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Limiting Reagents and Percent Yield

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

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

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

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

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We cant make anymore than 1 double cheeseburger with our ingredients.We cant 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)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.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 isnt enough of one of the reactants.It is equally impossible for a chemist to make a certain amount of a desired compound if there isnt enough of one of the reactants.

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As weve been learning, a balanced chemical rxn is a chemists recipe.As weve been learning, a balanced chemical rxn is a chemists recipe. –Which allows the chemist to predict the amount of product formed from the amounts of ingredients available Lets look at the reaction equation for the formation of ammonia:Lets look at the reaction equation for the formation of ammonia: N 2 (g) + 3H 2 (g) 2NH 3 (g) When 1 mole of N 2 reacts with 3 moles of H 2, 2 moles of NH 3 are produced.When 1 mole of N 2 reacts with 3 moles of H 2, 2 moles of NH 3 are produced. How much NH 3 could be made if 2 moles of N 2 were reacted with 3 moles of H 2 ?How much NH 3 could be made if 2 moles of N 2 were reacted with 3 moles of H 2 ?

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

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How much N 2 will be left over after the reaction?How much N 2 will be left over after the reaction? –In our rxn it takes 1 mol of N 2 to react all of 3 mols of H 2, so there must be 1 mol of N 2 that remains unreacted. We can use our new stoich calculation skills to determine 3 possible types of LR type calculations.We can use our new stoich calculation skills to determine 3 possible types of LR type calculations. 1.Determine which of the reactants will run out first (limiting reactant) 2.Determine amount of product 3.Determine how much excess reactant is wasted

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

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Our 1 st goal is to calculate how much S would react if all of the Cu was reacted.Our 1 st 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).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.Then we can use the Limiting Reactant to calculate the amount of product formed and the amount of excess reactant left over. 82g Cu 82g Cu mol Cu mol Cu mol S mol S g S

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2Cu + S Cu 2 S 82.0gCu 1molCu 63.5gCu 1mol S 2molCu 32.1g S 1mol S =20.7 g S So if all of our 82.0g of Copper were reacted completely it would require only 20.7 grams of Sulfur.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

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Copper being our Limiting Reactant is then used to determine how much product is produced.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. 82.0gCu 1molCu 63.5gCu 1molCu 2 S 2molCu 2 S ________ 159gCu 2 S 1molCu 2 S = 103 g Cu 2 S

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So the reaction between 82.0g of Cu and 25.0g of S can only produce 103g of Cu 2 S.So the reaction between 82.0g of Cu and 25.0g of S can only produce 103g of Cu 2 S. –The Cu runs out before the S and we will end up wasting 4.7 g of the S. 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 MgCl 2 + H 2 What is the LR when 6.0 g HCl reacts with 5.0 g Mg? What is the maximum amount of H 2 that can be formed? And how much of the other reactant is wasted?What is the LR when 6.0 g HCl reacts with 5.0 g Mg? What is the maximum amount of H 2 that can be formed? And how much of the other reactant is wasted?

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5.0g Mg 1molMg 24.3gMg 2molHCl 1molMg 36.5gHCl 1molHCl = 15.0g HCl 5.0g Mg mol Mg 2mol HCl g 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.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.Therefore, the 6.0g of HCl will run out before the 5.0g of Mg, so HCl is our Limiting Reactant.

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6.0g HCl 1molHCl 36.5gHCl 1molH 2 2molHCl 2.0gH 2 1molH 2 = 0.164 g H 2 produced 6.0g HCl 2mol HCl 1mol H 2 g H 2 6.0g HCl 1molHCl 36.5gHCl 1molMg 2molHCl 24.3gMg 1molMg = 1.997 g Mg 6.0g HCl 2mol HCl 1mol Mg g Mg - 5.0 g Mg = 3.01g Mg extra

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Calculating Percent Yield In theory, when a teacher gives an exam to the class, every student should get a grade of 100%.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 correctlyYour 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

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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.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.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.This assumption is as faulty as assuming that all students will score 100% on an exam.

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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.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.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.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.The actual yield is often less than the theoretical yield.

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The percent yield is the ratio of the actual yield to the theoretical yield as a percentThe 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 Percent yield= actual yield theoretical yield x 100 What causes a percent yield to be less than 100%?What causes a percent yield to be less than 100%?

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Rxns dont always go to completion; when this occurs, less than the expected amnt of product is formed.Rxns dont 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.

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Calcium carbonate is synthesized by heating,as shown in the following equation: CaO + CO 2 CaCO 3 What is the theoretical yield of CaCO 3 if 24.8 g of CaO is heated with 43.0 g of CO 2 ?What is the theoretical yield of CaCO 3 if 24.8 g of CaO is heated with 43.0 g of CO 2 ? What is the percent yield if 33.1 g of CaCO 3 is produced?What is the percent yield if 33.1 g of CaCO 3 is produced? Determine which reactant is the limiting and then decide what the theoretical yield is.

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24.8 g CaO 1molCaO 56g CaO 1mol CO 2 1mol CaO 44 g CO 2 1molCO 2 = 19.5gCO 2 24.8gCaO 24.8gCaO molCaO molCaO mol CO 2 mol CO 2 gCO 2 24.8 g CaO 1mol CaO 56g CaO 1molCaCO 3 1mol CaO 100g CaCO 3 1molCaCO 3 = 44.3 g CaCO 3 24.8gCaO 24.8gCaO molCaO molCaO mol CaCO 3 mol CaCO 3 gCaCO 3 LR

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CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO 3 (How efficient were we?)CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO 3 (How efficient were we?) Our percent yield is:Our percent yield is: Percent yield= 33.1 g CaCO 3 44.3 g CaCO 3 _____________ x 100 Percent yield = 74.7%

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