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Chapter 8 Quantities in Chemical Reactions. 2Tro's “Introductory Chemistry”, Chapter 8 Chemistry in the Kitchen: Making Pancakes The number of pancakes.

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Presentation on theme: "Chapter 8 Quantities in Chemical Reactions. 2Tro's “Introductory Chemistry”, Chapter 8 Chemistry in the Kitchen: Making Pancakes The number of pancakes."— Presentation transcript:

1 Chapter 8 Quantities in Chemical Reactions

2 2Tro's “Introductory Chemistry”, Chapter 8 Chemistry in the Kitchen: Making Pancakes The number of pancakes you can make depends on the amount of the ingredients you use. This relationship can be expressed mathematically. 1 cup flour  2 eggs  ½ tsp baking powder  5 pancakes 1 cup flour + 2 eggs + ½ tsp baking powder  5 pancakes

3 3Tro's “Introductory Chemistry”, Chapter 8 Making Pancakes, Continued Assume that you have enough flour and baking powder. You can use the number of eggs you have to determine the number of pancakes you can make.

4 4Tro's “Introductory Chemistry”, Chapter 8 Quantities in Chemical Reactions In Chemistry, the ‘recipe’ for making chemical compounds is called the balanced equation. For instance, the equation 3 H 2 (g) + N 2 (g)  2 NH 3 (g) tells us that in order to make 2 molecules of ammonia, 3 molecules of hydrogen have to react with 1 molecule of nitrogen. 3 molecules H 2  1 molecule N 2  2 molecules NH 3 Since we count molecules by moles: 3 moles H 2  1 mole N 2  2 moles NH 3

5 5Tro's “Introductory Chemistry”, Chapter 8 Stoichiometry: Mole-to-Mole Conversions The study of the numerical relationship between chemical quantities in a balanced chemical reaction is called stoichiometry. Example 8.1 —How Many Moles of NaCl Result from the Complete Reaction of 3.4 Mol of Cl 2 ? 2 Na(s) + Cl 2 (g) → 2 NaCl

6 6Tro's “Introductory Chemistry”, Chapter 8 Example 8.1 —How Many Moles of NaCl Result from the Complete Reaction of 3.4 Mol of Cl 2 ? 2 Na(s) + Cl 2 (g) → 2 NaCl Since the reaction makes 2 molecules of NaCl for every 1 mole of Cl 2, the number makes sense. 1 mol Cl 2  2 NaCl 3.4 mol Cl 2 mol NaCl Check: Solution: Solution Map: Relationships: Given: Find: mol Cl 2 mol NaCl

7 7Tro's “Introductory Chemistry”, Chapter 8 Practice According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose (C 6 H 12 O 6 )? C 6 H 12 O O 2  6 CO H 2 O

8 8 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.

9 9Tro's “Introductory Chemistry”, Chapter 8 Example 8.2—How Many Grams of Glucose Can Be Synthesized from 58.5 g of CO 2 in Photosynthesis? 6 CO 2 (g) + 6 H 2 O(g)  C 6 H 12 O 6 (s) + 6 O 2 (g) The equation for the reaction gives the mole relationship between amount of C 6 H 12 O 6 and CO 2, but we need to know the mass relationship. g C 6 H 12 O 6 mol CO 2 g CO 2 mol C 6 H 12 O 6

10 Example 8.2—How Many Grams of Glucose Can Be Synthesized from 58.5 g of CO 2 in Photosynthesis?, Continued 1 mol C 6 H 12 O 6 = 180.2g, 1 mol CO 2 = 44.01g, 1 mol C 6 H 12 O 6  6 mol CO g CO 2 g C 6 H 12 O 6 Solution: Solution Map: Relationships: Given: Find: g C 6 H 12 O 6 mol CO 2 g CO 2 mol C 6 H 12 O 6

11 11Tro's “Introductory Chemistry”, Chapter 8 Practice—How Many Grams of O 2 Can Be Made from the Decomposition of g of PbO 2 ? 2 PbO 2 (s) → 2 PbO(s) + O 2 (g) (PbO 2 = 239.2, O 2 = 32.00)

12 12Tro's “Introductory Chemistry”, Chapter 8 More Making Pancakes Recall that: What would happen if we had 3 cups of flour, 10 eggs, and 4 tsp of baking powder? 1 cup flour + 2 eggs + ½ tsp baking powder  5 pancakes

13 13Tro's “Introductory Chemistry”, Chapter 8 More Making Pancakes, Continued

14 14 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.

15 15Tro's “Introductory Chemistry”, Chapter 8 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.

16 Example 8.4—What Is the Limiting Reactant and Theoretical Yield When Mol of Al React with Mol of Cl 2 ? 2 Al(s) + 3 Cl 2 (g) → 2 AlCl 3 3 mol Cl 2  2 AlCl 3 ; 2 mol Al  2 mol AlCl mol Al, mol Cl 2 mol AlCl 3 Solution: Solution Map: Relationships: Given: Find: mol Cl 2 mol AlCl 3 mol Almol AlCl 3 Pick least amount Limiting reactant and theoretical yield Limiting Reactant and Theoretical Yield

17 17Tro's “Introductory Chemistry”, Chapter 8 Practice—How Many Moles of Si 3 N 4 Can Be Made from 1.20 Moles of Si and 1.00 Moles of N 2 in the Reaction 3 Si + 2 N 2  Si 3 N 4 ?

18 18 More Making Pancakes Assume that in making the pancake, you spill some of the pancake mix, 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 and this is known as the percent yield.

19 19Tro's “Introductory Chemistry”, Chapter 8 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. The actual yield of product will always be less than the theoretical yield.

20 20Tro's “Introductory Chemistry”, Chapter 8 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.

21 Example 8.6—When 11.5 g of C Are Allowed to React with g of Cu 2 O in the Reaction Below, 87.4 g of Cu Are Obtained. Cu 2 O(s) + C(s)  2 Cu(s) + CO(g) 1 mol C = 12.01g, 1 mol Cu 2 O = g, 1 mol Cu = g, 2 mol Cu = 1 mol Cu, 2 mol Cu = 1 mol Cu 2 O 11.5 g C, g Cu 2 O, 87.4 g Cu Limiting reactant, theoretical yield, percent yield Solution Map: Relationships: Given: Find: g Cmol Cumol Cg Cug Cu 2 Omol Cumol Cu 2 Og Cu Choose smallest

22 Since the percentage yield is < 100, the answer makes sense. Check: Solution: Example 8.6—When 11.5 g of C Are Allowed to React with g of Cu 2 O in the Reaction Below, 87.4 g of Cu Are Obtained. Cu 2 O(s) + C(s)  2 Cu(s) + CO(g), Continued The smallest amount is g Cu, therefore that is the theoretical yield. The reactant that produces g Cu is the Cu 2 O, Therefore, Cu 2 O is the limiting reactant.

23 23Tro's “Introductory Chemistry”, Chapter 8 Practice—How Many Grams of N 2 (g) Can Be Made from 9.05 g of NH 3 Reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l) If 4.61 g of N 2 Are Made, What Is the Percent Yield?

24 24Tro's “Introductory Chemistry”, Chapter 8

25 25Tro's “Introductory Chemistry”, Chapter 8 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.  H rxn

26 26 Sign of Enthalpy Change For exothermic reactions, the sign of the enthalpy change is negative when:  H = ─ Thermal energy is produced by the reaction. The surroundings get hotter. For the reaction CH 4 (s) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l), the  H rxn = −802.3 kJ per mol of CH 4. For endothermic reactions, the sign of the enthalpy change is positive when:  H = + Thermal energy is absorbed by the reaction. The surroundings get colder. For the reaction N 2 (s) + O 2 (g)  2 NO(g), the  H rxn = kJ per mol of N 2.

27 27Tro's “Introductory Chemistry”, Chapter 8 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 CH 4. Writing a reaction implies that amount of energy changes for the stoichiometric amount given in the equation. For the reaction C 3 H 8 (l) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g)  H rxn = −2044 kJ So 1 mol C 3 H 8  5 mol O 2  3 mol CO 2  4 mol H 2 O  −2044 kJ.

28 Example 8.7—How Much Heat Is Associated with the Complete Combustion of 11.8 x 10 3 g of C 3 H 8 (g)?  H = kJ/mol 1 mol C 3 H 8 = kJ, Molar mass = g/mol The sign makes sense x 10 3 g C 3 H 8, heat, kJ Check: Solution: Solution Map: Relationships: Given: Find: mol C 3 H 8 kJg C 3 H 8

29 29Tro's “Introductory Chemistry”, Chapter 8 Practice—How Much Heat Is Evolved When a g Diamond Is Burned? (  H combustion = −395.4 kJ/mol C)

30 30Tro's “Introductory Chemistry”, Chapter 8 Recommended Study Problems Chapter 8 NB: Study problems are used to check the student’s understanding of the lecture material. Students are EXPECTED TO BE ABLE TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS. If you encounter any problems, please talk to your professor or seek help at the HACC-Gettysburg learning center. Questions from text book Chapter 8, p 265 5, 9, 11, 17, 19, 23, 25, 31, 33, 35, 37, 41, 43, 45, 49, 53, 57, 59, 61, 65, 69, 71, 73, 91 ANSWERS -The answers to the odd-numbered study problems are found at the back of your textbook


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