Introductory Chemistry, 2nd Edition Nivaldo Tro

Introductory Chemistry, 2nd Edition Nivaldo Tro

Introductory Chemistry, 3rd Edition Nivaldo Tro Chapter 8 Quantities in Chemical Reactions Basic Principles of Chemistry Online Southeast Missouri State University Cape Girardeau, MO 2009, Prentice Hall Global Warming Scientists have measured an average 0.6 C rise in atmospheric temperature since 1860. During the same period atmospheric CO2 levels have risen 25%. Are the two trends causal? Tro's Introductory Chemistry, Chapter 8 2 The Source of Increased CO2

The primary source of the increased CO 2 levels are combustion reactions of fossil fuels we use to get energy. 1860 corresponds to the beginning of the Industrial Revolution in the U.S. and Europe. CH 4(g) 2 O 2(g) CO 2(g) 2 H 2O(g) 2 C8H18(l) 25 O 2(g) 16 CO 2(g) 18 H 2O(g) Tro's Introductory Chemistry, Chapter 8 3 Quantities in Chemical Reactions The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction. Law of Conservation of Mass. Balancing equations by balancing atoms. The study of the numerical relationship between chemical quantities in a chemical

reaction is called stoichiometry. Tro's Introductory Chemistry, Chapter 8 4 Making Pancakes The number of pancakes you can make depends on the amount of the ingredients you use. 1 cup flour + 2 eggs + tsp baking powder 5 pancakes This relationship can be expressed mathematically. 1 cup flour 2 eggs tsp baking powder 5 pancakes Tro's Introductory Chemistry, Chapter 8 5 Making Pancakes, Continued If you want to make more or less than 5 pancakes, you can use the number of eggs you have to determine the number of pancakes you can make. Assuming you have enough flour and baking powder.

5 pancakes 8 eggs 20 pancakes 2 eggs Tro's Introductory Chemistry, Chapter 8 6 Making Molecules Mole-to-Mole Conversions The balanced equation is the recipe for a chemical reaction. The equation 3 H2(g) + N2(g) 2 NH3(g) tells us that 3 molecules of H2 react with exactly 1 molecule of N2 and make exactly 2 molecules of NH3 or: 3 molecules H2 1 molecule N2 2 molecules NH3 Since we count molecules by moles: 3 moles H2 1 mole N2 2 moles NH3 Tro's Introductory Chemistry, Chapter 8

7 Example 8.1How Many Moles of NaCl Result from the Complete Reaction of 3.4 Mol of Cl2? 2 Na(s) + Cl2(g) 2 NaCl Given: Find: Solution Map: 3.4 mol Cl2 mol NaCl mol Cl2 2 mol NaCl 1 mol Cl2 mol NaCl Relationships: 1 mol Cl2 2 NaCl Solution: 2 mol NaCl 3.4 mol Cl2 1 mol Cl2

6.8 mol NaCl Check: Since the reaction makes 2 molecules of NaCl for every 1 mole of Cl2, the number makes sense. Tro's Introductory Chemistry, Chapter 8 8 Practice According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C6H12O6 + 6 O2 6 CO2 + 6 H2O Tro's Introductory Chemistry, Chapter 8 16

How Many Moles of Water Are Made in the Combustion of 0.10 Moles of Glucose? Given: Find: Solution Map: Relationships: 0.10 moles C6H12O6 moles H2O mol C6H12O6 mol H2O 6 mol H 2O 1 mol C6 H12 O6 C6H12O6 + 6 O2 6 CO2 + 6 H2O 1 mol C6H12O6 6 mol H2O Solution: 6 mol H 2 O 0.10 mol C 6 H12O 6 0.6 mol H 2 O 1 mol C 6 H12O 6

Check: 0.6 mol H2O = 0.60 mol H2O Since 6x moles of H2O as C6H12O6, the number makes sense. Making Molecules Mass-to-Mass Conversions We know there is a relationship between the mass and number of moles of a chemical. 1 mole = Molar Mass in grams. The molar mass of the chemicals in the reaction and the balanced chemical equation allow us to convert from the amount of any chemical in the reaction to the amount of any other. 18 Example 8.2How Many Grams of Glucose Can Be Synthesized from 58.5 g of CO2 in Photosynthesis? Photosynthesis: 6 CO2(g) + 6 H2O(g) C6H12O6(s) + 6 O2(g) The equation for the reaction gives the mole

relationship between amount of C6H12O6 and CO2, but we need to know the mass relationship, so the solution map will be: g CO2 mol CO2 mol C6H12O6 Tro's Introductory Chemistry, Chapter 8 g C6H12O6 19 Example 8.2How Many Grams of Glucose Can Be Synthesized from 58.5 g of CO2 in Photosynthesis?, Continued Given: Find: 58.5 g CO2 g C6H12O6

Solution Map: g CO2 1 mol 44.01 g Relationships: mol CO2 mol C6H12O6 1 mol C 6 H12O 6 6 mol CO 2 g C6H12O6 180.2 g 1 mol 1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6 6 mol CO2 Solution: 1 mol CO 2 1 mol C 6 H12O 6 180.2 g C 6 H12O 6 58.5 g CO 2

44.01 g CO 2 6 mol CO 2 1 mol C 6 H12 O 6 39.9 g C 6 H12O 6 Check: Since 6x moles of CO2 as C6H12O6, but the molar mass of C6H12O6 is 4x CO2, the number makes sense. PracticeHow Many Grams of O2 Can Be Made from the Decomposition of 100.0 g of PbO2? 2 PbO2(s) 2 PbO(s) + O2(g) (PbO2 = 239.2, O2 = 32.00) Tro's Introductory Chemistry, Chapter 8 28 PracticeHow Many Grams of O2 Can Be Made from the Decomposition of 100.0 g of PbO2? 2 PbO2(s) 2 PbO(s) + O2(g), Continued Given:

Find: 100.0 g PbO2, 2 PbO2 2 PbO + O2 g O2 Solution Map: g PbO2 mol PbO2 1 mol 239.2 g Relationships: mol O2 1 mol O 2 2 mol PbO 2 g O2 32.00 g 1 mol 1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 2 mol PbO2

Solution: 1 mol PbO 2 1 mol O 2 32.00 g O 2 100.0 g PbO 2 239.2 g PbO 2 2 mol PbO 2 1 mol O 2 6.689 g O 2 Check: Since moles of O2 as PbO2, and the molar mass of PbO2 is 7x O2, the number makes sense. More Making Pancakes We know that: 1 cup flour + 2 eggs + tsp baking powder 5 pancakes But what would happen if we had 3 cups of flour, 10 eggs, and 4 tsp of baking powder? Tro's Introductory Chemistry, Chapter 8

30 More Making Pancakes, Continued 5 pancakes 3 cups flour 15 pancakes 1 cups flour 5 pancakes 10 eggs 25 pancakes 2 eggs 5 pancakes 4 tsp baking powder 40 pancakes 1 tsp baking powder 2 Tro's Introductory Chemistry, Chapter 8 31

More Making Pancakes, Continued Each ingredient could potentially make a different number of pancakes. But all the ingredients have to work together! We only have enough flour to make 15 pancakes, so once we make 15 pancakes, the flour runs out no matter how much of the other ingredients we have. 32 More Making Pancakes, Continued The flour limits the amount of pancakes we can make. In chemical reactions we call this the limiting reactant. Also known as limiting reagent. The maximum number of pancakes we can make depends on this ingredient. In chemical reactions, we call this the theoretical yield. It also determines the amounts of the other ingredients we will use! Tro's Introductory Chemistry, Chapter 8

33 Example 8.4What Is the Limiting Reactant and Theoretical Yield When 0.552 Mol of Al React with 0.887 Mol of Cl2? 2 Al(s) + 3 Cl2(g) 2 AlCl3 Given: Find: Solution Map: 0.552 mol Al, 0.887 mol Cl2 mol AlCl3 mol Al mol AlCl3 2 mol AlCl 3 2 mol Al Relationships: mol Cl2 Pick least

amount mol AlCl3 2 mol AlCl 3 3 mol Cl2 Limiting reactant and theoretical yield 3 mol Cl2 2 AlCl3; 2 mol Al 2 mol AlCl3 2 mol AlCl 3 Solution: Limiting 0.552 mol Al 2 mol Al Reactant 0.552 mol AlCl 3 2 mol AlCl 3 0.877 mol Cl2 3 mol Cl2

0.58Theoretical 47 mol AlCl 3 Yield PracticeHow Many Moles of Si3N4 Can Be Made from 1.20 Moles of Si and 1.00 Moles of N2 in the Reaction 3 Si + 2 N2 Si3N4? Tro's Introductory Chemistry, Chapter 8 42 PracticeHow Many Moles of Si3N4 Can Be Made from 1.20 Moles of Si and 1.00 Moles of N 2 in the Reaction 3 Si + 2 N2 Si3N4?, Continued Given: Find: Solution Map: 1.20 mol Si, 1.00 mol N2 mol Si3N4

mol Si mol Si3N4 Pick least amount 1 mol Si3 N 4 3 mol Si Relationships: mol N2 mol Si3N4 1 mol Si3 N 4 2 mol N 2 2 mol N2 1 Si3N4; 3 mol Si 1 1 mol Si3 N 4 Solution: Limiting

1.20 mol Si 3 mol Si Reactant 0.400 mol Si3 N 4 Limiting reactant and theoretical yield Si3N4 1 mol Si3 N 4 1.00 mol N 2 2 mol N 2 0.500 mol Si3 N 4 Theoretical Yield More Making Pancakes Lets now assume that as we are making pancakes, we spill some of the batter, burn a pancake, drop one

on the floor, or other uncontrollable events happen so that we only make 11 pancakes. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of making pancakes by calculating the percentage of the maximum number of pancakes we actually make. In chemical reactions, we call this the percent yield. Actual Yield 100% Percent Yield Theoretical Yield 11 pancakes 100% 73% 15 pancakes 44 Theoretical and Actual Yield As we did with the pancakes, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make. The theoretical yield will always be the least possible amount of product.

The theoretical yield will always come from the limiting reactant. Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield. Tro's Introductory Chemistry, Chapter 8 45 Measuring Amounts in the Lab In the lab, our balances do not measure amounts in moles, unfortunately, they measure amounts in grams. This means we must add two steps to each of our calculations: first convert the amount of each reactant to moles, then convert the amount of product into grams. Tro's Introductory Chemistry, Chapter 8 46

Example 8.6When 11.5 g of C Are Allowed to React with 114.5 g of Cu2O in the Reaction Below, 87.4 g of Cu Are Obtained. Cu2O(s) + C(s) 2 Cu(s) + CO(g) Given: Find: Solution Map: 11.5 g C, 114.5 g Cu2O, 87.4 g Cu Limiting reactant, theoretical yield, percent yield gC mol C 1 mol 12.01 g g Cu2O mol Cu 2 mol Cu 1 mol C mol Cu2O 1 mol

143.02 g g Cu 1 mol Choose 63.54 g smallest mol Cu 2 mol Cu 1 mol Cu 2O g Cu 1 mol 63.54 g Actual Yield 100% Percent Yield Theoretica l Yield Relationships: 1 mol C = 12.01g, 1 mol Cu2O = 143.02g, 1 mol Cu = 63.54 g, 2 mol Cu = 1 mol Cu, 2 mol Cu = 1 mol Cu2O

Example 8.6When 11.5 g of C Are Allowed to React with 114.5 g of Cu2O in the Reaction Below, 87.4 g of Cu Are Obtained. Cu2O(s) + C(s) 2 Cu(s) + CO(g), Continued Solution: 1 mol C 2 mol Cu 63.54 g Cu 11.5 g C 122 g Cu 12.01 g C 1 mol C 1 mol Cu 1 mol Cu 2O 2 mol Cu 63.54 g Cu 114 .5 g Cu 2O 101.7 g Cu 143.02 g Cu 2O 1 mol Cu 2O 1 mol Cu Actual Yield The smallest amount is 101.7

g Cu, therefore that is the theoretical yield. 100% Percent Yield Theoretica l Yield 101.7 g Cu is the Cu2O, The reactant that produces Therefore, Cu2O is the limiting reactant. 87.4 g Cu 100 % 85.9% Yield 101.7 g Cu Check: Since the percentage yield is < 100, the answer makes sense. PracticeHow Many Grams of N2(g) Can Be Made from 9.05 g of NH3 Reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 Are Made, What Is the Percent Yield? Tro's Introductory Chemistry,

Chapter 8 59 PracticeHow Many Grams of N2(g) Can Be Made from 9.05 g of NH3 Reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 Are Made, What Is the Percent Yield?, Continued Given: Find: 9.05 g NH3, 45.2 g CuO g N2 Solution Map: g NH 3 mol NH3 1 mol 17.03 g g CuO 1 mol 79.55 g

mol N2 1 mol N 2 2 mol NH 3 mol CuO mol N2 1 mol N 2 3 mol CuO g N2 1 mol 28.02 gChoose smallest g N2 1 mol 28.02 g

Actual Yield 100% Percent Yield Theoretica l Yield Relationships: 1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g 2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2 PracticeHow Many Grams of N2(g) Can Be Made from 9.05 g of NH3 Reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 Are Made, What Is the Percent Yield?, Continued Solution: 1 mol NH 3 1 mol N 2 28.02 g N 2 9.05 g NH 3 7.42 g N 2 17.03 g NH 3 2 mol NH 3 1 mol N 2 1 mol CuO 1 mol N 2 28.02 g N 2

45.2 g CuO 5.30 g N 2 79.55 g CuO 3 mol CuO 1 mol N 2 Theoretical 4.61 g N 2 100 % 87.0% Yield 5.30 g N 2 Check: yield Since the percent yield is less than 100, the answer makes sense. Enthalpy Change We previously described processes as exothermic if they released heat, or endothermic if they absorbed heat. The enthalpy of reaction is the amount of thermal energy that flows through a process. At constant pressure. Hrxn

Tro's Introductory Chemistry, Chapter 8 62 Sign of Enthalpy Change For exothermic reactions, the sign of the enthalpy change is negative when: Thermal energy is produced by the reaction. The surroundings get hotter. H = For the reaction CH4(s) + 2 O2(g) CO2(g) + 2 H2O(l), the Hrxn = 802.3 kJ per mol of CH4. For endothermic reactions, the sign of the enthalpy change is positive when: Thermal energy is absorbed by the reaction. The surroundings get colder. H = + For the reaction N2(s) + O2(g) 2 NO(g), the Hrxn = +182.6 kJ per mol of N2. 63

Enthalpy and Stoichiometry The amount of energy change in a reaction depends on the amount of reactants. You get twice as much heat out when you burn twice as much CH4. Writing a reaction implies that amount of energy changes for the stoichiometric amount given in the equation. For the reaction C3H8(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g) Hrxn = 2044 kJ So 1 mol C3H8 5 mol O2 3 mol CO2 4 mol H2O 2044 kJ. Tro's Introductory Chemistry, Chapter 8 64 Example 8.7How Much Heat Is Associated with the Complete Combustion of 11.8 x 103 g of C3H8(g)? Given: Find: Solution Map:

11.8 x 103 g C3H8, heat, kJ g C3H8 mol C3H8 1 mol C3H 8 44.09 g Relationships: kJ - 2044 kJ 1 mol C 3 H 8 1 mol C3H8 = -2044 kJ, Molar mass = 44.11 g/mol Solution: 1 mol C3H8 - 2044 kJ 11.8 103 g C3H8 5.47 105 kJ

44.11 g C3H8 1 mol C3H8 Check: The sign is correct and the value is reasonable. PracticeHow Much Heat Is Evolved When a 0.483 g Diamond Is Burned? (Hcombustion = 395.4 kJ/mol C) Tro's Introductory Chemistry, Chapter 8 66 PracticeHow Much Heat Is Evolved When a 0.483 g Diamond Is Burned? (Hcombustion = 395.4 kJ/mol C), Continued Given: Find: Solution Map: Relationships: 0.483 g C heat, kJ gC

1 mol C 12.01 g mol C kJ - 395.4 kJ 1 mol C 1 mol C = 395.4 kJ, Molar mass = 12.01 g/mol Solution: 1 mol C - 395.4 kJ 0.483 g C 15.9 kJ 12.01 g 1 mol Check: The sign is correct and the value is reasonable

since there is less than 0.1 mol C.

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