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Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J.

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Presentation on theme: "Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J."— Presentation transcript:

1 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro You are given the number of moles of a reactant (Cl 2 ) 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. EXAMPLE 8.1 Mole-to-Mole Conversions The solution map begins with moles of chlorine and uses the stoichiometric conversion factor to obtain moles of sodium chloride Follow the solution map to solve the problem. There is enough Cl 2 to produce 6.8 mol of NaCl. Solution Map : 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. Solution :

2 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.1 Mole-to-Mole Conversions FOR MORE PRACTICE Example 8.8; Problems 17, 18, 19, 20. Continued Water is formed when hydrogen gas reacts explosively with oxygen gas according to the following balanced equation. How many moles of H 2 O result from the complete reaction of 24.6 mol of O 2 ? Assume that there is more than Enough H 2. SKILLBUILDER 8.1 Mole-to-Mole Conversions

3 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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. EXAMPLE 8.2 Mass-to-Mass Conversions In photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H 12 O 6 ) according to the following reaction. How many grams of glucose can be synthesized from 58.5 g of CO 2 ? Assume that there is more than enough water present to react with all of the CO 2. The solution map uses the general outline where A is carbon dioxide and B is glucose. Solution Map :

4 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.2 Mass-to-Mass Conversions FOR MORE PRACTICE Example 8.9; Problems 33, 34, 35, 36. 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. Solution : 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 ? SKILLBUILDER 8.2 Mass-to-Mass Conversions

5 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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. 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 rain- water according to the following reaction. Assuming that there is more than enough O 2 and H 2 O, how much HNO 3 in kilograms forms from 1.5  10 3 kg of NO 2 pollutant? Solution Map : The solution map follows the general format of:

6 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.3 Mass-to-Mass Conversions FOR MORE PRACTICE Problems 37, 38, 39, 40. Continued However, since the original quantity of NO 2 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. 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 Map : Solution : 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 O 2 and H 2 O how much H 2 SO 4 in kilograms forms from 2.6 × 10 3 kg of SO 2 ? SKILLBUILDER 8.3 Mass-to-Mass Conversions

7 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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. EXAMPLE 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants The solution map shows how to get from moles of each reactant to moles of AlCl 3. The reactant that makes the least amount of AlCl 3 is the limiting reactant.. Consider the following reaction. If we begin with mol of aluminum and mol of chlorine, what is the limiting reactant and theoretical yield of AlCl 3 in moles? Solution Map : Given: mol Al mol Cl 2 Find : limiting reactant theoretical yield

8 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants FOR MORE PRACTICE Problems 45, 46, 47, 48, 49, 50, 51, 52. Continued Solution : 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? SKILLBUILDER 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants Since the mol of Al makes the least amount of AlCl 3 is the limiting reactant. The theoretical yield is mol of AlCl 3.

9 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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. 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. EXAMPLE 8.5 Finding Limiting Reactant and Theoretical Yield Given: g NO, 12.4 g H 2 Find : maximum amount of NH 3 (theoretical yield) Conversion Factors : Ammonia, NH 3, 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 ?

10 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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. Solution Map :

11 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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. There is enough NO to make 26.0 g of NH 3 and enough H 2 to make 41.8 g of HN 3. Therefore, NO is the limiting reactant, and the maximum amount of ammonia that can possibly be made is 26.0 g, the theoretical yield. Solution : 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 H 2 ? SKILLBUILDER 8.5 Finding Limiting Reactant and Theoretical Yield

12 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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 N 2 ? FOR MORE PRACTICE Problems 55, 56, 57, 58.

13 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. 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 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 g of Cu 2 O, 87.4 g of Cu, are obtained. Find the limiting reactant, theoretical yield, and percent yield. Given: 11.5 g C g Cu 2 O 87.4 g Cu produced Find : limiting reactant theoretical yield percent yield

14 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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 Cu 2 O 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. Since Cu 2 O makes the least amount of product, Cu 2 O 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%. Solution :

15 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Continued FOR MORE PRACTICE Example 8.10; Problems 61, 62, 63, 64, 65, 66. The following reaction is used to obtain iron from iron ore: The reaction of 185 g of Fe 2 O 3 with 95.3 g of CO produces 87.4 g of Fe. Find the limiting reactant, theoretical yield, and percent yield. SKILLBUILDER 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield

16 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro You are given the mass of propane and asked to find the heat evolved (in kJ) in its combustion. EXAMPLE 8.7 Stoichiometry Involving ΔH Start with g C 3 H 8 and then use the molar mass of C 3 H 8 to find the number of moles. Next, use the stoichiometric relationship between mol C 3 H 8 and kJ to find the heat evolved. Solution Map : An LP gas tank in a home barbecue contains 11.8 × 10 3 g of propane (C 3 H 8 ). Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank. Given: 11.8 × 10 3 g C 3 H 8 Find : kJ Conversion Factors :

17 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.7 Stoichiometry Involving ΔH FOR MORE PRACTICE Example 8.11; Problems 69, 70, 71, 72, 73, 74. Continued Follow the solution map to solve the problem. Begin with 11.8 × 10 3 g C 3 H 8 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. Solution : 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 8.7 Stoichiometry Involving ΔH What mass of butane in grams is necessary to produce 1.5 × 10 3 kJ of heat? What mass of CO 2 is produced? SKILLBUILDER PLUS

18 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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 Na 2 O Conversion Factor : Solution Map : Solution :

19 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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 Na 2 O Conversion Factors : Molar mass Na = g/mol Molar mass Na 2 O = g/mol Solution Map :

20 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.9 Mass-to-Mass Conversions Continued Solution :

21 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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 Fe 2 S 3. 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 Fe 2 S 3 Find : limiting reactant theoretical yield percent yield

22 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.10 Limiting Reactant, Theoretical Yield, and Percent Yield Continued Solution Map :

23 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 8.10 Limiting Reactant, Theoretical Yield, and Percent Yield Continued The limiting reactant is Fe. The theoretical yield is 19.4 g of Fe 2 S 3. The percent yield is 73.2%. Solution :

24 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 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 CH 4 Find : kJ Conversion Factors : Solution Map : Solution :


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