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Chapter 9 - Section 3 Suggested Reading: Pages 312 - 318 Stoichiometry.

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Presentation on theme: "Chapter 9 - Section 3 Suggested Reading: Pages 312 - 318 Stoichiometry."— Presentation transcript:

1 Chapter 9 - Section 3 Suggested Reading: Pages Stoichiometry

2 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

3 Limiting Reactants Available Ingredients Copper Wire 0.5 g AgNO 3 Limiting Reactant 0.5 grams AgNO 3 Excess Reactants Copper Wire

4 Limiting Reactant The reactant that limits the amount of product that can be formed.

5 When quantities of reactants are available in the exact ratio described by the balanced equation, they are said to be in stoichiometric proportions.

6 Limiting Reactants Limiting Reactant used up in a reaction determines the amount of all products formed Excess Reactant added to ensure that the other reactant is completely used up usually cheaper & easier to recycle

7 Solving Problems – Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. The reactant that produces the smaller amount of product is the limiting reactant. Very similar to mass-mass problems!

8 Step 1: Write a balanced equation. Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water. O 2 + 2H 2  2 H 2 O

9 Step 2: Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water. O 2 + 2H 2  2 H 2 O For each reactant, calculate the amount of product formed.

10 Step 2: 1.22 g oxygen O 2 + 2H 2  2 H 2 O wanted given wanted = X H 2 O mol 2 1 given X = mol H 2 O 32 g 1 mole = mol O 2

11 Step 2: 1.05 g H 2 O 2 + 2H 2  2 H 2 O wanted given wanted = X H 2 O mol 2 2 given X= mol H 2 O 2 g 1 mole = mol H 2

12 Step 3: 1.22 g of O 2 would produce mol H 2 O The one that produces the smallest amount is your limiting reactant g of H 2 would produce mol H 2 O Oxygen is your limiting reactant!

13 Limiting Reactants Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.

14 Step 1: Write a balanced equation. Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride. 2Na + Cl 2  2NaCl

15 Step 2: Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride. 2Na + Cl 2  2NaCl For each reactant, calculate the amount of product formed.

16 Step 2: 1.7 g Na 2Na + Cl 2  2NaCl wanted given wanted = X NaCl mol 2 2 given X= mol NaCl 23 g 1 mole = mol Na

17 Step 2: 2.6 L Cl 2 2Na + Cl 2  2NaCl wanted given wanted = X NaCl mol 2 1 given X= mol NaCl 22.4 L 1 mole = mol Cl 2

18 *Step 3: 1.7 g Na would produce mol NaCl The one that produces the smallest amount is your limiting reactant. 2.6 L Cl 2 would produce mol NaCl Sodium is your limiting reactant!

19 Percent Yield calculated on paper measured in lab

20 Percent Yield When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical yield and % yield of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g? g actual: 46.3 g

21 Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO g K 2 CO 3 = 49.1 g KCl 2 mol KCl 1 mol K 2 CO 3 74 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g? g actual: 46.3 g Theoretical Yield:

22 Percent Yield Theoretical Yield = 49.1 g KCl % Yield = 46.3 g 49.1 g  100 = 94.3% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g49.1 g actual: 46.3 g


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