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Catalyst: 1 Double Cheeseburger needs 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon

Please TURN IN C.7/C.10 worksheet Objectives: Determine the limiting reagent in a reaction. Calculate the percent yield of a product in a reaction. Catalyst: 1 Double Cheeseburger needs 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon

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Catalyst: 1 Double Cheeseburger needs 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon

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  1. Please TURN IN C.7/C.10 worksheetObjectives: Determine the limiting reagent in a reaction.Calculate the percent yield of a product in a reaction Catalyst: 1 Double Cheeseburger needs 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon • For 5 Double Cheeseburgers how many units of each ingredient do I need? • Fill in the final column below with how many complete burgers I can make:

  2. A. The Concept of Limiting Reactants • Stoichiometric mixture • N2(g) + 3H2(g) 2NH3(g)

  3. A. The Concept of Limiting Reactants • Limiting reactant mixture • N2(g) + 3H2(g) 2NH3(g)

  4. A. The Concept of Limiting Reactants • For a Limiting reactant mixture the number of moles are not balanced to match the reaction equation N2(g) + 3H2(g) 2NH3(g) • Limiting reactant is the reactant that runs out first • When the limiting reactant is exhausted, then the reaction stops

  5. LIMITING REACTANT a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4Na (s)+ O2(g)2 Na2O(s) Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 5.00g Na(1 mole Na)(2 mole Na2O)( 62 g Na2O) = 6.74 g of Na2O 23 g Na 4 mole Na 1 mol Na2O 5.00g O2(1 mole O2)(2 mole Na2O)( 62 g Na2O) = 19.38 g of Na2O 32 g O2 1 mole O2 1 mol Na2O Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide. Wrong answer

  6. LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4Na (s)+ O2(g)2 Na2O(s) The amount of O2 used to make 6.74 g of Na2O is calculated by: 5.00g Na(1 mole Na)(1 mole O2)( 32 g O2) = 1.74 g of O2 was used 23 g Na 4 mole Na 1 mol O2 The amount of oxygen (O2) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

  7. PRACTICE PROBLEMS 2 C2H2 + 5 O2  4 CO2 + 2 H2O 1. How many moles of carbon dioxide could be produced from 220.0 g of C2H2 and 545.0 g of O2? How many grams of CO2 can be produced by the reaction of 35.5 grams of C2H2 and 45.9 grams of O2? 13.63 mol 50.5 g

  8. C. Percent Yield • Theoretical Yield • The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. • The actual yield (amount produced) of a reaction is usually less than the maximum expected (theoretical yield). • Percent Yield • The actual amount of a given product as the percentage of the theoretical yield.

  9. Calcium carbonate is synthesized by heating,as shown in the following equation: CaO + CO2 CaCO3 • What is the theoretical yield of CaCO3 if 24.8 g of CaO is heated with 43.0 g of CO2? • What is the percent yield if 33.1 g of CaCO3 is produced? Determine which reactant is the limiting and then decide what the theoretical yield is.

  10. 24.8gCaO • molCaO • mol CaCO3 • gCaCO3 • 24.8gCaO • molCaO • mol CO2 • gCO2 24.8 g CaO 1molCaO 1mol CO2 44 g CO2 56g CaO 1mol CaO 1molCO2 LR = 19.5gCO2 1mol CaO 100g CaCO3 24.8 g CaO 1molCaCO3 56g CaO 1mol CaO 1molCaCO3 = 44.3 g CaCO3

  11. _____________ • CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO3. We only produced 33.1 g. • Our percent yield is: 33.1 g CaCO3 Percent yield= x 100 44.3 g CaCO3 Percent yield = 74.7%

  12. Percent Yield Questions • The electrolysis of water forms H2 and O2. 2H2O  2H2 + O2 What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? 2. Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2 2Fe2O3 + 8SO2 If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide?

  13. actual 1 mol H2O 1 mol O2 12.3 g O2 x = x theoretical 18.02 g H2O 2 mol H2O 12.43 g O2 32 g O2 x 1 mol O2 = = 12.43 g 98.9% • The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? • Actual yield is given: 12.3 g O2 • Next, calculate theoretical yield # g O2= 14.0 g H2O Finally, calculate % yield % yield = x 100% x 100%

  14. actual 143 g Fe2O3 = theoretical 181.48 g Fe2O3 = = = 181.48 g Fe2O3 78.8% 199.7 g Fe2O3 1 mol O2 1 mol FeS2 2 mol Fe2O3 2 mol Fe2O3 159.7 g Fe2O3 159.7 g Fe2O3 x x x x x x 32 g O2 119.97 g FeS2 4 mol FeS2 11 mol O2 1 mol Fe2O3 1 mol Fe2O3 2. 4FeS2+ 11O2 2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3? First, determine limiting reagent # g Fe2O3= 300 g FeS2 200 g O2 % yield = x 100% x 100%

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