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Section 9-3: Limiting Reactants and Percent Yield

Section 9-3: Limiting Reactants and Percent Yield. Coach Kelsoe Chemistry Pages 312–318. Section 9-3 Objectives. Describe a method for determining which of two reactants is a limiting reactant.

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Section 9-3: Limiting Reactants and Percent Yield

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  1. Section 9-3:Limiting Reactants and Percent Yield Coach Kelsoe Chemistry Pages 312–318

  2. Section 9-3 Objectives • Describe a method for determining which of two reactants is a limiting reactant. • Calculate the amount in moles or mass in grams of a product, given the amounts in moles or masses in grams of two reactants, one of which is in excess. • Distinguish between theoretical yield, actual yield, and percent yield. • Calculate percent yield, given the actual yield, and quantity of a reactant.

  3. Limiting Reactants and Percent Yield • In the lab, a reaction is rarely carried out with exactly the required amounts of each of the reactants. In most cases, one or more reactants is present in excess. • Think of it like cooking hot dogs. Hot dog buns come in packs of 8, while the actual hot dogs are packaged in tens.

  4. Limiting Reactants and Percent Yield • Once one of the reactants is used up, no more product can be formed. The substance that is completely used up is called the limiting reactant. • The limiting reactant is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction. • The substance that is not completely used up in a reaction is sometimes called the excess reactant.

  5. Limiting Reactants and Percent Yield • Limiting reactant may also be referred to as a limiting reagent. • Your book compares limiting reactants and excess reactants to passengers on a plane. If you have 400 people wanting to travel and a plane that can carry 350, what is the limiting reactant?

  6. Limiting Reactants and Percent Yield • The same reasoning can be applied to chemical reactions. Consider the reaction between carbon and oxygen to form carbon dioxide: C(s) + O2(g)  CO2(g) • According to the equation, one mole of carbon reacts with one mole of oxygen to form one mole of carbon dioxide. However, if you tried to react 5 mol of C with 10 mol of O, you would only get 5 mol of CO2.

  7. Sample Problem 9-6 • Silicon dioxide (also known as quartz) is usually unreactive, but reacts readily with hydrogen fluoride according to the following equation: SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l) • If 2 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant? • Given: amount of HF = 2.0 mol, amount of SiO2 = 4.5 mol • Unknown: limiting reactant

  8. Sample Problem 9-6 • To solve this problem, we’ll have to use a mole ratio, but which amount do we need? • Let’s try one: • 2.0 mol HF x 1 mol SiO2/4 mol HF = 0.50 mol SiO2 • Under ideal conditions, the 2.0 mol of HF will require 0.5 mol of SiO2 for complete reaction. Because the amount of SiO2 available (4.5 mol) is more than the amount required (0.5 mol), the limiting reactant is HF.

  9. Sample Problem • Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, H2O2, as the propellant. The reaction is given by the following equation: N2H4(l) + 2H2O2(l)  N2(g) + 4H2O(g) • Which is the limiting reactant in this reaction when 0.750 mol of N2H4 is mixed with 0.500 mol of H2O2? • How much of the excess reactant, in moles, remains unchanged?

  10. Sample Problem • Given: amount of N2H4: 0.750 mol, amount of H2O2: 0.500 mol • Unknown: limiting reactant, how much remains unchanged, how much of each is formed • 0.750 mol N2H4 x 2 mol H2O2/1 mol N2H4= 1.50 mol • H2O2 is our limiting reactant. • L.R. (0.500) x mol excess (1)/mol limit (2) = 0.500

  11. Sample Problem 9-7 • The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. This substance can also be made in the laboratory by the reaction between red-hot iron and steam according to the following equation: 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g) • When 36.0 g of H2O is mixed with 167 g of Fe, which is the limiting reactant? • What mass in grams of black iron oxide is produced? • What mass in grams of excess reactant remains when the reaction is completed?

  12. Sample Problem 9-7 • Given: mass of H2O: 36 g, mass of Fe: 167 g • Unknown: limiting reactant, mass of Fe3O4 in grams, mass of excess reactant remaining • 36.0 g H2O x 1 mol H2O/18.02 g = 2.00 mol167 g Fe x 1 mol Fe/55.85 g = 2.99 mol2.99 mol Fe x 4 mol H2O/3 mol Fe = 3.99 mol; H2O is limiting reactant • 2.00 mol x 1 mol Fe3O4/4 mol H2O x 231.55 g/1 mol Fe3O4= 116 g Fe3O4 • 2 mol H2O x 3 mol Fe/4 mol H2O x 55.85 g/1 mol Fe = 83.2 g Fe consumed  167 g originally present – 83.8 g consumed = 83.2 g Fe left

  13. Percent Yield • The amount of products calculated in the stoichiometric problems in this chapter so far represent theoretical yields. • The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. • In most chemical reactions, the amount of product obtained is less than the theoretical yield.

  14. Percent Yield • There are several reasons we don’t get a theoretical yield when we perform a reaction: • Some of the reactant may be used in competing side reactions that reduce the amount of the desired product. • Once a product is formed, it often is usually collected in impure form, and some of the product is often lost during the purification process. • The measured amount of a product obtained from a reaction is called the actual yield of that product.

  15. Percent Yield • Chemists are usually interested in the efficiency of a reaction. The efficiency is expressed by comparing the actual and theoretical yields. • The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100. • Percent yield = actual yield x 100 theoretical yield

  16. Sample Problem 9-8 • Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C6H6, with chlorine, as represented by the following equation: C6H6(l) + Cl2(g)  C6H5Cl(s) + HCl(g) • When 36.8 g of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percent yield of C6H5Cl?

  17. Sample Problem 9-8 • Given: mass of C6H6= 36.8 g, mass of Cl2 = excess, actual yield of C6H5Cl = 38.8 g • Unknown: percent yield of C6H5Cl • Find theoretical yield of C6H5Cl • 36.8 g C6H6x 1 mol C6H6/78.12g x 1 mol C6H5Cl/1 mol C6H6x 112.56 g/1 mol C6H5Cl = 53.0 g C6H5Cl (theoretical • Find percent yield • % yield = actual/theor. X 100  38.8/53.0 x 100 = 73.2%

  18. Vocabulary • Actual yield • Excess reactant • Limiting reactant • Percent yield • Theoretical yield

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