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Transcript
EXAMPLE 8.1 Mole-to-Mole Conversions Sodium chloride, NaCl, forms by the following reaction between sodium and chlorine. How many moles of NaCl result from the complete reaction of 3.4 mol of Cl 2 ? Assume that there is more than enough Na. You are given the number of moles of a reactant (Cl2) and asked to find how many moles of product (NaCl) will result if the reactant completely reacts. The conversion factor comes from the balanced chemical equation. The solution map begins with moles of chlorine and uses the stoichiometric conversion factor to obtain moles of sodium chloride Solution Map: Follow the solution map to solve the problem. There is enough Cl2 to produce 6.8 mol of NaCl. Solution: Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Mole-to-Mole Conversions EXAMPLE 8.1 Continued SKILLBUILDER 8.1 Mole-to-Mole Conversions Water is formed when hydrogen gas reacts explosively with oxygen gas according to the following balanced equation. How many moles of H2O result from the complete reaction of 24.6 mol of O2? Assume that there is more than Enough H2. FOR MORE PRACTICE Introductory Chemistry, Third Edition By Nivaldo J. Tro Example 8.8; Problems 17, 18, 19, 20. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.2 Mass-to-Mass Conversions In photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H12O6) according to the following reaction. How many grams of glucose can be synthesized from 58.5 g of CO2 ? Assume that there is more than enough water present to react with all of the CO2 . Set up the problem in the normal way. The main conversion factor is the stoichiometric relationship between moles of carbon dioxide and moles of glucose. This conversion factor comes from the balanced equation. The other conversion factors are simply the molar masses of carbon dioxide and glucose. The solution map uses the general outline Solution Map: where A is carbon dioxide and B is glucose. Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.2 Mass-to-Mass Conversions Continued Follow the solution map to solve the problem. Begin with grams of carbon dioxide and multiply by the appropriate factors to arrive at grams of glucose. SKILLBUILDER 8.2 Solution: Mass-to-Mass Conversions Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl, according to the following reaction. How much HCl in grams can be neutralized by 5.50 g of Mg(OH)2? FOR MORE PRACTICE Introductory Chemistry, Third Edition By Nivaldo J. Tro Example 8.9; Problems 33, 34, 35, 36. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.3 Mass-to-Mass Conversions One of the components of acid rain is nitric acid, which forms when NO 2 a pollutant, reacts with oxygen and rainwater according to the following reaction. Assuming that there is more than enough O2 and H2O, how much HNO3 in kilograms forms from 1.5 103 kg of NO2 pollutant? Set up the problem in the normal way. The main conversion factor is the stoichiometric relationship between moles of nitrogen dioxide and moles of nitric acid. This conversion factor comes from the balanced equation. The other conversion factors are simply the molar masses of nitrogen dioxide and nitric acid and the relationship between kilograms and grams. The solution map follows the general format of: Introductory Chemistry, Third Edition By Nivaldo J. Tro Solution Map: Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.3 Mass-to-Mass Conversions Continued However, since the original quantity of NO2 is given in kilograms, you must first convert to grams. Since the final quantity is requested in kilograms, you must convert back to kilograms at the end. Solution Map: Follow the solution map to solve the problem. Begin with kilograms of nitrogen dioxide and multiply by the appropriate conversion factors to arrive at kilograms of nitric acid. Solution: SKILLBUILDER 8.3 Mass-to-Mass Conversions Another component of acid rain is sulfuric acid, which forms when SO 2 also a pollutant, reacts with oxygen and rainwater according to the following reaction. Assuming that there is plenty of O2 and H2O how much H2SO4 in kilograms forms from 2.6 × 103 kg of SO2? FOR MORE PRACTICE Introductory Chemistry, Third Edition By Nivaldo J. Tro Problems 37, 38, 39, 40. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants Consider the following reaction. If we begin with 0.552 mol of aluminum and 0.887 mol of chlorine, what is the limiting reactant and theoretical yield of AlCl3 in moles? Set up the problem in the normal way. The conversion factors are the stoichiometric relationships (from the balanced equation) between the moles of each reactant and the moles of product. The solution map shows how to get from moles of each reactant to moles of AlCl3. The reactant that makes the least amount of AlCl3 is the limiting reactant.. Introductory Chemistry, Third Edition By Nivaldo J. Tro Given: 0.552 mol Al 0.887 mol Cl2 Find: limiting reactant theoretical yield Solution Map: Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants Continued Solution: Since the 0.552 mol of Al makes the least amount of AlCl 3 is the limiting reactant. The theoretical yield is 0.552 mol of AlCl3. SKILLBUILDER 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants Consider the following reaction. If you begin with 4.8 mol of sodium and 2.6 mol of fluorine, what is the limiting reactant and theoretical yield of NaF in moles? FOR MORE PRACTICE Introductory Chemistry, Third Edition By Nivaldo J. Tro Problems 45, 46, 47, 48, 49, 50, 51, 52. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.5 Finding Limiting Reactant and Theoretical Yield Ammonia, NH3, can be synthesized by the following reaction. What is the maximum amount of ammonia in grams that can be synthesized from 45.8 g of NO and 12.4 g of H 2? Although this problem does not specifically ask for the limiting reactant, it must be found to determine the theoretical yield, which is the maximum amount of ammonia that can be synthesized. Begin by setting up the problem in the normal way. Given: 545.8 g NO, 12.4 g H2 Find: maximum amount of NH3 (theoretical yield) Conversion Factors: The main conversion factors are the stoichiometric relationship between moles of each reactant and moles of ammonia. The other conversion factors are simply the molar masses of nitrogen monoxide, hydrogen gas, and ammonia. Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.5 Finding Limiting Reactant and Theoretical Yield Continued Find the limiting reactant by calculating how much product can be made from each reactant. The reactant that makes the least amount of product is the limiting reactant. Introductory Chemistry, Third Edition By Nivaldo J. Tro Solution Map: Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.5 Finding Limiting Reactant and Theoretical Yield Continued Follow the solution map, beginning with the actual amount of each reactant given, to calculate the amount of product that can be made from each reactant. Solution: There is enough NO to make 26.0 g of NH3 and enough H2 to make 41.8 g of HN3. Therefore, NO is the limiting reactant, and the maximum amount of ammonia that can possibly be made is 26.0 g, the theoretical yield. SKILLBUILDER 8.5 Finding Limiting Reactant and Theoretical Yield Ammonia can also be synthesized by the following reaction. What is the maximum amount of ammonia in grams that can be synthesized from 25.2 g of N 2 and 8.42 g of H2? Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.5 Finding Limiting Reactant and Theoretical Yield Continued SKILLBUILDER PLUS What is the maximum amount of ammonia in kilograms that can be synthesized from 5.22 kg of H 2 and 31.5 kg of N2? FOR MORE PRACTICE Introductory Chemistry, Third Edition By Nivaldo J. Tro Problems 55, 56, 57, 58. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Consider the following reaction. When 11.5 g of C are allowed to react with 114.5 g of Cu2O, 87.4 g of Cu, are obtained. Find the limiting reactant, theoretical yield, and percent yield. Set up the problem in the normal way. Given: 11.5 g C 114.5 g Cu2O 87.4 g Cu produced Find: limiting reactant theoretical yield percent yield The main conversion factors are the stoichiometric relationships between moles of each reactant and moles of copper. The other conversion factors are simply the molar masses of copper(I) oxide, carbon, and copper Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Continued The solution map shows how to find the mass of Cu formed by the initial masses of Cu2O and C. The reactant that makes the least amount of product is the limiting reactant and determines the theoretical yield. Solution Map: Follow the solution map, beginning with the actual amount of each reactant given, to calculate the amount of product that can be made from each reactant. Solution: Since Cu2O makes the least amount of product, Cu2O is the limiting reactant. The theoretical yield is simply the amount of product made by the limiting reactant. The percent yield is the actual yield (87.4 g Cu) divided by the theoretical yield (101.7 g Cu) multiplied by 100%. Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Continued SKILLBUILDER 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield The following reaction is used to obtain iron from iron ore: The reaction of 185 g of Fe2O3 with 95.3 g of CO produces 87.4 g of Fe. Find the limiting reactant, theoretical yield, and percent yield. FOR MORE PRACTICE Introductory Chemistry, Third Edition By Nivaldo J. Tro Example 8.10; Problems 61, 62, 63, 64, 65, 66. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.7 Stoichiometry Involving ΔH An LP gas tank in a home barbecue contains 11.8 × 103 g of propane (C3H8). Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank. You are given the mass of propane and asked to find the heat evolved (in kJ) in its combustion. Given: 11.8 × 103 g C3H8 Find: kJ Conversion Factors: Start with g C3H8 and then use the molar mass of C3H8 to find the number of moles. Next, use the stoichiometric relationship between mol C3H8 and kJ to find the heat evolved. Solution Map: Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.7 Stoichiometry Involving ΔH Continued Follow the solution map to solve the problem. Begin with 11.8 × 103 g C3H8 and multiply by the appropriate conversion factors to arrive at kJ. The answer is negative, as it should be for heat evolved by the reaction. SKILLBUILDER 8.7 Solution: Stoichiometry Involving ΔH Ammonia reacts with oxygen according to the following equation: Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH 3. SKILLBUILDER PLUS What mass of butane in grams is necessary to produce 1.5 × 103 kJ of heat? What mass of CO2 is produced? FOR MORE PRACTICE Introductory Chemistry, Third Edition By Nivaldo J. Tro Example 8.11; Problems 69, 70, 71, 72, 73, 74. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.8 Mole-to-Mole Conversions How many moles of sodium oxide can be synthesized from 4.8 mol of sodium? Assume that there is more than enough oxygen present. The balanced equation is: Given: 4.8 mol Na Find: mol Na2O Conversion Factor: Solution Map: Solution: Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.9 Mass-to-Mass Conversions How many grams of sodium oxide can be synthesized from 17.4 g of sodium? Assume that there is more than enough oxygen present. The balanced equation is: Given: 17.4 g Na Find: g Na2O Conversion Factors: Molar mass Na = 22.99 g/mol Molar mass Na2O = 61.98 g/mol Solution Map: Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.9 Mass-to-Mass Conversions Continued Solution: Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.10 Limiting Reactant, Theoretical Yield, and Percent Yield 10.4 g of Fe reacts with 11.8 g of S to produce 14.2 g of Fe2S3. Find the limiting reactant, theoretical yield, and percent yield for this reaction. The balanced chemical equation is: Given: 10.4 g Fe 11.8 g S 14.2 g Fe2S3 Find: limiting reactant theoretical yield percent yield Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.10 Limiting Reactant, Theoretical Yield, and Percent Yield Continued Solution Map: Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.10 Limiting Reactant, Theoretical Yield, and Percent Yield Continued Solution: The limiting reactant is Fe. The theoretical yield is 19.4 g of Fe2S3 . The percent yield is 73.2%. Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. EXAMPLE 8.11 Stoichiometry Involving ΔH Calculate the heat evolved (in kJ) upon complete combustion of 25.0 g of CH 4. Given: 25 g CH4 Find: kJ Conversion Factors: Solution Map: Solution: Introductory Chemistry, Third Edition By Nivaldo J. Tro Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.