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Limiting Reagents and Percent Yield

Limiting Reagents and Percent Yield. What Is a Limiting Reagent?. Many cooks follow a recipe when making a new dish. When a cook prepares to cook he/she needs to know that sufficient amounts of all the ingredients are available.

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Limiting Reagents and Percent Yield

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  1. Limiting Reagents and Percent Yield

  2. What Is a Limiting Reagent? • Many cooks follow a recipe when making a new dish. • When a cook prepares to cook he/she needs to know that sufficient amounts of all the ingredients are available. • Let’s look at a recipe for the formation of a double cheeseburger:

  3. 1 hamburger bun 1 tomato slice 1 lettuce leaf 2 slices of cheese 2 burger patties

  4. How many hamburger buns do you need? 5 • If you want to make 5 double cheese burgers: • How many hamburger patties do you need? 10 • How many slices of cheese do you need? 10 • How many slices of tomato do you need? 5

  5. 1 bun, 2 patties, 2 slices of cheese, 1 tomato slice 1 • 2 buns, 4 patties, 4 slices of cheese, 2 tomato slices • How many double cheeseburgers can you make if you start with: 2 • 1 mole of buns, 2 moles of patties, 2 moles of cheese, 1 mole of tomato slices 1 mole • 10 buns, 20 patties, 2 slices of cheese, 10 tomato slices 1

  6. We can’t make anymore than 1 double cheeseburger with our ingredients. • The slices of cheese limits the number of cheeseburgers we can make. • If one of our ingredients gets used up during our preparation it is called the limiting reactant (LR) • The LR limits the amount of product we can form; in this case double cheeseburgers. • It is equally impossible for a chemist to make a certain amount of a desired compound if there isn’t enough of one of the reactants.

  7. As we’ve been learning, a balanced chemical rxn is a chemist’s recipe. • Which allows the chemist to predict the amount of product formed from the amounts of ingredients available • Let’s look at the reaction equation for the formation of ammonia: N2(g) + 3H2(g)  2NH3(g) • When 1 mole of N2 reacts with 3 moles of H2, 2 moles of NH3 are produced. • How much NH3 could be made if 2 moles of N2 were reacted with 3 moles of H2? 2 mols of ammonia

  8. N2(g) + 3H2(g)  2NH3(g) • The amount of H2 limits the amount of NH3 that can be made. • From the amount of N2 available we can make 4 moles of NH3 • From the amount of H2 available we can only make 2 moles of NH3. • H2 is our limiting reactant here. • It runs out before the N2 is used up. • Therefore, at the end of the reaction there should be N2 left over. • When there is reactant left over it is said to be in excess.

  9. How much N2 will be left over after the reaction? • In our rxn it takes 1 mol of N2 to react all of 3 mols of H2, so there must be 1 mol of N2 that remains unreacted. • We can use our new stoich calculation skills to determine 3 possible types of LR type calculations. • Determine which of the reactants will run out first (limiting reactant) • Determine amount of product • Determine how much is in excess (is wasted)

  10. Limiting Reactant Problems: Given the following reaction: 2Cu + S  Cu2S • What is the limiting reactant when 82.0 g of Cu reacts with 25.0 g S? • What is the maximum amount of Cu2S that can be formed? • How much of the other reactant is in excess?

  11. Our 1st goal is to calculate how much S would react if all of the Cu was reacted. • From that we can determine the limiting reactant (LR). • Then we can use the Limiting Reactant to calculate the amount of product formed and the amount of excess reactant left over. • 82g Cu • mol Cu • mol S • g S Mole/Mole Ratio Molar Mass Molar Mass

  12. 2Cu + S  Cu2S 1mol S 1molCu 32.1g S 82.0gCu • So if all of our 82.0g of Copper were reacted completely it would require only 20.7 grams of Sulfur. • Since we initially had 25g of S, we are going to run out of the Cu, the limiting reactant) & end up with 4.3 grams of S 2molCu 1mol S 63.5gCu =20.7 g S

  13. ________ • Copper being our Limiting Reactant is then used to determine how much product is produced. • The amount of Copper we initially start with limits the amount of product we can make. 1molCu2S 1molCu 159gCu2S 82.0gCu 2molCu2S 1molCu2S 63.5gCu = 103 g Cu2S

  14. So the reaction between 82.0g of Cu and 25.0g of S can only produce 103g of Cu2S. • The Cu runs out before the S and we will end up with 4.7 g of the S in excess. Ex 2: Hydrogen gas can be produced in the lab by the rxn of Magnesium metal with HCl according to the following rxn equation: Mg + 2HCl  MgCl2 + H2 • What is the LR when 6.0 g HCl reacts with 5.0 g Mg? What is the maximum amount of H2 that can be formed? And how much of the other reactant is in excess?

  15. 5.0g Mg  • mol Mg  • 2mol HCl • g HCl 36.5gHCl 2molHCl 1molMg 5.0g Mg 1molHCl 1molMg 24.3gMg = 15.0g HCl • So if 5.0g of Mg were used up it would take 15.0g HCl, but we only had 6.0g of HCl to begin with. • Therefore, the 6.0g of HCl will run out before the 5.0g of Mg, so HCl is our Limiting Reactant.

  16. 6.0g HCl • 2mol HCl  • 1mol H2  • g H2 2.0gH2 1molH2 1molHCl 6.0g HCl 1molH2 2molHCl 36.5gHCl = 0.164 g H2 produced • 6.0g HCl • 2mol HCl  • 1mol Mg • g Mg 24.3gMg 1molMg 1molHCl 6.0g HCl 1molMg 2molHCl 36.5gHCl =1.997 g Mg - 5.0 g Mg = 3.01g Mg in excess

  17. Calculating Percent Yield • In theory, when a teacher gives an exam to the class, every student should get a grade of 100%. • Your exam grade, expressed as a perc-ent, is a quantity that shows how well you did on the exam compared with how well you could have done if you had answered all questions correctly

  18. This calc is similar to the percent yield calc that you do in the lab when the product from a chemical rxn is less than you expected based on the balanced eqn. • You might have assumed that if we use stoich to calculate that our rxn will produce 5.2 g of product, that we will actually recover 5.2 g of product in the lab. • This assumption is as faulty as assuming that all students will score 100% on an exam.

  19. When an equation is used to calculate the amount of product that is possible during a rxn, a value representing the theoretical yield is obtained. • The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. • In contrast, the amount of product that forms when the rxn is carried out in the lab is called the actual yield. • The actual yield is often less than the theoretical yield.

  20. The percent yield is the ratio of the actual yield to the theoretical yield as a percent • It measures the measures the efficiency of the reaction actual yield Percent yield= x 100 theoretical yield • What causes a percent yield to be less than 100%?

  21. Rxns don’t always go to completion; when this occurs, less than the expected amnt of product is formed. • Impure reactants and competing side rxns may cause unwanted products to form. • Actual yield can also be lower than the theoretical yield due to a loss of product during filtration or transferring between containers. • If a wet precipitate is recovered it might weigh heavy due to incomplete drying, etc.

  22. 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.

  23. 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

  24. _____________ • CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO3 (How efficient were we?) • Our percent yield is: 33.1 g CaCO3 Percent yield= x 100 44.3 g CaCO3 Percent yield = 74.7%

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