Limiting Reactants 2.02 g H 2 1 mol H 2 68.1 g HCl 1 mol HCl 36.46 g HCl = 1.89 g H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 68.1 g

9 Limiting Reactants Zn: 2.44 g H 2 HCl: 1.89 g H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 1.89 g H 2 left over zinc

10 Limiting Reactants #2 Limiting Reactants #2 b 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? 2Mg + O 2  2MgO 5.42 g ? g 4.00 g

11 Limiting Reactants #2 5.42 g Mg 1 mol Mg 24.31 g Mg = 8.99 g MgO 2 mol MgO 2 mol Mg 40.31 g MgO 1 mol MgO 2Mg + O 2  2MgO 5.42 g ? g 4.00 g

12 Limiting Reactants #2 40.31 g MgO 1 mol MgO 4.00 g O 2 1 mol O 2 32.00 g O 2 = 10.1 g MgO 2 mol MgO 1 mol O 2 2Mg + O 2  2MgO 5.42 g ? g 4.00 g

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

14 Limiting Reactants b 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

15 Percent Yield calculated on paper measured in lab

16 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. 1 mol K 2 CO 3 138.2 g K 2 CO 3 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl 45.8 g K 2 CO 3 K 2 CO 3 + 2 HCl  2 KCl + H 2 CO 3 = 49.4 grams KCl

17 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. 46.3 grams KCl 49.4 grams KCl x 100 % yield = 93.7 %


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