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C. Johannesson I. I.Stoichiometric Calculations (p. 275-287) Stoichiometry – Ch. 9.

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Presentation on theme: "C. Johannesson I. I.Stoichiometric Calculations (p. 275-287) Stoichiometry – Ch. 9."— Presentation transcript:

1 C. Johannesson I. I.Stoichiometric Calculations (p ) Stoichiometry – Ch. 9

2 C. Johannesson

3 A. Proportional Relationships b I have 5 eggs and unlimited quantity of everything else. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

4 C. Johannesson A. Proportional Relationships b Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

5 C. Johannesson B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass -moles  grams Molarity - moles  liters soln Molar volume -moles  liters gas Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

6 C. Johannesson 1 mol of a gas=22.4 L at STP C. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

7 C. Molar Volume at STP Molar Mass (g/mol) 6.02  particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molar Volume (22.4 L/mol) LITERS OF GAS AT STP Molarity (mol/L)

8 C. Johannesson D. Stoichiometry Problems b How many moles of KClO 3 must decompose (into KCl and O 2 ) in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

9 C. Johannesson b How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O g KClO 3 1 mol KClO 3 ? g9.00 L D. Stoichiometry Problems 2KClO 3  2KCl + 3O 2

10 C. Johannesson D. Stoichiometry Problems b How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu g Ag 1 mol Ag 12.0 g? g

11 C. Johannesson g Cu 1 mol Cu D. Stoichiometry Problems b How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M

12 C. Johannesson II. Stoichiometry in the Real World (p ) Stoichiometry – Ch. 9

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

14 C. Johannesson A. Limiting Reactants b Limiting Reactant used up in a reaction determines the amount of product b Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

15 C. Johannesson A. 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

16 C. Johannesson A. Limiting Reactants b When 36.0 g H 2 O is mixed with 67.0 g Fe, which is the limiting reactant? 3Fe (s) + 4H 2 O (g)  Fe 3 O 4 + 4H 2(g)

17 C. Johannesson B. Percent Yield calculated on paper measured in lab

18 C. Johannesson B. 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

19 C. Johannesson B. 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:

20 C. Johannesson B. 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


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