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. 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. 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. Distinguish between theoretical yield, actual yield, and percent yield. Calculate percent yield, given the actual yield, and quantity of a reactant. 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. 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. 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. 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 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. 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. 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? 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) + O 2 (g)  CO 2 (g) The same reasoning can be applied to chemical reactions. Consider the reaction between carbon and oxygen to form carbon dioxide: C(s) + O 2 (g)  CO 2 (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 CO 2. 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 CO 2.

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: SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l) Silicon dioxide (also known as quartz) is usually unreactive, but reacts readily with hydrogen fluoride according to the following equation: SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l) If 2 mol of HF are exposed to 4.5 mol of SiO 2, which is the limiting reactant? If 2 mol of HF are exposed to 4.5 mol of SiO 2, which is the limiting reactant? Given: amount of HF = 2.0 mol, amount of SiO 2 = 4.5 mol Given: amount of HF = 2.0 mol, amount of SiO 2 = 4.5 mol Unknown: limiting reactant 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? To solve this problem, we’ll have to use a mole ratio, but which amount do we need? Let’s try one: Let’s try one: 2.0 mol HF x 1 mol SiO 2 /4 mol HF = 0.50 mol SiO 2 2.0 mol HF x 1 mol SiO 2 /4 mol HF = 0.50 mol SiO 2 Under ideal conditions, the 2.0 mol of HF will require 0.5 mol of SiO 2 for complete reaction. Because the amount of SiO 2 available (4.5 mol) is more than the amount required (0.5 mol), the limiting reactant is HF. Under ideal conditions, the 2.0 mol of HF will require 0.5 mol of SiO 2 for complete reaction. Because the amount of SiO 2 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, N 2 H 4, and hydrogen peroxide, H 2 O 2, as the propellant. The reaction is given by the following equation: N 2 H 4 (l) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 O(g) Some rocket engines use a mixture of hydrazine, N 2 H 4, and hydrogen peroxide, H 2 O 2, as the propellant. The reaction is given by the following equation: N 2 H 4 (l) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 O(g) a) Which is the limiting reactant in this reaction when 0.750 mol of N 2 H 4 is mixed with 0.500 mol of H 2 O 2 ? b) How much of the excess reactant, in moles, remains unchanged?

10. Sample Problem Given: amount of N 2 H 4 : 0.750 mol, amount of H 2 O 2 : 0.500 mol Given: amount of N 2 H 4 : 0.750 mol, amount of H 2 O 2 : 0.500 mol Unknown: limiting reactant, how much remains unchanged, how much of each is formed Unknown: limiting reactant, how much remains unchanged, how much of each is formed 0.750 mol N 2 H 4 x 2 mol H 2 O 2 /1 mol N 2 H 4 = 1.50 mol 0.750 mol N 2 H 4 x 2 mol H 2 O 2 /1 mol N 2 H 4 = 1.50 mol a) H 2 O 2 is our limiting reactant. b) L.R. (0.500) x mol excess (1)/mol limit (2) = 0.500

11. Sample Problem 9-7 The black oxide of iron, Fe 3 O 4, 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) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) The black oxide of iron, Fe 3 O 4, 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) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) a) When 36.0 g of H 2 O is mixed with 167 g of Fe, which is the limiting reactant? b) What mass in grams of black iron oxide is produced? c) What mass in grams of excess reactant remains when the reaction is completed?

12. Sample Problem 9-7 Given: mass of H 2 O: 36 g, mass of Fe: 167 g Given: mass of H 2 O: 36 g, mass of Fe: 167 g Unknown: limiting reactant, mass of Fe 3 O 4 in grams, mass of excess reactant remaining Unknown: limiting reactant, mass of Fe 3 O 4 in grams, mass of excess reactant remaining a) 36.0 g H 2 O x 1 mol H 2 O/18.02 g = 2.00 mol 167 g Fe x 1 mol Fe/55.85 g = 2.99 mol 2.99 mol Fe x 4 mol H 2 O/3 mol Fe = 3.99 mol; H 2 O is limiting reactant b) 2.00 mol x 1 mol Fe 3 O 4 /4 mol H 2 O x 231.55 g/1 mol Fe 3 O 4 = 116 g Fe 3 O 4 c) 2 mol H 2 O x 3 mol Fe/4 mol H 2 O 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 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. 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. 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: 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. 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. 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. 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. 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. The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100. Percent yield = actual yield x 100 Percent yield = actual yield x 100 theoretical yield

16. Sample Problem 9-8 Chlorobenzene, C 6 H 5 Cl, 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, C 6 H 6, with chlorine, as represented by the following equation: C 6 H 6 (l) + Cl 2 (g)  C 6 H 5 Cl(s) + HCl(g) Chlorobenzene, C 6 H 5 Cl, 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, C 6 H 6, with chlorine, as represented by the following equation: C 6 H 6 (l) + Cl 2 (g)  C 6 H 5 Cl(s) + HCl(g) When 36.8 g of C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percent yield of C 6 H 5 Cl? When 36.8 g of C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percent yield of C 6 H 5 Cl?

17. Sample Problem 9-8 Given: mass of C 6 H 6 = 36.8 g, mass of Cl 2 = excess, actual yield of C 6 H 5 Cl = 38.8 g Given: mass of C 6 H 6 = 36.8 g, mass of Cl 2 = excess, actual yield of C 6 H 5 Cl = 38.8 g Unknown: percent yield of C 6 H 5 Cl Unknown: percent yield of C 6 H 5 Cl Find theoretical yield of C 6 H 5 Cl Find theoretical yield of C 6 H 5 Cl 36.8 g C 6 H 6 x 1 mol C 6 H 6 /78.12g x 1 mol C 6 H 5 Cl/1 mol C 6 H 6 x 112.56 g/1 mol C 6 H 5 Cl = 53.0 g C 6 H 5 Cl (theoretical 36.8 g C 6 H 6 x 1 mol C 6 H 6 /78.12g x 1 mol C 6 H 5 Cl/1 mol C 6 H 6 x 112.56 g/1 mol C 6 H 5 Cl = 53.0 g C 6 H 5 Cl (theoretical Find percent yield Find percent yield % yield = actual/theor. X 100  38.8/53.0 x 100 = 73.2% % yield = actual/theor. X 100  38.8/53.0 x 100 = 73.2%

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