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Quantities in Chemical Reactions

Quantities in Chemical Reactions
Chapter 8 Quantities in Chemical ReactionsGlobal Warming • Scientists have measured an average 0.6 °C rise in atmospheric temperature since 1860. • During the same period atmospheric CO 2 levels have risen 25. • Are the two trends causal www.ThesisScientist.comThe Source of Increased CO 2 • 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 (g)  2 O (g)  CO (g)  2 H O(g) 2 C H (l)  25 O (g)  16 CO (g)  18 H O(g) 4 2 2 2 8 18 2 2 2 www.ThesisScientist.comQuantities 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. www.ThesisScientist.comCounting and ratio’s It takes me .2 gal of gas to get to IVC. It is a very simple ratio: = What if I only had .1 gal 200 2X4’s, 3 sinks, 2 showers, you can make a house with 3 bathrooms and 3 bedrooms. What if you had 12 sinks…how many houses could you make. 200 3 2 = 1 3 3 www.ThesisScientist.comMaking Pancakes • The number of pancakes you can make depends on the amount of the ingredients you use. 1 cup flour + 2 eggs + ½ tsp baking powder  5 pancakes • This relationship can be expressed mathematically. 1 cup flour  2 eggs  ½ tsp baking powder  5 pancakes www.ThesisScientist.comMaking 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 www.ThesisScientist.comMaking Molecules MoletoMole Conversions • The balanced equation is the “recipe” for a chemical reaction. • The equation 3 H (g) + N (g)  2 NH (g) tells us 2 2 3 that 3 molecules of H react with exactly 1 molecule 2 of N and make exactly 2 molecules of NH or: 2 3 3 molecules H 1 molecule N 2 molecules NH 2 2 3 • Since we count molecules by moles: 3 moles H 1 mole N 2 moles NH 2 2 3 www.ThesisScientist.comExample 8.1: • 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 there is more than enough Na. 2 Na(s) + Cl (g)  2 NaCl(s) 2 www.ThesisScientist.comExample: How many moles of NaCl result from the complete reaction of 3.4 mol of Cl in 2 the reaction below 2 Na(s) + Cl (g)  2 NaCl(s) 2 • Write down the given quantity and its units. Given: 3.4 mol Cl 2 www.ThesisScientist.comExample: Information: How many moles of NaCl Given: 3.4 mol Cl 2 result from the complete reaction of 3.4 mol of Cl in 2 the reaction below 2 Na(s) + Cl (g)  2 NaCl(s) 2 • Write down the quantity to find and/or its units. Find: moles NaCl www.ThesisScientist.comExample: Information: How many moles of NaCl Given: 3.4 mol Cl 2 result from the complete Find: moles NaCl reaction of 3.4 mol of Cl in 2 the reaction below 2 Na(s) + Cl (g)  2 NaCl(s) 2 • Collect needed conversion factors: According to the equation: 1 mole Cl 2 moles NaCl 2 www.ThesisScientist.comExample: Information: How many moles of NaCl Given: 3.4 mol Cl 2 result from the complete Find: moles NaCl reaction of 3.4 mol of Cl in 2 Conversion Factor: the reaction below 1 mol Cl 2 mol NaCl 2 Na(s) + Cl (g)  2 NaCl(s) 2 2 • Write a solution map for converting the units: mol Cl mol NaCl 2 2 mol NaCl 1 mol Cl 2 www.ThesisScientist.comInformation: Example: Given: 3.4 mol Cl How many moles of NaCl 2 Find: moles NaCl result from the complete Conversion Factor: reaction of 3.4 mol of Cl in 2 1 mol Cl 2 mol NaCl the reaction below 2 Solution Map: mol Cl mol NaCl 2 Na(s) + Cl (g)  2 NaCl(s) 2 2 • Apply the solution map: 2 mol NaCl 3.4 mol Cl moles NaCl 2 1 mol Cl 2 = 6.8 mol NaCl • Significant figures and round: = 6.8 moles NaCl www.ThesisScientist.comInformation: Example: Given: 3.4 mol Cl How many moles of NaCl 2 Find: moles NaCl result from the complete Conversion Factor: reaction of 3.4 mol of Cl in 2 1 mol Cl 2 mol NaCl the reaction below 2 Solution Map: mol Cl mol NaCl 2 Na(s) + Cl (g)  2 NaCl(s) 2 2 • Check the solution: 3.4 mol Cl 6.8 mol NaCl 2 The units of the answer, moles NaCl, are correct. The magnitude of the answer makes sense because the equation tells us you make twice as many moles of NaCl as the moles of Cl . 2 www.ThesisScientist.comPractice • According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose C H O + 6 O 6 CO + 6 H O 6 12 6 2 2 2 www.ThesisScientist.comHow Many Moles of Water Are Made in the Combustion of 0.10 Moles of Glucose Given: 0.10 moles C H O 6 12 6 Find: moles H O 2 Solution Map: mol C H O mol H O 6 12 6 2 6 mol H O 2 1 mol C H O 6 12 6 C H O + 6 O → 6 CO + 6 H O  1 mol C H O 6 mol H O Relationships: 6 12 6 2 2 2 6 12 6 2 Solution: 6 mol H O 2 0.10 mol C H O 0.6 mol H O 6 12 6 2 1 mol C H O 6 12 6 Check: 0.6 mol H O = 0.60 mol H O 2 2 Since 6x moles of H O as C H O , the number makes sense. 2 6 12 6 www.ThesisScientist.comMaking Molecules MasstoMass 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. www.ThesisScientist.comExample 8.2—How Many Grams of Glucose Can Be Synthesized from 58.5 g of CO in 2 Photosynthesis • Photosynthesis: 6 CO (g) + 6 H O(g)  C H O (s) + 6 O (g) 2 2 6 12 6 2 • The equation for the reaction gives the mole relationship between amount of C H O and 6 12 6 CO , but we need to know the mass relationship, 2 so the solution map will be: mol C H O g C H O g CO mol CO 6 12 6 6 12 6 2 2 www.ThesisScientist.comExample 8.2: • In photosynthesis, plants convert carbon dioxide and water into glucose (C H O ), according to the following 6 12 6 reaction. How many grams of glucose can be synthesized from 58.5 g of CO Assume there is more than enough 2 water to react with all the CO . 2 sunlight 6 CO (g)  6 H O(l)  6 O (g)  C H O (aq) 2 2 2 6 12 6 www.ThesisScientist.comExample: How many grams of glucose can be synthesized from 58.5 g of CO in the reaction 2 6 CO (g) + 6 H O(l)  2 2 6 O (g) + C H O (aq) 2 6 12 6 • Write down the given quantity and its units. Given: 58.5 g CO 2 www.ThesisScientist.comExample: Information: How many grams of glucose Given: 55.4 g CO 2 can be synthesized from 58.5 g of CO in the reaction 2 6 CO (g) + 6 H O(l)  2 2 6 O (g) + C H O (aq) 2 6 12 6 • Write down the quantity to find and/or its units. Find: g C H O 6 12 6 www.ThesisScientist.comExample: Information: How many grams of glucose Given: 55.4 g CO 2 can be synthesized from 58.5 g Find: g C H O 6 12 6 of CO in the reaction 2 6 CO (g) + 6 H O(l)  2 2 6 O (g) + C H O (aq) 2 6 12 6 • Collect needed conversion factors: Molar mass C H O = 6(mass C) + 12(mass H) + 6(mass O) 6 12 6 = 6(12.01) + 12(1.01) + 6(16.00) = 180.2 g/mol Molar mass CO = 1(mass C) + 2(mass O) 2 = 1(12.01) + 2(16.00) = 44.01 g/mol 1 mole CO = 44.01 g CO 2 2 1 mole C H O = 180.2 g C H O 6 12 6 6 12 6 1 mole C H O 6 mol CO (from the chem. equation) 6 12 6 2 www.ThesisScientist.comInformation: Example: How many grams of glucose Given: 58.5 g CO 2 can be synthesized from 58.5 g Find: g C H O 6 12 6 of CO in the reaction 2 Conversion Factors: 6 CO (g) + 6 H O(l)  2 2 1 mol C H O = 180.2 g 6 12 6 6 O (g) + C H O (aq) 2 6 12 6 1 mol CO = 44.01 g 2 1 mol C H O 6 mol CO 6 12 6 2 • Write a solution map: g mol mol g CO CO C H O C H O 2 2 6 12 6 6 12 6 1 mol CO 1 mol C H O 180.2 g C H O 2 6 12 6 6 12 6 44.01 g CO 6 mol CO 1 mol C H O 2 2 6 12 6 www.ThesisScientist.comInformation: Example: Given: 58.5 g CO 2 How many grams of glucose Find: g C H O 6 12 6 can be synthesized from 58.5 g Conversion Factors: of CO in the reaction 2 1 mol C H O = 180.2 g 6 12 6 6 CO (g) + 6 H O(l)  2 2 1 mol CO = 44.01 g 2 6 O (g) + C H O (aq) 2 6 12 6 1 mol C H O 6 mol CO 6 12 6 2 Solution Map: g CO mol CO 2 2 mol C H O g C H O 6 12 6 6 12 6 • Apply the solution map: 1 mole CO 1 mol C H O 180.2 g C H O 2 6 12 6 6 12 6 58.5 g CO 2 44.01 g CO 6 mol CO 1 mole C H O 2 2 6 12 6 = 39.9216 g C H O 6 12 6 • Significant figures and round: = 39.9 g C H O 6 12 6 www.ThesisScientist.comInformation: Example: Given: 58.5 g CO 2 How many grams of glucose Find: g C H O 6 12 6 can be synthesized from 58.5 g Conversion Factors: of CO in the reaction 2 1 mol C H O = 180.2 g 6 12 6 6 CO (g) + 6 H O(l)  2 2 1 mol CO = 44.01 g 2 6 O (g) + C H O (aq) 2 6 12 6 1 mol C H O 6 mol CO 6 12 6 2 Solution Map: g CO mol CO 2 2 mol C H O g C H O 6 12 6 6 12 6 • Check the solution: 58.5 g CO = 39.9 g C H O 2 6 12 6 The units of the answer, g C H O , are correct. 6 12 6 It is hard to judge the magnitude. www.ThesisScientist.comPractice—How Many Grams of O Can Be Made from 2 the Decomposition of 100.0 g of PbO 2 2 PbO (s) → 2 PbO(s) + O (g) 2 2 (PbO = 239.2, O = 32.00) 2 2 www.ThesisScientist.comPractice—How Many Grams of O Can Be Made from the 2 Decomposition of 100.0 g of PbO 2 2 PbO (s) → 2 PbO(s) + O (g), Continued 2 2 Given: 100.0 g PbO , 2 PbO → 2 PbO + O 2 2 2 Find: g O 2 Solution Map: g PbO mol PbO mol O g O 2 2 2 2 1 mol 1 mol O 32.00 g 2 239.2 g 2 mol PbO 1 mol 2 1 mol O = 32.00g, 1 mol PbO = 239.2g, 1 mol O ≡ 2 mol PbO Relationships: 2 2 2 2 Solution: 1 mol PbO 1 mol O 32.00 g O 2 2 2 100.0 g PbO 2 239.2 g PbO 2 mol PbO 1 mol O 2 2 2  6.689 g O 2 Check: Since ½ moles of O as PbO , and the molar mass of PbO is 7x 2 2 2 O , the number makes sense. 2 www.ThesisScientist.comMore Making Pancakes • We know that: 1 cup flour + 2 eggs + ½ tsp baking powder  5 pancakes • But what would happen if we had 3 cups of flour, 10 eggs, and 4 tsp of baking powder www.ThesisScientist.comMore 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 www.ThesisScientist.comMore 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. www.ThesisScientist.comMore 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 www.ThesisScientist.comExample 8.4: • What is the limiting reactant and theoretical yield when 0.552 mol of Al react with 0.887 mol of Cl 2 2 Al(s) + 3 Cl (g) → 2 AlCl 2 3 www.ThesisScientist.comExample: What is the limiting reactant and theoretical yield when 0.552 mol of Al react with 0.887 mol of Cl 2 2 Al(s) + 3 Cl (g) → 2 AlCl 2 3 • Write down the given quantity and its units. Given: 0.552 mol Al 0.877 mol Cl 2 www.ThesisScientist.comExample: Information: What is the limiting reactant Given: 0.552 mol Al, 0.877 mol Cl 2 and theoretical yield when 0.552 mol of Al react with 0.887 mol of Cl 2 2 Al(s) + 3 Cl (g) → 2 AlCl 2 3 • Write down the quantity to find and/or its units. Find: limiting reactant theoretical yield www.ThesisScientist.comInformation: Example: Given: 0.552 mol Al, 0.877 mol Cl What is the limiting reactant 2 Find: limiting reactant, theor. yield and theoretical yield when 0.552 mol of Al react with 0.887 mol of Cl 2 2 Al(s) + 3 Cl (g) → 2 AlCl 2 3 • Collect needed conversion factors: 2 mol AlCl 2 mol Al (from the chem. equation) 3 2 mol AlCl 3 mol Cl (from the chem. equation) 3 2 www.ThesisScientist.comInformation: Example: Given: 0.552 mol Al, 0.877 mol Cl What is the limiting reactant 2 Find: limiting reactant, theor. yield and theoretical yield when Conversion Factors: 0.552 mol of Al react with 0.887 mol of Cl 2 mol AlCl 2 mol Al, 3 2 2 Al(s) + 3 Cl (g) → 2 AlCl 2 mol AlCl 3 mol Cl 2 3 3 2 • Write a solution map: mol mol Smallest Al AlCl 2 mol AlCl 3 3 amount is 2 mol Al theoretical mol mol yield 2 mol AlCl AlCl Cl 3 3 2 3 mol Cl 2 www.ThesisScientist.comInformation: Example: Given: 0.552 mol Al, 0.877 mol Cl 2 What is the limiting reactant Find: limiting reactant, theor. yield Conversion Factors: and theoretical yield when 2 mol AlCl 2 mol Al, 0.552 mol of Al react with 3 2 mol AlCl 3 mol Cl 3 2 0.887 mol of Cl 2 Solution Map: 2 Al(s) + 3 Cl (g) → 2 AlCl 2 3 mol each reactant  mol AlCl 3 • Apply the solution map: Limiting reactant = Al 2 mol AlCl 2 mol AlCl 3 3 0.552 mol Al 0.887 mol Cl 2 3 mol Cl 2 mol Al 2  0.591 mol AlCl  0.552 mol AlCl 3 3 Smallest Theoretical yield = 0.552 mol AlCl 3 amount www.ThesisScientist.comInformation: Example: Given: 0.552 mol Al, 0.877 mol Cl 2 What is the limiting reactant Find: limiting reactant, theor. yield Conversion Factors: and theoretical yield when 2 mol AlCl 2 mol Al, 0.552 mol of Al react with 3 2 mol AlCl 3 mol Cl 3 2 0.887 mol of Cl 2 Solution Map: 2 Al(s) + 3 Cl (g) → 2 AlCl 2 3 mol each reactant  mol AlCl 3 • Check the solution: Limiting reactant = Al Theoretical yield = 0.552 mol AlCl 3 Usually hard to judge as there are multiple factors, but because Al resulted in smallest amount of AlCl , 3 the answer makes sense. www.ThesisScientist.comMore Making Pancakes • Let’s 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. 11 pancakes Actual Yield 100 73 100 Percent Yield 15 pancakes Theoretical Yield www.ThesisScientist.comTheoretical 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. www.ThesisScientist.comMeasuring 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. www.ThesisScientist.comExample 8.6: • When 11.5 g of C are allowed to react with 114.5 g of Cu O in the reaction below, 87.4 g of Cu are obtained. 2 Find the limiting reactant, theoretical yield, and percent yield. Cu O(s)  C(s)  2 Cu(s)  CO(g) 2 www.ThesisScientist.comExample: When 11.5 g of C reacts with 114.5 g of Cu O, 87.4 g of Cu are 2 obtained. Find the limiting reactant, theoretical yield, and percent yield. Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 • Write down the given quantity and its units. Given: 11.5 g C 114.5 g Cu O 2 87.4 g Cu produced www.ThesisScientist.comExample: Information: When 11.5 g of C reacts with Given: 11.5 g C, 114.5 g Cu O 2 114.5 g of Cu O, 87.4 g of Cu are 2 87.4 g Cu produced obtained. Find the limiting reactant, theoretical yield, and percent yield. Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 • Write down the quantity to find and/or its units. Find: limiting reactant theoretical yield percent yield www.ThesisScientist.comInformation: Example: Given: 11.5 g C, 114.5 g Cu O When 11.5 g of C reacts with 2 87.4 g Cu produced 114.5 g of Cu O, 87.4 g of Cu are 2 Find: lim. rct., theor. yld., yld. obtained. Find the limiting reactant, theoretical yield, and percent yield. Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 • Collect needed conversion factors: Molar mass Cu O = 143.02 g/mol 2 Molar mass Cu = 63.54 g/mol Molar mass C = 12.01 g/mol 1 mol Cu O  2 mol Cu (from the chem. equation) 2 1 mol C  2 mol Cu (from the chem. equation) www.ThesisScientist.comInformation: Example: Given: 11.5 g C, 114.5 g Cu O 2 When 11.5 g of C reacts with 87.4 g Cu produced 114.5 g of Cu O, 87.4 g of Cu are 2 Find: lim. rct., theor. yld., yld. obtained. Find the limiting Conversion Factors: 1 mol C = 12.01 g; reactant, theoretical yield, and 1 mol Cu = 63.54 g; 1 mol Cu O = 2 143.08 g; 1 mol Cu O  2 mol Cu; percent yield. 2 1 mol C  2 mol Cu Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 • Write a solution map: g mol mol g Smallest C C Cu Cu 1 mol C 2 mol Cu 63.54 g Cu amount is 12.01 g C 1 mol C 1 mol Cu theoretical Yield. g mol mol g Cu O Cu O Cu Cu 2 2 1 mol Cu O 2 mol Cu 63.54 g Cu 2 143.08 g Cu O 1 mol Cu O 1 mol Cu 2 2 www.ThesisScientist.comInformation: Example: Given: 11.5 g C, 114.5 g Cu O 2 When 11.5 g of C reacts with 87.4 g Cu produced 114.5 g of Cu O, 87.4 g of Cu are 2 Find: lim. rct., theor. yld., yld. obtained. Find the limiting Conversion Factors: 1 mol C = 12.01 g; 1 mol Cu = 63.54 g; 1 mol Cu O = 2 reactant, theoretical yield, and 143.08 g; 1 mol Cu O  2 mol Cu; 2 percent yield. 1 mol C  2 mol Cu Solution Map: Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 g rct  mol rct  mol Cu  g Cu • Apply the solution map: 1 mole C 2 mol Cu 63.54 g Cu 11.5 g C122 g Cu 12.01 g C 1 mol C 1 mole Cu 1 mole Cu O 2 mol Cu 63.54 g Cu 2 114.5 g Cu O101.7 g Cu 2 143.08 g Cu O 1 mol Cu O 1 mole Cu 2 2 www.ThesisScientist.comInformation: Example: Given: 11.5 g C, 114.5 g Cu O 2 When 11.5 g of C reacts with 87.4 g Cu produced 114.5 g of Cu O, 87.4 g of Cu are 2 Find: lim. rct., theor. yld., yld. obtained. Find the limiting Conversion Factors: 1 mol C = 12.01 g; 1 mol Cu = 63.54 g; 1 mol Cu O = 2 reactant, theoretical yield, and 143.08 g; 1 mol Cu O  2 mol Cu; 2 percent yield. 1 mol C  2 mol Cu Solution Map: Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 g rct  mol rct  mol Cu  g Cu • Apply the solution map: 11.5 g C can make 122 g Cu Theoretical yield = 101.7 g Cu 1 114.5 14.5 g g C Cu u O c O ca an m n ma ake ke 101.7 g 101.7 g Cu Cu Limiting reactant = Cu O 2 2 2 Least amount www.ThesisScientist.comInformation: Example: Given: 11.5 g C, 114.5 g Cu O 2 When 11.5 g of C reacts with 87.4 g Cu produced 114.5 g of Cu O, 87.4 g of Cu are 2 Find: lim. rct., theor. yld., yld. obtained. Find the limiting Conversion Factors: 1 mol C = 12.01 g; reactant, theoretical yield, and 1 mol Cu = 63.54 g; 1 mol Cu O = 2 143.08 g; 1 mol Cu O  2 mol Cu; percent yield. 2 1 mol C  2 mol Cu Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 • Write a solution map: Actual Yield 100 Percent Yield Theoretical Yield www.ThesisScientist.comInformation: Example: Given: 11.5 g C, 114.5 g Cu O 2 When 11.5 g of C reacts with 87.4 g Cu produced 114.5 g of Cu O, 87.4 g of Cu are 2 Find: lim. rct., theor. yld., yld. obtained. Find the limiting Conversion Factors: 1 mol C = 12.01 g; 1 mol Cu = 63.54 g; 1 mol Cu O = 2 reactant, theoretical yield, and 143.08 g; 1 mol Cu O  2 mol Cu; 2 percent yield. 1 mol C  2 mol Cu Solution Map: Actual Yield Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 100 Percent Yield Theoretical Yield • Apply the solution map: Actual Yield 100 Percent Yield Theoretical Yield 87.4 g Cu 100 85.9 101.7 g Cu www.ThesisScientist.comInformation: Example: Given: 11.5 g C, 114.5 g Cu O 2 When 11.5 g of C reacts with 87.4 g Cu produced 114.5 g of Cu O, 87.4 g of Cu are 2 Find: lim. rct., theor. yld., yld. obtained. Find the limiting Conversion Factors: 1 mol C = 12.01 g; reactant, theoretical yield, and 1 mol Cu = 63.54 g; 1 mol Cu O = 2 143.08 g; 1 mol Cu O  2 mol Cu; percent yield. 2 1 mol C  2 mol Cu Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 • Check the solutions: Limiting reactant = Cu O 2 Theoretical yield = 101.7 g Percent yield = 85.9 The percent yield makes sense as it is less than 100. www.ThesisScientist.comExample 8.6—When 11.5 g of C Are Allowed to React with 114.5 g of Cu O in the Reaction Below, 87.4 g of Cu Are Obtained. 2 Cu O(s) + C(s)  2 Cu(s) + CO(g) 2 Given: 11.5 g C, 114.5 g Cu O, 87.4 g Cu 2 Find: Limiting reactant, theoretical yield, percent yield Solution Map: g C mol C mol Cu g Cu 1 mol 1 mol 2 mol Cu Choose 1 mol C 63.54 g 12.01 g smallest g Cu O mol Cu O mol Cu g Cu 2 2 1 mol 2 mol Cu 1 mol 1 mol Cu O 143.02 g 63.54 g 2 Actual Yield 100 Percent Yield Theoretica l Yield 1 mol C = 12.01g, 1 mol Cu O = 143.02g, 1 mol Cu = 63.54 g, Relationships: 2 2 mol Cu = 1 mol Cu, 2 mol Cu = 1 mol Cu O 2 www.ThesisScientist.comExample 8.6—When 11.5 g of C Are Allowed to React with 114.5 g of Cu O in the Reaction Below, 87.4 g of Cu Are Obtained. 2 Cu O(s) + C(s)  2 Cu(s) + CO(g), Continued 2 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 O 2 mol Cu 63.54 g Cu 2 114.5 g Cu O 101.7 g Cu 2 143.02 g Cu O 1 mol Cu O 1 mol Cu 2 2 Actual Yield The smallest amount is 101.7 g Cu, therefore that is the theoretical yield. 100 Percent Yield Theoretica l Yield The reactant that produces 101.7 g Cu is the Cu O, 2 Therefore, Cu O is the limiting reactant. 2 87.4 g Cu 100 85.9 Yield 101.7 g Cu Check: Since the percentage yield is 100, the answer www.ThesisScientist.com makes sense.Practice—How Many Grams of N (g) Can Be Made from 2 9.05 g of NH Reacting with 45.2 g of CuO If 4.61 g of 3 N Are Made, What Is the Percent Yield 2 2 NH (g) + 3 CuO(s) → N (g) + 3 Cu(s) + 3 H O(l) 3 2 2 www.ThesisScientist.comPractice—How Many Grams of N (g) Can Be Made from 9.05 g of NH Reacting with 2 3 45.2 g of CuO 2 NH (g) + 3 CuO(s) → N (g) + 3 Cu(s) + 3 H O(l) 3 2 2 If 4.61 g of N Are Made, What Is the Percent Yield, Continued 2 Given: 9.05 g NH , 45.2 g CuO 3 Find: g N 2 Solution Map: g NH mol NH mol N g N 3 3 2 2 1 mol 1 mol N 1 mol 2 Choose 28.02 g 2 mol NH 17.03 g 3 smallest g CuO mol CuO mol N g N 2 2 1 mol 1 mol 1 mol N 2 28.02 g 79.55 g 3 mol CuO Actual Yield 100 Percent Yield Theoretica l Yield 1 mol NH = 17.03g, 1 mol CuO = 79.55g, 1 mol N = 28.02 g Relationships: 3 2 2 mol NH = 1 mol N , 3 mol CuO = 1 mol N 3 2 2 www.ThesisScientist.comPractice—How Many Grams of N (g) Can Be Made from 9.05 g of NH Reacting with 2 3 45.2 g of CuO 2 NH (g) + 3 CuO(s) → N (g) + 3 Cu(s) + 3 H O(l) 3 2 2 If 4.61 g of N Are Made, What Is the Percent Yield, Continued 2 Solution: 1 mol NH 1 mol N 28.02 g N 3 2 2 9.05 g NH 7.42 g N 3 2 17.03 g NH 2 mol NH 1 mol N 3 3 2 1 mol CuO 1 mol N 28.02 g N 2 2 45.2 g CuO 5.30 g N 2 79.55 g CuO 3 mol CuO 1 mol N 2 Theoretical yield 4.61 g N 2 100 87.0 Yield 5.30 g N 2 Check: Since the percent yield is less than 100, the answer makes sense. www.ThesisScientist.comEnthalpy 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. DH rxn www.ThesisScientist.comSign 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. DH = ─  For the reaction CH (s) + 2 O (g)  CO (g) + 2 H O(l), the 4 2 2 2 DH = −802.3 kJ per mol of CH . rxn 4 • For endothermic reactions, the sign of the enthalpy change is positive when:  Thermal energy is absorbed by the reaction.  The surroundings get colder. DH = +  For the reaction N (s) + O (g)  2 NO(g), the 2 2 DH = +182.6 kJ per mol of N . rxn www.ThesisScientist.2 comEnthalpy 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 H (l) + 5 O (g)  3 CO (g) + 4 H O(g) 3 8 2 2 2 DH = −2044 kJ rxn So 1 mol C H 5 mol O 3 mol CO 4 mol H O  3 8 2 2 2 −2044 kJ. www.ThesisScientist.comExample 8.7—How Much Heat Is Associated with the 4 Complete Combustion of 1.18 x 10 g of C H (g) 3 8 3 Given: 11.8 x 10 g C H , 3 8 heat, kJ Find: Solution Map: g C H mol C H kJ 3 8 3 8 2044 kJ 1 mol C H 3 8 1 mol C H 3 8 44.09 g Relationships: 1 mol C H = 2044 kJ, Molar mass = 44.11 g/mol 3 8 Solution: 1 mol C H 2044 kJ 4 5 3 8 1.1810 g C H 5.4710 kJ 3 8 44.11 g C H 1 mol C H 3 8 3 8 Check: The sign is correct and the value is reasonable. www.ThesisScientist.comPractice—How Much Heat Is Evolved When a 0.483 g Diamond Is Burned (DH = −395.4 kJ/mol C) combustion www.ThesisScientist.comPractice—How Much Heat Is Evolved When a 0.483 g Diamond Is Burned (DH = −395.4 kJ/mol C), Continued combustion Given: 0.483 g C Find: heat, kJ Solution Map: g C mol C kJ 1 mol C 395.4 kJ 12.01 g 1 mol C Relationships: 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. www.ThesisScientist.com
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