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Chapter 11 Section 3

Chapter 11 Section 3. Limiting Reagent & Percent Yield. Limiting Reagent (= Reactant). What does it mean? Consider making cars. If you had 100 car bodies and 300 tires, how many cars can you make given the equation? 1 car body + 4 tires → 1 car Animation of Limiting Reagent

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Chapter 11 Section 3

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  1. Chapter 11 Section 3 Limiting Reagent & Percent Yield

  2. Limiting Reagent (= Reactant) • What does it mean? • Consider making cars. If you had 100 car bodies and 300 tires, how many cars can you make given the equation? • 1 car body + 4 tires → 1 car • Animation of Limiting Reagent • Khan Academy- Limiting Reagent

  3. Limiting Reagent (=Reactant) 2H2(g) + O2(g)  2H2O(g) Which reactant is used up in Case II? Which reactant is left over in Case II? How much left over? How many mol of the product are produced?

  4. Why LR Important? • Because the amounts of products produced depend on the amount of the LR • Once the LR is determined, use the amount of the LR for all stoichiometry problems

  5. An easy way to determine the LR • If given in grams, convert the mass of each reactant to mol • Divide the mol by its coefficient • The smallest quotient is the LR (Ex) 2H2(g) + O2(g)  2H2O(g) Given 8.0 g of H2 and 50. g O2, which is the LR? 1) • 3.96 mol H2 ÷ 2 = 1.98 mol H2 1.56 mol O2 ÷ 1 = 1.56 mol O2(LR) **Must use the given amounts for the calculations

  6. Example 2 Cu(s) + S(s) → Cu2S(s) What is the maximum number of grams of Cu2S(s) that can be formed when 80.0 g of Cu react with 25.0 g S? How much of S is left over? (Sol) 1.26 mol Cu ÷ 2 = 0.63 mol Cu 0.78 mol S ÷ 1 = 0.78 mol S Cu is the LR

  7. Percent Yield • Theoretical yield: • the maximum amount of a product produced based on the assumption that 100% of the limiting reactant reacts and according to the balanced equation • All the stoichiometry problems we did so far • Actual yield: • the amount of a product actually obtained • always less than the theoretical amount • Percent yield • Always less than 100%

  8. Example Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is: CaCO3(s) → CaO(s) + CO2(g) What is the theoretical yield of CaO if 28.4 g of CaCO3 is heated? If 12.7 g of CaO is obtained, what is the percent yield?

  9. Problem Solving Strategy • Read the question thoroughly • Need to determine LR or not? • If the amount of only one reactant is given, that’s the LR • Once the LR is determined, use it for all setups. • Follow the “flow-chart” to solve for the theoretical amount. • The actual amount is always less than the theoretical amount. • The actual amount is usually given as the amount “produced” or “obtained” • The % yield is always less than 100%. • To get the excess amount, subtract the “used” amount from the “starting” amount.

  10. Example Solid calcium carbonate, CaCO3, is able to remove sulphur dioxide from waste gases by the reaction: CaCO3  +  SO2  + other reactants ------>  CaSO3 + other products In a particular experiment, 255 g of CaCO3 was exposed to 135 g of SO2 in the presence of an excess amount of the other chemicals required for the reaction. a)  What is the theoretical yield of CaSO3? b)  If only 198 g of CaSO3 was isolated from the products, what was the percentage yield of CaSO3 in this experiment?

  11. Solution • Determine the LR. • molar mass of CaCO3 = 100.09 g/mol; CaSO3 = 120.2 g/mol; SO2 = 64.07 g/mol • Determine the theoretical yield of CaSO3. • Determine the % yield of CaSO3 if 198 g CaSO3 are produced.

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