7 10 Classifying Chemical Reactions by What Atoms

Chapter 8 Quantities in Chemical Reactions

8. 2 Stoichiometry: Relationships between Ingredients �The numerical relationship between chemical quantities in a balanced chemical equation is called reaction stoichiometry. �We can predict the amounts of products that form in a chemical reaction based on the amounts of reactants. �We can predict how much of the reactants are necessary to form a given amount of product. �We can predict how much of one reactant is required to completely react with another reactant. © 2012 Pearson Education, Inc.

8. 2 Making Pancakes: Relationships between Ingredients � A recipe gives numerical relationships between

8. 2 Making Pancakes: Relationships between Ingredients � The recipe shows the numerical relationships between the pancake ingredients. � If we have 2 eggs—and enough of everything else —we can make 5 pancakes. � We can write this relationship as a ratio. � 2 eggs: 5 pancakes © 2012 Pearson Education, Inc.

What if we have 8 eggs? Assuming that we have enough of everything else,

8. 3 Making Molecules: Mole-to-Mole Conversions � In a balanced chemical equation, we have a “recipe” for how reactants combine to form products. � The following equation shows how hydrogen and nitrogen combine to form ammonia (NH 3). © 2012 Pearson Education, Inc.

3 H 2(g) + N 2(g) 2 NH 3(g) � The balanced equation shows that 3 H 2 molecules react with 1 N 2 molecule to form 2 NH 3 molecules. � We can express these relationships as ratios. 3 H 2 molecules : 1 N 2 molecule : 2 NH 3 molecules � Since we do not ordinarily deal with individual molecules, we can express the same ratios in moles. 3 mol H 2 : 1 mol N 2 : 2 mol NH 3 © 2012 Pearson Education, Inc.

3 H 2(g) + N 2(g) 2 NH 3(g) � If we have 3

Example 8. 1 Mole-to-Mole Conversions Sodium chloride, Na. Cl, forms by this reaction between sodium and chlorine. 2 Na(s) + Cl 2(g) 2 Na. Cl(s) How many moles of Na. Cl result from the complete reaction of 3. 4 mol of Cl 2? Assume that there is more than enough Na. SORT GIVEN: 3. 4 mol Cl 2 You are given the number of moles of a reactant (Cl 2) and asked to find the number of moles of product (Na. Cl) that will form if the reactant completely reacts. FIND: mol Na. Cl STRATEGIZE SOLUTION MAP Draw the solution map beginning with moles of chlorine and using the stoichiometric conversion factor to calculate moles of sodium chloride. The conversion factor comes from the balanced chemical equation. RELATIONSHIPS USED 1 mol Cl 2 : 2 mol Na. Cl (from balanced chemical equation)

Example 8. 1 Mole-to-Mole Conversions Continued SOLVE SOLUTION Follow the solution map to solve the problem. There is enough Cl 2 to produce 6. 8 mol of Na. Cl. CHECK Check your answer. Are the units correct? Does the answer make physical sense? The answer has the correct units, moles. The answer is reasonable because each mole of Cl 2 makes two moles of Na. Cl. SKILLBUILDER 8. 1 Mole-to-Mole Conversions Water is formed when hydrogen gas reacts explosively with oxygen gas according to the balanced equation: O 2(g) + 2 H 2(g) 2 H 2 O(g) 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. Answer: 49. 2 mol H 20 For More Practice Example 8. 8; Problems 15, 16, 17, 18.

8. 4 Making Molecules: Mass-to-Mass Conversions �A chemical equation contains conversion factors between moles of reactants and moles of products. �We are often interested in relationships between mass of reactants and mass of products. �The general outline for this type of calculation is: © 2012 Pearson Education, Inc.

2 C 8 H 18(l) + 25 O 2(g) 16 CO 2(g) + 18 H 2 O(g) � What mass of carbon dioxide is emitted by an automobile per 5. 0 x 102 g pure octane used? � The balanced chemical equation gives us a relationship between moles of C 8 H 18 and moles of CO 2. � Before using that relationship, we must convert from grams to moles. © 2012 Pearson Education, Inc.

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 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. SORT You are given the mass of carbon dioxide and asked to find the mass of glucose that can form if the carbon dioxide completely reacts. STRATEGIZE The solution map uses the general outline Mass A Moles B Mass B where A is carbon dioxide and B is glucose. GIVEN: 58. 5 g CO 2 FIND: g C 6 H 12 O 6 SOLUTION MAP

Example 8. 2 Mass-to-Mass Conversions Continued 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. RELATIONSHIPS USED SOLVE SOLUTION 6 mol CO 2 : 1 mol C 6 H 12 O 6 (from balanced chemical equation) Molar mass CO 2 = 44. 01 g/mol Molar mass C 6 H 12 O 6 = 180. 2 g/mol 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. CHECK Are the units correct? Does the answer make physical sense? The units, g C 6 H 12 O 6, are correct. The magnitude of the answer seems reasonable because it is of the same order of magnitude as the given mass of carbon dioxide. An answer that is orders of magnitude different would immediately be suspect.

Example 8. 2 Mass-to-Mass Conversions Continued SKILLBUILDER 8. 2 Mass-to-Mass Conversions Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl, according to the reaction: Mg(OH)2(aq) + 2 HCl(aq) 2 H 2 O(l) + Mg. Cl 2(aq) How much HCl in grams can be neutralized by 5. 50 g of Mg(OH)2? Answer: 6. 88 g HCl For More Practice Example 8. 9; Problems 31, 32, 33, 34.

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 simplified reaction. 4 NO 2(g) + 2 H 2 O(l) 4 HNO 3(aq) Assuming that there is more than enough O 2 and H 2 O, how much HNO 3 in kilograms forms from 1. 5 103 kg of NO 2 pollutant? SORT You are given the mass of nitrogen dioxide (a reactant) and asked to find the mass of nitric acid that can form if the nitrogen dioxide completely reacts. GIVEN: 1. 5 103 kg NO 2 FIND: kg HNO 3

Example 8. 3 Mass-to-Mass Conversions Continued SOLUTION MAP STRATEGIZE The solution map follows the general format of: Mass Moles Mass 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. 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. RELATIONSHIPS USED 4 mol NO 2 : 4 mol HNO 3 (from balanced chemical equation) Molar mass NO 2 = 46. 01 g/mol Molar mass HNO 3 = 63. 02 g/mol 1 kg = 1000 g

Example 8. 3 Mass-to-Mass Conversions Continued SOLVE SOLUTION 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. CHECK Are the units correct? Does the answer make physical sense? The units, kg HNO 3 are correct. The magnitude of the answer seems reasonable because it is of the same order of magnitude as the given mass of nitrogen dioxide. An answer that is orders of magnitude different would immediately be suspect. 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. 2 SO 2(g) + 2 H 2 O(l) 2 H 2 SO 4(aq) Assuming that there is more than enough O 2 and H 2 O, how much in kilograms forms from H 2 SO 4 2. 6 103 kg of SO 2? Answer: 4. 0 103 kg H 2 SO 4 For More Practice Problems 35, 36, 37, 38.

Administrative Stuff � Silence your electronic devices and make sure your clickers are ready. � Chapter 7 Homework is due TODAY at 5 pm. � Continue learning the polyatomic ions and solubility rules from chapters 5 and 6. � Exam 2 is 10/14/13 and covers Chapters 6, 7, 8, and 9. � We have 5 classes between now and then including today. © 2012 Pearson Education, Inc.

8. 5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield � Suppose we have 3 cups flour, 10 eggs, and 4 tsp baking powder. � How many pancakes can we make? We have enough flour for 15 pancakes, enough eggs for 25 pancakes, and enough baking powder for 40 pancakes. © 2012 Pearson Education, Inc.

8. 5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield If this were a chemical reaction, the flour would be the limiting reactant and 15 pancakes would be theoretical yield. © 2012 Pearson Education, Inc.

8. 5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield �Suppose we cook our pancakes. We accidentally burn three of them and one falls on the floor. �So even though we had enough flour for 15 pancakes, we finished with only 11 pancakes. �If this were a chemical reaction, the 11 pancakes would be our actual yield, the amount of product actually produced by a chemical reaction. © 2012 Pearson Education, Inc.

8. 5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield � Our percent yield, the percentage of theoretical yield that was actually attained, is: Since four of the pancakes were ruined, we got only 73% of our theoretical yield. © 2012 Pearson Education, Inc.

Actual Yield and Percent Yield � The actual yield of a chemical reaction must be determined experimentally and depends on the reaction conditions. � The actual yield is almost always less than 100%. � Some of the product does not form. � Product is lost in the process of recovering it. © 2012 Pearson Education, Inc.

Limiting Reactant, Theoretical Yield, Actual Yield, and Percent Yield To summarize: • Limiting reactant (or limiting reagent)—the reactant that is completely consumed in a chemical reaction. • Theoretical yield—the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. • Actual yield—the amount of product actually produced by a chemical reaction. • Percent yield—(actual yield/theoretical yield)*100%. © 2012 Pearson Education, Inc.

Example: Ti(s) + 2 Cl 2(g) Ti. Cl 4(s) Given (moles): 1. 8 mol

Limiting Reactant, Theoretical Yield, Actual Yield, and Percent Yield � In many industrial applications, the more costly reactant or the reactant that is most difficult to remove from the product mixture is chosen to be the limiting reactant. � When working in the laboratory, we measure the amounts of reactants in grams. � To find limiting reactants and theoretical yields from initial masses, we must add two steps to our calculations. © 2012 Pearson Education, Inc.

Example: Na(s) + Cl 2(g) 2 Na. Cl(s) Given (grams): 53. 2 g Na

Example: Na(s) + Cl 2(g) 2 Na. Cl(s) Given (grams): actual yield 86. 4 g Na. Cl Find: percent yield �The actual yield is usually less than theoretical yield because at least a small amount of product is lost or does not form during a reaction. © 2012 Pearson Education, Inc.

Example 8. 6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Consider the reaction: Cu 2 O(s) + C(s) 2 Cu(s) + CO(g) When 11. 5 g of C are allowed to react with 114. 5 g of Cu 2 O, 87. 4 g of Cu are obtained. Find the limiting reactant, theoretical yield, and percent yield. SORT You are given the mass of the reactants, carbon and copper(I) oxide, as well as the mass of copper formed by the reaction. You are asked to find the limiting reactant, theoretical yield, and percent yield. GIVEN: 11. 5 g C 114. 5 g Cu 2 O 87. 4 g Cu produced FIND: limiting reactant theoretical yield percent yield

Example 8. 6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Continued STRATEGIZE SOLUTION MAP 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 theoretical yield. The main conversion factors are the stoichiometric relationships between moles of each reactant and moles of copper. The other conversion factors are the molar masses of copper(I) oxide, carbon, and copper. RELATIONSHIPS USED 1 mol Cu 2 O: 2 mol Cu 1 mol C : 2 mol Cu Molar mass Cu = 63. 55 g/mol Molar mass C = 12. 01 g/mol Molar mass Cu 2 O = 143. 10 g/mol

Example 8. 6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Continued SOLVE SOLUTION 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 then the amount of product made by the limiting reactant. The percent yield is the actual yield (87. 4 g Cu) divided by theoretical yield (101. 7 g Cu) multiplied by 100%. CHECK Are the units correct? Does the answer make physical sense? The theoretical yield has the right units (g Cu). The magnitude of theoretical yield seems reasonable because it is of the same order of magnitude as the given masses of C and Cu 2 O. The theoretical yield is reasonable because it is less than 100%. Any calculated theoretical yield above 100% would be suspect.

Example 8. 6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield Continued SKILLBUILDER 8. 6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield This reaction is used to obtain iron from iron ore: Fe 2 O 3(s) + 3 CO(g) 2 Fe(s) + 3 CO 2(g) 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. Answer: Limiting reactant is CO; theoretical yield = 127 g Fe; percent yield = 68. 8% For More Practice Example 8. 10; Problems 61, 62, 63, 64, 65, 66.

8. 7 Enthalpy: A Measure of the Heat Evolved or Absorbed in a Reaction � Chemical reactions can be exothermic (they emit thermal energy when they occur). � Chemical reactions can be endothermic (they absorb thermal energy when they occur). � The amount of thermal energy emitted or absorbed by a chemical reaction, under conditions of constant pressure (which are common for most everyday reactions), can be quantified with a function called enthalpy. © 2012 Pearson Education, Inc.

8. 7 Enthalpy: A Measure of the Heat Evolved or Absorbed in a Reaction � We define the enthalpy of reaction, ΔHrxn, as the amount of thermal energy (or heat) that flows when a reaction occurs at constant pressure. � The sign of ΔHrxn (positive or negative) depends on the direction in which thermal energy flows when the reaction occurs. � Energy flowing out of the chemical system is like a withdrawal and carries a negative sign. � Energy flowing into the system is like a deposit and carries a positive sign. © 2012 Pearson Education, Inc.

Figure 8. 3 Exothermic and Endothermic reactions � (a) In an exothermic reaction, energy is released into the surroundings. (b) In an endothermic reaction, energy is absorbed from the surroundings. © 2012 Pearson Education, Inc.

Sign of ΔHrxn �When thermal energy flows out of the reaction and into the surroundings (as in an exothermic reaction), then ΔHrxn is negative. �The enthalpy of reaction for the combustion of CH 4, the main component in natural gas: �This reaction is exothermic and therefore has a negative enthalpy of reaction. �The magnitude of ΔHrxn tells us that 802. 3 k. J of heat are emitted when 1 mol CH 4 reacts with 2 mol O 2. © 2012 Pearson Education, Inc.

Sign of ΔHrxn �When thermal energy flows into the reaction and out of the surroundings (as in an endothermic reaction), then ΔHrxn is positive. �The enthalpy of reaction for the reaction between nitrogen and oxygen gas to form nitrogen monoxide: �This reaction is endothermic and therefore has a positive enthalpy of reaction. �The magnitude of ΔHrxn tells us that 182. 6 k. J of heat are absorbed from the surroundings when 1 mol N 2 reacts with 1 mol O 2. © 2012 Pearson Education, Inc.

Stoichiometry of ΔHrxn � The amount of heat emitted or absorbed when a chemical reaction occurs depends on the amounts of reactants that actually react. � We usually specify ΔHrxn in combination with the balanced chemical equation for the reaction. � The magnitude of ΔHrxn is for the stoichiometric amounts of reactants and products for the reaction as written. © 2012 Pearson Education, Inc.

Stoichiometry of ΔHrxn �For example, the balanced equation and ΔHrxn for the combustion of propane (the fuel used in LP gas) is: �When 1 mole of C 3 H 8 reacts with 5 moles of O 2 to form 3 moles of CO 2 and 4 moles of H 2 O, 2044 k. J of heat are emitted. �These ratios can be used to construct conversion factors between amounts of reactants or products and the quantity of heat exchanged. © 2012 Pearson Education, Inc.

Stoichiometry of ΔHrxn � To find out how much heat is emitted upon the

Example 8. 7 Stoichiometry Involving ΔHrxn � An LP gas tank in a home barbecue contains 11. 8 x 103 g of propane (C 3 H 8). � Calculate the heat (in k. J) associated with the complete combustion of all of the propane in the tank. © 2012 Pearson Education, Inc.

Example: Complete combustion of 11. 8 x 103 g of propane (C 3 H 8) SOLUTION MAP: RELATIONSHIPS USED: 1 mol C 3 H 8 : – 2044 k. J (from balanced equation) Molar mass C 3 H 8 = 44. 11 g/mol © 2012 Pearson Education, Inc.

Example: Complete combustion of 11. 8 x 103 g of propane (C 3 H 8) SOLUTION: Often in the homework, the absolute value of Q, |Q|, is requested and words are used to convey the sign of the heat absorbed or given off in the reaction. © 2012 Pearson Education, Inc.

Everyday Chemistry Bunsen Burners � Most Bunsen burners have a mechanism to adjust the amount of air (and therefore of oxygen) that is mixed with the methane. � If you light the burner with the air completely closed off, you get a yellow, smoky flame that is not very hot. � As you increase the amount of air going into the burner, the flame becomes bluer, less smoky, and hotter. � When you reach the optimum adjustment, the flame has a sharp, inner blue triangle, gives off no smoke, and is hot enough to melt glass easily. � Continuing to increase the air beyond this point causes the flame to become cooler again and may actually extinguish it. © 2012 Pearson Education, Inc.