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Chapter 9 - Section 3 Suggested Reading: Pages 312 - 318
Stoichiometry Chapter 9 - Section 3 Suggested Reading: Pages
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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
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Limiting Reactants Available Ingredients Copper Wire 0.5 g AgNO3
0.5 grams AgNO3 Excess Reactants Copper Wire
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Limiting Reactant The reactant that limits the amount of product that can be formed.
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When quantities of reactants are available in the exact ratio described by the balanced equation, they are said to be in stoichiometric proportions.
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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
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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!
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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 H2 H2O
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Step 2: For each reactant, calculate the amount of product formed. Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water. O H2 H2O
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= O2 + 2H2 2 H2O wanted 2 X H2O = 1 given 0.038 mol X =
Step 2: 1.22 g oxygen 1 mole = 0.038 mol O2 32 g O H2 H2O given wanted wanted 2 X H2O = 1 given 0.038 mol X = 0.076 mol H2O
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= O2 + 2H2 2 H2O wanted 2 X H2O = 2 given 0.525 mol X =
Step 2: 1.05 g H2 1 mole = 0.525 mol H2 2 g O H2 H2O given wanted wanted 2 X H2O = 2 given 0.525 mol X = 0.525 mol H2O
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The one that produces the smallest amount is your limiting reactant.
Step 3: The one that produces the smallest amount is your limiting reactant. 1.22 g of O2 would produce mol H2O Oxygen is your limiting reactant! 1.05 g of H2 would produce mol H2O
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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.
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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 Cl2 2NaCl
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Step 2: For each reactant, calculate the amount of product formed. 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 Cl2 2NaCl
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= 2Na + Cl2 2NaCl wanted 2 X NaCl = 2 given 0.0739 mol X =
Step 2: 1.7 g Na 1 mole = mol Na 23 g 2Na Cl2 2NaCl wanted given wanted 2 X NaCl = 2 given mol X = mol NaCl
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= 2Na + Cl2 2NaCl wanted 2 X NaCl = 1 given 0.116 mol X =
Step 2: 2.6 L Cl2 1 mole = 0.116 mol Cl2 22.4 L 2Na Cl2 2NaCl wanted given wanted 2 X NaCl = 1 given 0.116 mol X = 0.232 mol NaCl
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The one that produces the smallest amount is your limiting reactant.
*Step 3: The one that produces the smallest amount is your limiting reactant. 1.7 g Na would produce mol NaCl Sodium is your limiting reactant! 2.6 L Cl2 would produce mol NaCl
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Percent Yield measured in lab calculated on paper
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Percent Yield K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical yield and % yield of KCl. K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g
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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 g 2 mol KCl 1 mol K2CO3 74 g KCl 1 mol KCl = 49.1 g KCl
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Percent Yield K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.1 g
actual: 46.3 g Theoretical Yield = 49.1 g KCl 46.3 g 49.1 g % Yield = 100 = 94.3%
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