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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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**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:**1 hamburger bun**1 tomato slice 1 lettuce leaf 2 slices of cheese 2 burger patties**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**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**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.**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**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.**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 excess reactant is wasted**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 wasted?**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**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**________**• 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**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 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 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 wasted?**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.**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 extra**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**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.**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.**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%?**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.**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.**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**_____________**• 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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