Chapter 11 Assessment

CHAPTER11 SOLUTIONS MANUAL Stoichiometry Section 11.1 Defining Stoichiometry pages 368–372 Practice Problems pages 371–372 1. Interpret the following balanced chemical equations in terms of particles, moles, and mass. Show that the law of conservation of mass is observed. a. N2(g) 3H2(g) → 2NH3(g) 1 molecule N2 3 molecules H2 → 2 molecules NH3 1 mole N2 3 moles H2 → 2 moles NH3 Mass: N2: (2 mol)(14.007 g/mol) 28.014 g 3H2: (6 mol)(1.008 g/mol) 6.048 g Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2NH3: (2 mol)(14.007 g/mol) (6 mol)(1.008 g/mol) 34.062 g c. 2Mg(s) O2(g) → 2MgO(s) 2 atoms Mg 1 molecule O2 0 2 formula units MgO 2 moles Mg 1 mole O2 → 2 moles MgO Mass: 2Mg: (2 mol)(24.305 g/mol) 48.610 g O2: (2 mol)(15.999 g/mol) 31.998 g 2MgO: (2 mol)(24.305 g/mol) (2 mol)(15.999 g/mol) 80.608 g 48.610 g Mg 31.998 g O2 → 80.608 g MgO 80.608 g reactants 80.608 g products 2. Challenge For each of the following, balance the chemical equation; interpret the equation in terms of particles, moles, and mass; and show that the law of conservation of mass is observed. a. ___Na(s) ___H2O(l) → ___NaOH(aq) ___H2(g) 28.014 g N2 6.048 g H2 0 34.062 g NH3 2Na(s) 2H2O(l) → 2NaOH(aq) 1H2(g) 34.062 g reactants 34.062 g products 2 atoms Na 2 molecules H2O → 2 formula units NaOH 1 molecule H2 b. HCl(aq) KOH(aq) → KCl(aq) H2O(l) 1 molecule HCl 1 formula unit KOH → 1 formula unit KCl 1 molecule H2O 1 mole HCl 1 mole KOH → 1 mole KCl 1 mole H2O Mass: HCl: (1 mol)(1.008 g/mol) (1 mol)(35.453 g/mol) 36.461 g KOH: (1 mol)(39.098 g/mol) (1 mol)(15.999 g/mol) (1 mol)(1.008 g/mol) 56.105 g 2 mol Na 2 mol H2O → 2 mol NaOH 1 mol H2 Mass: 2Na: (2 mol)(22.990 g/mol) 45.980 g 2H2O: (4 mol)(1.008 g/mol) (2 mol)(15.999 g/mol) 36.030 g 2NaOH: (2 mol)(22.990 g/mol) (2 mol)(15.999 g/mol) (2 mol)(1.008 g/mol) 79.994 g H2: (2 mol)(1.008 g/mol) 2.016 g KCl: (1 mol)(39.098 g/mol) (1 mol)(35.453 g/mol) 74.551 g 45.980 g Na 36.030 g H2O → 79.994 g NaOH 2.016 g H2 H2O: (2 mol)(1.008 g/mol) (1 mol)( 15.999 g/mol) 18.015 g 82.01 g reactants 82.01 g products 36.461 g HCl 56.105 g KOH → 74.551 g KCl 18.015 g H2O 92.566 g reactants 92.566 g products Solutions Manual Chemistry: Matter and Change • Chapter 11 209 11 SOLUTIONS MANUAL b. ___Zn(s) ___HNO3(aq) → ___Zn(NO3)2(aq) ___N2O(g) ___H2O(l) b. 3Fe(s) 4H2O(l) → Fe3O4(s) 4H2(g) 3 mol Fe 3 mol Fe 3 mol Fe __ __ _ 4 mol H2O 4Zn(s) 10HNO3(aq) → 4Zn(NO3)2(aq) 1N2O(g) 5H2O(l) 4 mol H2 1 mol Fe O 4 mol H 4 mol H O _ __ __ 2 4 atoms Zn 10 molecules HNO3 → 4 formula units Zn(NO3)2 1 molecule N2O 5 molecules H2O 4 mol Zn 10 mol HNO3 → 4 mol Zn(NO3)2 1 mol N2O 5 mol H2O Mass: 1 mol Fe3O4 3 2 3 mol Fe 3 mol Fe 4 3 mol Fe 1 mol Fe O 1 mol Fe O 4 mol H O __ __ __ 3 4 4 mol H2 3 4 2 4 mol H2O 4 mol H2 4 mol H O 4 mol H 4 mol H __ __ __ 2 2 1 mol Fe3O4 2 1 mol Fe3O4 4 mol H2O c. 2HgO(s) → 2Hg(l) O2(g) 4Zn: (4 mol)(65.39 g/mol) 261.56 g 1 mol O 1 mol O 2 mol HgO _ __ __ 10HNO3: (10 mol)(1.008 g/mol) (10 mol)(14.007 g/mol) (30 mol)(15.999 g/mol) 630.12 g 2 2 mol Hg 2 2 mol Hg 2 mol HgO 2 mol HgO 2 mol Hg __ 2 mol Hg __ _ 4Zn(NO3)2: (4 mol)(65.39 g/mol) (8 mol)(14.007 g/mol) (24 mol)(15.999 g/mol) 757.592 g 2 mol HgO 1 mol O2 1 mol O2 4. Challenge Balance the following equations, and determine the possible mole ratios. a. ZnO(s) HCl(aq) → ZnCl2(aq) H2O(l) N2O: (2 mol)(14.007 g/mol) (1 mol)(15.999 g/mol) 44.013 g ZnO(s) 2HCl(aq) → ZnCl2(aq) H2O(l) 5H2O: (10 mol)(1.008 g/mol) (5 mol)(15.999 g/mol) 90.075 g 1 mol ZnO __ 1 mol ZnO 1 mol ZnO __ __ 2 mol HCl 1 mol ZnCl2 1 mol H2O 261.56 g Zn 630.12 g HNO3 → 757.592 g Zn(NO3)2 44.013 g N2O 90.075 g H2O 2 mol HCl 2 mol HCl 2 mol HCl __ __ __ 891.68 g reactants 891.68 g products 1 mol ZnCl 1 mol ZnCl 1 mol ZnCl __ __ __ 3. Determine all possible mole ratios for the following balanced chemical equations. a. 4Al(s) 3O2(g) → 2Al2O3(s) 2 mol Al O 3 mol O 4 mol Al __ _ __ 2 2 3 mol O2 2 mol Al2O3 3 4 mol Al 2 mol Al O 3 mol O 4 mol Al _ __ __ 2 4 mol Al 2 3 mol O2 3 2 mol Al2O3 1 mol ZnO 1 mol ZnCl2 2 1 mol ZnO 1 mol H2O 2 2 mol HCl 2 1 mol H2O 1 mol H O __ 1 mol H O 1 mol H O __ __ 2 2 1 mol ZnO 2 2 mol HCl 1 mol ZnCl2 b. butane (C4H10) oxygen → carbon dioxide water 2C4H10(g) 13O2(g) 0 8CO2(g) 10H2O(l) 2 mol C H 2 mol C H 2 mol C H __ __ __ 4 10 13 mol O2 4 10 8 mol CO2 4 10 10 mol H2O 8 mol CO 10 mol H O 13 mol O __ __ __ 2 2 mol C4H10 2 2 mol C4H10 2 2 mol C4H10 10 mol H O _ 8 mol CO 10 mol H O __ __ 2 2 13 mol O2 2 8 mol CO2 13 mol O2 8 mol CO 13 mol O 13 mol O __ __ _ 2 10 mol H2O 210 Chemistry: Matter and Change • Chapter 11 2 10 mol H2O 2 8 mol CO2 Solutions Manual Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. CHAPTER Model Write the mole ratios for the reaction of hydrogen gas and oxygen gas. 1 mol O2/2 mol H2O Section 11. Show the water molecules produced. 2H2(g) O2(g) → 2H2O. and mass 8.1 Assessment page 372 5. 2CH4(g) S8(s) → 2CS2(l) 4H2S(g) b.00 mol CS 2 1 mol S8 Chemistry: Matter and Change • Chapter 11 2 211 . a division of The McGraw-Hill Companies. molecules.CHAPTER 11 Section 11. formula units). ___CH4(g) ___S8(s) → ___CS2(l) ___H2S(g) a. y. Calculate the moles of CS2 produced when 1. 2 mol H2O/1 mol O2. Methane and sulfur react to produce carbon disulfide (CS2). Make a sketch of six hydrogen molecules reacting with the correct number of oxygen molecules. Inc. 2H2O2 → 2H2O O2 2 mol H2O2/2 mol H2O. 1 mol O2/2 mol H2O2.50 mol S8 Solutions Manual 2 mol CS _ 3.2 Stoichiometric Calculations pages 373–378 Practice Problems pages 375–377 11. 2 mol H2O2/1 mol O2. 2 mol H2O/2 mol H2O2. moles. and explain how these masses are related. The masses of reactants and products are equal. State how many mole ratios can be written for a chemical reaction involving three substances. Categorize the ways in which a balanced chemical equation can be interpreted. thus (n)(n 1) (3)(2) 6 mole ratios 7. n 3. and determine the possible mole ratios. The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products. Apply The general form of a chemical reaction is xA yB → zAB. Copyright © Glencoe/McGraw-Hill. Apply Hydrogen peroxide (H2O2) decomposes to produce water and oxygen. Balance the equation. In the equation.50 moles of S8 is used. yB/xA and yB/zAB zAB/xA and zAB/yB 9. State the mole ratios for this reaction. SOLUTIONS MANUAL 10. xA/yB and xA/zAB 6H2 3O2 6H2O 2H2/O2 and 2H2/2H2O O2/2H2 and O2/2H2O 2H2O/2H2 and 2H2O/O2 Sketches should show six hydrogen atoms combining with three oxygen atoms to form six water molecules. 1. Compare the mass of the reactants and the mass of the products in a chemical reaction. and z are coefficients. 6. Write a balanced chemical equation for this reaction. a liquid often used in the production of cellophane. particles (atoms. A and B are elements and x. 5 mol H2SO4 produced 12. 1 mol Cl __ 1.00 mol H S 2 2 1 mol S8 12. What mass of Cl2 gas is needed to react with 1. a. What is the mass of all of the products formed by reaction with 1. Step 2: Make mole → mole conversion. mass of products 292 g 2 1 mol Cl2 2 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.25 mol TiO2 13.25 mol Cl 99. 212 2 1 mol TiO2 70.50 mol Step 1: Make mole → mole conversion. 1.8 g TiO 2 1 mol TiO2 2 Calculate the total mass of the reactants. 1.25 mol of TiO2? Cl2 ? g Calculate the mass to TiO2 used.25 mol TiO2 2NaCl(s) → 2Na(s) Cl2(g) 79. Titanium tetrachloride (TiCl4) is extracted from titanium oxide (TiO2) using chlorine and coke (carbon). 2 mol H2SO4 2 mol SO2 12.865 g TiO __ 99.5 mol SO2 __ elements sodium and chlorine by means of electrical energy. How many moles of H2S are produced? 1. is obtained from the process? Na 2 2 1 mol Cl2 2 1 mol C __ 1. Sodium chloride is decomposed into the 1. How much chlorine gas.11 SOLUTIONS MANUAL c. How many moles of H2SO4 are produced from 12.0 g 292 g 2 2 2 mol NaCl Step 3: Make mole → mass conversion. What mass of C is needed to react with 1 mol O _ NaCl TiO2(s) C(s) 2Cl2(g) → TiCl4(s) CO2(g) a. in grams.50 mol NaCl 1.25 mol of TiO2? 2 2.90 g Cl __ 177 g Cl 1. b.011 g C __ 15.8 g 177 g 15. a division of The McGraw-Hill Companies.9 g Cl _ 88.25 mol of TiO2? 2. Write the balanced chemical equation for the reaction.25 mol C 12. Inc.25 mol TiO2 1. as shown below. 2SO2(g) O2(g) 2H2O(l) → 2H2SO4(aq) b.50 mol Cl2 c.5 mol SO2 14.0 g C 1 mol C c.25 mol Cl2 2 mol Cl __ 2.5 moles of SO2? 12.50 mol S8 4 mol H S _ 6.50 mol Cl Step 1: Make mole → mole conversion.25 mol O2 needed 2. Step 1: Balance the chemical equation.25 mol C 1 mol TiO2 Step 2: Make mole → mass conversion. 2 mol SO2 6. Step 2: Make mole → mass conversion. the mass of the products must equal the mass of reactants. Challenge Titanium is a transition metal used 70. How many moles of O2 are needed? Electric energy in many alloys because it is extremely strong and lightweight. CHAPTER .6 g Cl Because mass is conserved. Challenge Sulfuric acid (H2SO4) is formed when sulfur dioxide (SO2) reacts with oxygen and water. Determine the mass of N2 produced from the decomposition of NaN3 shown below. Describe how a mole ratio is correctly expressed when it is used to solve a stoichiometric problem.307 mol N 2 2 2 mol NaN3 Step 3: Make mole → mass conversion. Apply How can you determine the mass of liquid bromine (Br2) needed to react completely with a given mass of magnesium? Write a balanced equation. 2.83 g H2SO4 98. is produced? Step 1: Balance the chemical equation. a division of The McGraw-Hill Companies.0390 mol SO2 2 mol H SO __ 2 4 2 mol SO2 0. Use the mole ratio from the balanced equation to convert moles of magnesium to moles of bromine. how much H2SO4. If 2.0 g NaN3 → ? g N2(g) 2NaN3(s) → 2Na(s) 3N2(g) Step 1: Make mass → mole conversion.0390 mol H2SO4 Solutions Manual Chemistry: Matter and Change • Chapter 11 213 .CHAPTER 11 SOLUTIONS MANUAL Step 4: Make mole → mass conversion.0 g NaN3 18.50 g of SO2 reacts with excess oxygen and water. Convert the mass of the known substance to moles of known substance. 2SO2(g) O2(g) 2H2O(l) 0 2H2SO4(aq) Step 2: Make mass → mole conversion. Convert moles of unknown to mass of the unknown. The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products. moles of unknown/moles of known 20.538 mol NaN3 3 mol N __ 2. 1. Convert the given mass of magnesium to moles.50 g SO2 1. Challenge In the formation of acid rain.538 mol NaN 1 mol NaN3 3 65. 3. Explain why a balanced chemical equation is needed to solve a stoichiometric problem.02 g NaN3 Step 2: Make mole → mole conversion.307 mol N2 28. 0.64 g N 2 16.07 g SO2 Step 3: Make mole → mole conversion. One of the reactions used to inflate automobile air bags involves sodium azide (NaN3): 2NaN3(s) → 2Na(s) 3N2(g). 4. 2 1 mol N2 metric problems.02 g N __ 64. Copyright © Glencoe/McGraw-Hill. Write the balanced chemical equation for the reaction. in grams. Inc. List the four steps used in solving stoichio- __ 1. sulfur dioxide (SO2) reacts with oxygen and water in the air to form sulfuric acid (H2SO4). 0.09 g H SO __ 2 4 1 mol H2SO4 Section 11. 2 64. 1 mol SO __ 0. 2.0390 mol H2SO4 3. 15. 100. Convert from moles of bromine to mass of bromine. 2. Balance the equation. Use the mole ratio to convert from moles of the known to moles of the unknown. 100.0390 mol SO 2 19.2 Assessment N2 gas page 378 17. mass of solid iron produced 15.031 g NH3 1 mol NH3 0. limiting reactant Make mass → mole conversion. the molar mass.99 g Na 100.0 g of water available for photosynthesis.0 g Na given 86.0 g of Na and 100.6 g Na in excess 24.893 mol NH3 17. If 100.6261 mol Fe2O3 2 mol Fe __ 1 mol Fe2O3 reaction. Inc.9 g Na _ 1 mol Na 86.0 g Na 3 6 mol Na 100. Design a concept map for the following Make mole → mole conversion. c.7 g Fe2O3 0. Calculate Hydrogen reacts with excess nitrogen as follows: 0.757 mol Na needed Make mole → mass conversion.252 mol Fe 1 mol Fe the reaction is complete Make mole → mole conversion.85 g Fe _ 69.70 g H2) Make mole ratio comparison. Write the balanced chemical equation for the reaction.1667 The actual ratio is less than the needed ratio. The concept map should explain how to determine the mass of CaCl2 produced from a given mass of HCl. 1.2 g NH3 22. The reaction between solid sodium and iron(III) oxide is one in a series of reactions that inflates an automobile airbag: 6Na(s) Fe2O3(s) → 3Na2O(s) 2Fe(s).34 mol H2 1 mol 1.1439 compared to 0. so iron(III) oxide is the limiting reactant. the mole ratio. a. reactant in excess Sodium is the excess reactant. a division of The McGraw-Hill Companies.350 mol Na N2(g) 3H2(g) → 2NH3(g) (2. 6CO2(g) 6H2O(l) 0 C6H12O6(aq) 6O2(g) 1 mol Fe O __ 2 3 159.893 mol NH3) 3 2 4.757 mol Na Practice Problems 100.6261 mol Fe2O3 214 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.34 mol H2 2. mass of excess reactant that remains after 3.0 g of carbon dioxide and 64.3 Limiting Reactants pages 379–384 1.0 g Fe2O3 55.016 g H2 (_) ( __ ) 2 mol NH3 3 mol H2 (0. a. Section 11.37 g Na needed page 383 23. 0. 0.6261 mol Fe2O3 6 mol Na __ 1 mol Fe2O3 3.350 mol Na 22. Concept maps will vary.37 g Na needed 13. A plant has 88.11 SOLUTIONS MANUAL 21.0 g of Fe2O3 are used in this reaction. determine the following. how many grams of NH3 is formed? ( __ ) 1. 0.70 g of H2 reacts. b.92 g Fe d. 22. but all should show the use of these conversion factors: the inverse of molar mass.252 mol Fe CaCO3(s) 2HCl(aq) → CaCl2(aq) H2O(l) CO2(g) Make mole → mass conversion. 1 mol Na __ 4. Challenge Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen.6261 mol Fe O 1 mol Fe O __ compared to __ 2 If 2. CHAPTER . Determine the mass of glucose produced. Airborne sulfur reacts with the silver plating on a teapot to produce tarnish (silver sulfide). Inc.00 mol CO2 __ 2. 27.11 CHAPTER SOLUTIONS MANUAL Section 11. and rewrite the incorrect statements to make them correct. d. __ is used in the match heads of some matches. b.00 mol CO2 produce carbon dioxide. correct Solutions Manual Chemistry: Matter and Change • Chapter 11 215 . Make mole ratio comparison. 2. 2. page 384 Make mass → mole conversion.55 mol H2O 6 mol H2O 26.00 mol CO 2 2 44.333 mol C6H12O6 Make mole 0 mass conversion. Determine the excess reactant. Analyze Tetraphosphorous trisulphide (P4S3) 2 1 mol H2O 36. It is produced in the reaction 8P4 3S8 → 8P4S3. Determine the limiting reactant. Make mole → mole conversion. Determine which of the following statements are incorrect. Wood burns in a campfire. 4 mol P4 reacts with 1.563 compared to 1.24 g C6H12O6 1 mol C6H12O6 c. Oxygen is in excess. Silver is the limiting reactant. Phosphorus is the limiting reactant. c.0 g H2O needed 2. 0. 1 mol C H O __ 6 12 6 6 mol CO2 0. it prevents the sulfur in the air from reacting.01 g CO2 1 mol H O 64. a division of The McGraw-Hill Companies.02 g H O __ 64. Identify the limiting and the excess reactant in The wood limits.333 mol C6H12O6 60.0 g H2O needed 28. Describe the reason why a reaction between two elements comes to an end. Sulfur is the limiting reactant when 4 mol P4 and 4 mol S8 react. Determine the mass in excess. 0.55 mol H O 25. Water is the excess reactant.00 mol CO 6 mol CO __ compared to __ .0 g H2O in excess e. Make mole → mass conversion.3 Assessment b.02 g H2O each reaction. a. When a layer of tarnish covers the silver surface. The fire will burn only while wood is present.00 mol H2O 18. 88. Sulfur is in excess. c. 6 mol P4 reacts with 6 mol S8. correct b.0 g CO2 1 mol CO __ 2.0 g H2O given 36. Baking powder in batter decomposes to Make mole → mole conversion.5 mol S8 to form 4 mol P4S3. 2 2 3. 180.0 g H O __ 3. so carbon dioxide is the limiting reactant. forming 1320 g P4S3. 2. a.0 g C6H12O6 A decomposition reaction usually has only one reactant. The reaction is limited by the amount of baking powder present.00 The actual ratio is less than the needed ratio. one of the reactants is used up 2 2 2 18.00 mol H O 6 mol H2O 2 6 mol CO2 Copyright © Glencoe/McGraw-Hill. 3 g ZnI 319.179 mol AlCl3 silver nitrate solution (AgNO3). Determine the theoretical yield if 1.0 g Ag _ 100 68. Zinc reacts with iodine in a synthesis reaction: Zn I2 0 ZnI2. percent yield Solutions Manual Copyright © Glencoe/McGraw-Hill.2% yield of Ag Zn(s) I2(s) → ZnI2(s) Make mole → mole conversion. Identify which type of yield—theoretical Make mole → mass conversion.3 g of ZnI2 is the theoretical yield.912 mol of zinc is used.3 g AlCl __ 23.630 mol Ag 1 mol Cu Make mole → mass conversion.179 mol Al(OH)3 0.9 g of AlCl3 is the theoretical yield.179 mol AlCl3 30. % yield Write the balanced chemical equation. The reaction occurs as follows: Al(OH)3(s) 3HCl(aq) → AlCl3(aq) 3H2O(l). If 60. determine the theoretical yield of AlCl3 produced when the tablet reacts with HCl.11 SOLUTIONS MANUAL Section 11. determine the percent yield of the reaction. Challenge When copper wire is placed into a Cu(s) 2AgNO3(aq) → 2Ag(s) Cu(NO3)2(aq) Make mass → mole conversion. If a 20. 3 1 mol AlCl3 1 mol Cu __ 0.6 g of pages 385–388 product is recovered. 1 mol Ag 68. actual yield. 0.48% yield of ZnI2 page 387 14. Write the balanced chemical equation for the reaction. 29.179 mol Al(OH)3 2 610.0 g of Ag is the theoretical yield. Make mole → mole conversion. 1.6 g ZnI __ 100 3 0.0 g Al(OH)3 515. silver crystals and copper(II) nitrate (Cu(NO3)2) solution form.912 mol ZnI2 __ 610. 107.630 mol Ag 23.0 g Ag 88.912 mol Zn Section 11. 20.315 mol Cu Make mole → mole conversion.315 mol Cu 133. or percent yield—is a measure of the efficiency of a chemical reaction. a.0 g of Al(OH)3 is present in an antacid tablet.3 g ZnI2 84. Make mass → mole conversion.0 g of silver is recovered from the reaction.4 Assessment 1 mol ZnI _ 1.0 g Al(OH)3 0. c. Determine the percent yield if 515. 216 page 388 Chemistry: Matter and Change • Chapter 11 yield. 1 mol Al(OH) __ determine the theoretical yield of silver. CHAPTER . a division of The McGraw-Hill Companies. 3 78.0 g Ag 60.2 g ZnI2 1 mol ZnI2 2 610.55 g Cu 0. 1.4 Percent Yield b. 0. Make mole → mass conversion.9 g AlCl 2 mol Ag _ 0. If 14. Inc.912 mol ZnI 2 1 mol Zn 2 31. b.0 g Cu __ 1 mol AlCl3 1 mol Al(OH)3 63. a. Aluminum hydroxide (Al(OH)3) is often present in antacids to neutralize stomach acid (HCl).9 g Ag __ 68. Practice Problems % yield 28.0g sample of copper is used. Explain how percent yield is calculated. Subscripts give the numbers of different kinds of atoms within a molecule or formula unit. moles. Not all reactions go to completion. 37. What is the theoretical yield. 0.67 g 2.500 mol Fe 1 mol Fe 1.39 g mass of MgO 39. 1. of iron(III) sulfide? ( __ ) Copyright © Glencoe/McGraw-Hill.31 g MgO 2 mol MgO __ __ 2 mol Mg 1 mol MgO 3. List several reasons why the actual yield from a chemical reaction is not usually equal to the theoretical yield. What relationships can be determined from a balanced chemical equation? The relationships among particles. and mass for all reactants and products. Apply In an experiment. 33.77 g Fe) Chapter 11 Assessment 38.67 g 3. the relationship between reactants and products cannot be determined. Other unexpected products form from competing reactions.06 g 35. Explain why mole ratios are central to stoichiometric calculations.885 g Fe2S3 2 mol Fe 1 mol Fe2S3 155. you combine 83. If the equation is not balanced. Calculate the percent yield of the reaction of 2Mg(s) O2(g) → 2MgO(s) 40.67 g Mass of crucible and Mg 38.31 g Mg The coefficients in the balanced chemical equation show the numbers of representative particles involved in a reaction. divide the actual yield by the theoretical yield and multiply the quotient by 100 34.48 g MgO __ 100 3. Inc. 39.500 mol Fe 55.48 g Theoretical yield: 2. Why must a chemical equation be balanced before you can determine mole ratios? Mole ratios are determined by the coefficients in a balanced equation.39 g Mg moles B _ moles A magnesium with excess oxygen: 1 mol Mg __ 24.9% yield of MgO Solutions Manual Chemistry: Matter and Change • Chapter 11 217 .15 g 33. a division of The McGraw-Hill Companies. 41. 2Fe(s) 3S(s) → Fe2S3(s) (83. Explain how the conservation of mass allows you to interpret a balanced chemical equation in terms of mass.9 g Fe2S3 pages 392–397 Section 11. Some of the reactants or products stick to the surface of the container and are not massed or transferred. Mole ratios allow for the conversion from moles of one substance in a balanced chemical equation to moles of another substance in the same equation. in grams.96 g MgO % yield 3.845 g Fe ( __ ) ( __ ) 1 mol Fe2S3 207. What is the mole ratio that can convert from moles of A to moles of B? 35.15 g mass of Mg 38.983 mol Mg 40. The mass of the reactants will always equal the mass of the products.11 CHAPTER SOLUTIONS MANUAL 32.06 g Mass of crucible and MgO (after heating) 39.77 g of iron with an excess of sulfur and then heat the mixture to obtain iron(III) sulfide.0983 mol Mg 0.96 g MgO 87.1 Mastering Concepts 36. Why are coefficients used in mole ratios instead of subscripts? Reaction Data Mass of empty crucible 35. 982 g/mol) 6 mol O(15.999 g/mol) 95. the element tin can be extracted. producing nitrogen gas.92 g 107.71 g SnO2 24.99 g O2 → 203. Write a balanced equation to represent the picture shown. 43. Inc. using smallest whole-number ratios. Interpret the following equation in terms of particles. 4 2 2 7 4 2 1 mol N2 2 7 1 mol Cr2O3 1 mol (NH ) Cr O __ 4 2 2 Mass: 4Al 4 mol Al(26. and mass.10 depicts an equation with squares . Write mole ratios for this equation. Particles: 4 atoms Al 3 molecules O2 → 2 formula units Al2O3 1 mol (NH ) Cr O 1 mol (NH ) Cr O __ __ Moles: 4 mol Al 3 mol O2 → 2 mol Al2O3.011 g/mol) 24.02 g 2 2 2 mol M2N 1 mol M4 218 Interpret the chemical equation in terms of particles. and mass.71 g Sn 56.11 CHAPTER SOLUTIONS MANUAL 42. a division of The McGraw-Hill Companies.999 g/mol) 203. 4 4 Moles: 1 mol SnO2 2 mol C → 1 mol Sn 2 mol CO SnO2: 1 mol(118.710 g/mol) 118.99 g 2Al2O3 4 mol Al(26. Particles: 1 formula unit SnO2 2 atoms C → 1 atom Sn 2 molecules CO 2M2N → M4 N2 Mass: 2 mol M N __ 2 mol M N __ . and water vapor. solid chromium(III) oxide.02 g C → 118. When heated by a flame. moles. Chemistry: Matter and Change • Chapter 11 150.02 g CO Solutions Manual Copyright © Glencoe/McGraw-Hill.999 g/mol) 56.011 g/mol) 2 mol (15.93 g Al 95. ammonium dichro- mate decomposes. Mastering Problems 44. (NH4)2Cr2O7 → N2 Cr2O3 4H2O 4Al(s) 3O2(g) → 2Al2O3(s) Write the mole ratios for this reaction that relate ammonium chromate to the products._ 2CO: 2 mol(12. 2 1 mol M4 2 1 mol N2 1 mol M 1 mol M __ _ . moles. Figure 11.02 g 1 mol N2 2 mol M2N Sn: 1 mol(118. SnO2(s) 2C(s) → Sn(l) 2CO(g) representing Element M and circles representing Element N.982 g/mol) 107. . .710 g/mol) 2 mol (15.999 g/mol) 150.92 g Al2O3 2 1 mol (NH4)2Cr2O7 45.93 g 7 4 mol H2O __ __ 1 mol N2 1 mol Cr2O3 1 mol (NH4)2Cr2O7 1 mol (NH4)2Cr2O7 4 mol H O __ 3O2 6 mol O(15.71 g 1 mol N 1 mol N __ . Smelting When tin(IV) oxide is heated with carbon in a process called smelting.71 g 2C: 2 mol(12. 9 g HCl 331. What mole ratio would you use to determine moles of Fe if moles of Fe2O3 are known? Cu(s) 4HNO3(aq) → Cu(NO3)2(aq) 2NO2(g) 2H2O(l). Particles: 2 molecules HCl 1 formula unit Pb(NO3)2 → 1 formula unit PbCl2 2 molecules HNO3 Moles: 2 mol HCl 1 mol Pb(NO3)2 → 1 mol PbCl2 2 mol HNO3. 2 mol H2O/1 mol Cu(NO3)2. 2 mol Fe __ 1 mol Fe2O3 49. 4 mol HNO3/2 mol H2O. Chrome The most important commercial ore of chromium is chromite (FeCr2O4). List three mole ratios.1 g 2HNO3: 2 mol(1. 1 mol Cu/1 mol Cu(NO3)2. 2 mol H2O/2 mol NO2 Fe2O3(s) 2Al(s) → 2Fe(s) Al2O3(s) heat 47. 1 mol Cu(NO3)2/2 mol NO2. SOLUTIONS MANUAL 2HCl(aq) Pb(NO3)2(aq) → PbCl2(s) 2HNO3(aq) b. and explain how you would use them in stoichiometric calculations. 2 mol H2O/1 mol Cu. SiO2(s) 4HF(aq) → SiF4(g) 2H2O(l) b. Write the balanced chemical equation for the reaction. 2 mol H2O/1 mol SiF4. 4 mol HNO3/1 mol Cu. 1 mol Cu(NO3)2/2 mol H2O. a. used to find the amount of HF that will react with a known amount of SiO2. Write the balanced chemical equation for the reaction. When aluminum is mixed with iron(III) oxide.9 g What mole ratio would you use to convert from moles of chromite to moles of ferrochrome? Pb(NO3)2: 1 mol(207.999 g/mol) 126. lead(II) chloride precipitates and a solution of nitric acid is produced.008 g/mol) 2 mol (35. 1 mol Cu(NO3)2/1 mol Cu. Examples might be the following. 1 mol SiF4/1 mol SiO2. One of the steps in the process used to extract chromium from the ore is the reaction of chromite with coke (carbon) to produce ferrochrome (FeCr2).453 g/mol) 278.007 g/mol) 6 mol(15. and mass.008 g/mol) 2 mol (14. 4 mol HNO3/2 mol NO2. answers should include any six of the following ratios: 1 mol Cu/4 mol HNO3. Students may write any of 12 ratios. 2 mol NO2/4 mol HNO3. 1 mol Cu/2 mol NO2. used to find the amount of SiF4 that can be formed from a known amount of SiO2. Solid silicon dioxide.0 g 72. Inc.999 g/mol) 331. iron metal and aluminum oxide are produced along with a large quantity of heat. a. often called silica. 1 mol Cu(NO3)2/4 mol HNO3.453 g/mol) 72.1 g PbCl2 126. When solid copper is added to nitric acid.2 g 1 mol FeCr __ 2 1 mol FeCr2O4 PbCl2: 1 mol(207. 1 mol Cu/2 mol H2O. 2 mol H2O/4 mol HNO3. and water are produced. List six mole ratios for the reaction. When hydrochloric acid solution reacts with lead(II) nitrate solution. Mass: 2C(s) FeCr2O4(s) → FeCr2(s) 2CO2(g) 2HCl: 2 mol(1. 2 mol NO2/2 mol H2O. 4 mol HNO3/1 mol Cu(NO3)2. 48. moles. 2 mol NO2/1 mol Cu(NO3)2.CHAPTER 11 46.0 g HNO3 Solutions Manual Chemistry: Matter and Change • Chapter 11 219 . copper(II) nitrate.2 g Pb(NO3)2 → 278. nitrogen dioxide. Copyright © Glencoe/McGraw-Hill. used to find the amount of H2O that will be produced with the SiF4 50. reacts with hydrofluoric acid (HF) solution to produce the gas silicon tetrafluoride and water. 4 mol HF/1 mol SiO2.2 g/mol) 2 mol (14. a division of The McGraw-Hill Companies. 2 mol NO2/1 mol Cu.2 g/mol) 2 mol (35.007 g/mol) 6 mol(15. Interpret the equation in terms of molecules and formula units. Write the balanced chemical equation for the reaction. 30 0. the least common denominator: The balanced equation provides the relationship between reactants and products. How is molar mass used in some stoichiometric calculations? W: 0.30 4 58. The calculations are used to determine the mass of reactants and products.90 0. 1Mg(OH)2 2HCl → 1MgCl2 2H2O b. Determine the mole ratio you would use to convert moles of SO2 to moles of CaSO4. a division of The McGraw-Hill Companies. CHAPTER .30 1 Y: 0. Table 11. and the coefficients in the equation are used to write mole ratios relating reactants and products. Balance the reaction of Mg(OH)2 with HCl. What information does a balanced equation 2 mol SO2 provide? 52. The products of this reaction are calcium sulfate and carbon dioxide.20 Divide each molar quantity by 0. Antacids Magnesium hydroxide is an ingre- dient in some antacids. W and X. 57.20/0. Antacids react with excess hydrochloric acid in the stomach to relieve indigestion. Two substances. and how do the calculations support this law? Stoichiometry is based on the law of conservation of mass. 2SO2 2CaCO3 O2 → 2CaSO4 2CO2 2 mol CaSO __ 4 Section 11.30 2 Z: 1. What is the first step in all stoichiometric calculations? Write a balanced chemical equation for the reaction. verifying the law of conservation of mass. 55. Use the data to determine the coefficients that will balance the equation W X → Y Z.90/0. Air Pollution The pollutant SO2 is removed from the air in a reaction that also involves calcium carbonate and oxygen. Inc.60/0.30/0.30 mol. Once found. Reaction Data Moles of Reactants Moles of Products W X Y Z 0. react to form the products Y and Z.2 Mastering Concepts 54. 56. the sum of reactants will equal the sum of products.30 3 Molar mass is a conversion factor for converting moles of a given substance to mass or mass of a given substance to moles.2 shows the moles of the reactants and products involved when the reaction was carried out.60 1.11 SOLUTIONS MANUAL 51. What information must you have in order to calculate the mass of product formed in a chemical reaction? 3W X 0 2Y 4Z 53. Write the mole ratio that would be used to determine the number of moles of MgCl2 produced when HCl reacts with Mg(OH)2. 1 mol MgCl __ 2 1 mol Mg(OH)2 220 or 1 mol MgCl __ 2 2 mol HCl Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. ___Mg(OH)2 ___HCl → ___ MgCl2 ___ H2O a. On what law is stoichiometry based. You must have the balanced chemical equation and know the quantity of one substance in the reaction other than the product you are to determine. X: 0. 2 mol C6H12O6 4.16 g/mol) 4. a reaction occurs and water vapor and sulfur are produced.4 mol C2H5OH 8. Write the balanced chemical equation for the reaction. thus. 1 mol C H COOC H __ 1 mol C H OH 116. Mastering Problems 60. 2H2S(g) O2(g) → 2H2O(g) 2S(s) b. Student sketches should show the formation of six water molecules (H2O) and six sulfur atoms (S). will be produced? + CaC2(s) 2H2O(l) → Ca(OH)2(aq) C2H2(g) 59.50 mol CaC2.11 represents the contents of a flask. In the figure. Copyright © Glencoe/McGraw-Hill.50 mol of calcium carbide (CaC2) reacts with an excess of water.18 g C H COOC H ___ 3 7 4.0119 mol NaHCO3 (1 mol Na3C6H5O7/3 mol NaHCO3) 0.00397 mol 63. and blue circles represent hydrogen. the yellow circles represent sulfur. When the contents of the flasks are mixed. and the other contains oxygen. also called sodium bicarbonate. Inc. Each box in Figure 11. One flask contains hydrogen sulfide. Using the same color code. Ethyl butanoate (C3H7COOC2H5). a. Antacid Fizz When an antacid tablet dissolves in water. and citric acid (H3C6H5O7). a gas used in welding. the fizz is due to a reaction between sodium hydrogen carbonate (NaHCO3).50 mol of ethanol is used.2 mol C6H12O6 The mole ratio of CaC2:C2H2 is 1:1. Ethanol (C2H5OH ). also known as grain alcohol. Welding If 5.0119 mol of NaHCO3 is dissolved? 0. 5.50 mol C2H2 will be produced from 5.07 g __ 1 mol C2H5OH 62. C2H5OH(l) C3H7COOH(l) → C3H7COOC2H5(l) H2O(l) Determine the mass of ethyl butanoate produced if 4. a division of The McGraw-Hill Companies. C6H12O6 → 2C2H5OH 2CO2 ( 2 mol C H OH __ 8. an ester. ) ) 3 7 2 5 1 mol C3H7COOC2H5 523 g C3H7COOC2H5 390 g C2H5OH Chemistry: Matter and Change • Chapter 11 221 . The unbalanced chemical equation for the reaction is shown below. sketch a representation of the flask after the reaction occurs. is formed when the alcohol ethanol (C2H5OH) and butanoic acid (C3H7COOH) are heated in the presence of sulfuric acid.CHAPTER 11 SOLUTIONS MANUAL 61. 3NaHCO3(aq) H3C6H5O7(aq) → 3CO2(g) 3H2O(l) Na3C6H5O7(aq) How many moles of Na3C6H5O7 can be produced if one tablet containing 0. ___C6H12O6 → ___C2H5OH ___CO2 Balance the chemical equation and determine the mass of C2H5OH produced from 750 g of C6H12O6. how many moles of acetylene (C2H2). the red circles represent oxygen. can be made from the fermentation of sugar (C6H12O6). Esterification The process in which an organic acid and an alcohol that forms as ester and water is known as esterification.50 mol C2H5OH 2 5 2 5 750 g C6H12O6(180.4 mol C2H5OH Solutions Manual ( 2 5 1 mol C6H12O6 46. 0 mol N2 1 mol N H 30. a division of The McGraw-Hill Companies.00 g O2) (4 mol KO2/3 mol O2) (71.00 mol CO2 0.625 mol C8H18 __ 0. A solution of potassium chromate reacts with a 67. Inc.4 g C H 8 18 1 mol C8H18 solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate.37 g CHCl3 Oxygen Generation Reaction Data b.2 g PbCrO __ 80. CHAPTER . is needed to produce 10.00 g O2) (4 mol CO2/3 mol O2) (44. a. 4KO2 2H2O 4CO2 0 4 KHCO3 3O2 Complete Table 11.0 102 g 380 g O2 (1 mol O2/32.72 g CH 8 18 65.1 g KO2/1 mol KO2) 1100 g 4 1 mol K2CrO4 323.0 mol of nitrogen gas? 10.04 g CH __ 6.8 g PbCrO 4 4 1 mol PbCrO4 3 119. is 66.0 grams of CHCl3? 50.00 mol of CO2. determine the mass of lead chromate formed. Chloroform (CHCl3).03 g N H __ __ 2 2 1 mol N2 3.11 SOLUTIONS MANUAL 64. Write the balanced chemical equation. CH4(g) 3Cl2(g) 0 CHCl3(g) 3HCl(g) How much CH4 in grams is needed to produce 50.250 mol K2CrO4 1 mol CHCl 1 mol CH __ __ 16.625 mol C H 2 mol C8H18 produced by a reaction between methane and chlorine. How much hydrazine. Mass KO2 Mass H2O Mass CO2 Mass KHCO3 Mass O2 1100 g 140 g 7. a. 1 mol PbCrO __ 4 1 mol CHCl3 4 4 1 mol CH4 68. Greenhouse Gas Carbon dioxide is a greenhouse gas that is linked to global warming. in grams. N2H2(l) H2O2(l) 0 N2(g) 2H2O(g) 380 g O2 (1 mol O2/32.0 g CHCl3 8 18 16 mol CO2 114. The products of this reaction are nitrogen gas and water. Rocket Fuel The exothermic reaction between liquid hydrazine (N2H2) and liquid hydrogen peroxide (H2O2) is used to fuel rockets.28 g C H __ 71.250 mol of potassium 0.00 g O2) (4 mol KHCO3/3 mol O2) (100.01 g CO2/1 mol CO2) 7. K2CrO4(aq) Pb(NO3)2(aq) 0 PbCrO4(s) 2KNO3(aq) chromate. Oxygen Production The Russian Space Agency uses potassium superoxide (KO2) for the chemical oxygen generators in their space suits. Write the balanced chemical equation. an important solvent. It is released into the atmosphere through the combustion of octane (C8H18) in gasoline.00 102 g N2H2 222 2 2 1 mol N2H2 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.00 g O2) (2 mol H2O/3 mol O2) (18.02 g H2O/1 mol H2O) 140 g 380 g O2 (1 mol O2/32. Starting with 0. Write the balanced chemical equation for the combustion of octane and calculate the mass of octane needed to release 5.3. 2C8H18(l) 25O2(g) 0 16CO2(g) 18H2O(l) 5.12 g KHCO3/1 mol KHCO3) 1600 g b.0 102 g 1600 g 380 g 380 g O2 (1 mol O2/32. Write the balanced chemical equation for this reaction. 74. Copyright © Glencoe/McGraw-Hill. a division of The McGraw-Hill Companies.46 103 mol AgBr 1.23 g PbSO4 73. If the mass of the ore from which the gold Balance the equation.46 103 mol AgBr 1.170 mol C2H5OH 4.46 103 mol Na3Ag(S2O3)2 401. How is a mole ratio used to find the limiting reactant? The actual mole ratio of reactants from the chemical equation is compared to the mole ratio determined from the given quantities. and determine the mass of CO2 produced from the combustion of 100. the ore is treated with sodium cyanide solution in the presence of oxygen and water.0 g.340 mol CO2 4.08 g C2H5OH 2. To extract gold from its ore. Mass does not determine the limiting reactant.2 g Au 1 mol Au __ Solutions Manual 8 mol NaCN Mastering Concepts 73. 25. Explain why the statement “the limiting reactant is the reactant with the lowest mass” is incorrect. but the number of moles. 2 mol PbSO 1 mol Pb 25 g Pb __ __ Determine the mass of Na3Ag(S2O3)2 produced if 0. Soluble sodium silver thiosulfate (Na3Ag(S2O3)2) is produced.2 g Au __ 100 33. what percentage of the ore is gold? C2H5OH(l) O2(g) 0 CO2(g) H2O(g) 150.0 g of sodium cyanide is used. some of the silver bromide decomposes. a.97 g Au 50. Pb(s) PbO2(s) 2H2SO4(aq) 0 2PbSO4(aq) 2H2O(l) b. AgBr(s) 2Na2S2O3(aq) 0 Na3Ag(S2O3)2(aq) NaBr(aq) 44.0 g CO2 produced 70. Determine the mass of gold that can be extracted if 25.77 g AgBr 1.275 g AgBr 1 mol AgBr __ 187. Gasohol is a mixture of ethanol and gasoline. The products of the reaction are lead(II) sulfate in solution and water.0 g C H OH __ 2 5 2 5 46.3 4 207. Film Photographic film contains silver bromide in gelatin. 4Au(s) 8NaCN(aq) O2(g) 2H2O(l) 0 4NaAu(CN)2(aq) 4NaOH(aq) a.275 g of AgBr is removed. Chemistry: Matter and Change • Chapter 11 223 .01 g NaCN 196.0 g NaCN 1 mol NaCN 4 mol Au __ __ 49. Determine the mass of lead(II) sulfate produced when 25.587 g Na3Ag(S2O3)2 1 mol Na3Ag(S2O3)2 ___ Section 11.46 103 1 mol Na Ag(S O ) __ 3 2 3 2 1 mol AgBr mol Na3Ag(S2O3)2 1.170 mol C2H5OH 2. b. was extracted is 150.340 mol CO2 2 mol CO __ 2 1 mol C2H5OH 50. and a sulfuric acid solution to produce an electric current. Car Battery Car batteries use lead.5% gold in ore 72.0 g of ethanol.2 g Pb 1 mol Pb 303. Inc.0 g ore C2H5OH(l) 3O2(g) 0 2CO2(g) 3H2O(l) 1 mol C H OH 100. lead(IV) oxide.01 g CO __ 2 1 mol CO2 191. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. The limiting reactant is the reactant that produces the lowest number of moles of product.2 g PbSO4 1 mol PbSO4 __ 71.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid. producing fine grains of silver. 0. Once exposed.CHAPTER 11 SOLUTIONS MANUAL 69.12 g Na3Ag(S2O3)2 0. 0 mol Fe reacts.0 mol CsXeF + cially by the reaction of hematite (Fe2O3) with carbon monoxide.00 mol Fe in excess. Figure 11. Thomas Edison invented the nickel-iron battery.0 mol of hematite would require 75. Mastering Problems 76. how many moles of M and N were present at the start of the reaction? 6 moles of element M (in the form of 3 moles of M2) and 6 moles of element N (likewise. Nickel-Iron Battery In 1901. For each mole of Fe that reacts. 3 moles of N2) c. two moles of NiO(OH) react with each mole of Fe. a. How many moles of CsXeF7 can be produced from the reaction of 12.00 mol of NiO(OH) react? According to the balanced equation. Inc. If each square represents 1 mol of M and each circle represents 1 mol of N.12 uses squares to represent Element Fe(s) 2NiO(OH)(s) 2H2O(l) 0 Fe(OH)2(s) 2Ni(OH)2(aq) . 78. 4 mol Fe react with 8.13.85 g Fe 2 mol Fe _ _ 3 mol CO 1 mol Fe 1120 g Fe Ethyne Hydrogen Ethane Ethyne Hydrogen is limiting.0 mol are available. 30. So. 10.11 CHAPTER SOLUTIONS MANUAL 77. 3M2 N2 0 2M3N b.0 mol CO 55. leaving 1. ethyne is the excess reactant. 75.0 mol of xenon hexafluoride? CsF(s) XeF6(s) 0 CsXeF7(s).5 mol of cesium fluoride with 10.0 mol of carbon monoxide? Fe2O3(s) 3CO(g) 0 2Fe(s) 3CO2(g) According to the balanced equation. Because 4. 1 mole of hematite reacts with 3 moles of carbon monoxide. 224 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. How many grams of iron are produced if 25. M and circles to represent Element N. Write the balanced equation for the reaction.00 mol NiO(OH). CO is the limiting reactant.0 mol CO but only 30.00 mol of Fe and 8. Identify the limiting reactant and excess reactant. 7 1 mol XeF6 7 79.0 mol of hematite react with 30. one mole of Fe(OH)2(s) is produced. The following reaction takes place in the battery. → + 1 mol CsXeF __ 10. Since 25.0 mol XeF6 d. The reaction between ethyne (C2H2) and hydrogen (H2) is illustrated in Figure 11. a division of The McGraw-Hill Companies. The product is ethane (C2H6).0 mol Fe(OH)2 is produced. Iron Production Iron is obtained commer- M2 is the limiting reactant and N2 is the excess reactant. How many moles of product form? How many moles of M and N are unreacted? 2 moles of M3N form with 2 moles of N2 unreacted (4 total moles of element N) How many mol of Fe(OH)2 are produced when 5. Which is the limiting reactant? Which is the excess reactant? Explain. 4. One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). One mol of ethyne is left over. b.60 mol Li 6.380.1 g Zn(OH) 2 1 mol Zn(OH)2 2 1 mol Br2 1 mol LiBr c. a division of The McGraw-Hill Companies. Solutions Manual Chemistry: Matter and Change • Chapter 11 225 . Zn(s) 2MnO2(s) H2O(l) 0 Zn(OH)2(s) Mn2O3(s) a. Br2 is the limiting reactant.226 mol Cl2 1 mol P _ 4 10 mol CI2 0. MnO2 reacts with Zn in a two-to-one ratio. b. the excess reactant and the excess mass. the ratio is 1:1. In the reaction.0 g Zn 1 mol Zn _ 0.312 mol LiBr 0.88 g P4 According to the balanced equation. Theoretical yield is the amount of product predicted by a stoichiometric calculation. which reactant is limiting? Which reactant is in excess? produce lithium bromide.0226 mol P4 needed 25.345 mol MnO2 0. determine 10Cl2(g) P4(s) 0 4PCl5(s) 2Li(s) Br2(l) 0 2LiBr(s) 1 mol CI __ 0. Cl2 is limiting reactant.94 g Li _ 22. the mass of lithium bromide produced. 0.156 mol Br2 2 mol LiBr _ 0.380 mol Zn 65.0 g of bromine are present at the beginning of the reaction.39 g Zn(OH) __ 17.29 mol Li 86.1 or 0. Lithium reacts spontaneously with bromine to phosphorus (P4) produces solid phosphorus pentachloride.156 mol Br2 2 mol Li _ 0. MnO2 is the limiting reactant. Determine the mass of Zn(OH)2 produced.312 mol Li used 3.60 mol Li 0. produces electrical energy according to this equation.0 g MnO2 1 mol MnO __ 2 According to the balanced equation. Alkaline Battery An alkaline battery Copyright © Glencoe/McGraw-Hill. 1 mol Zn(OH) __ 2 2 mol MnO2 99. P4 is in excess. Only 2 mol Li are required for 1 mol Br2.0 g of Zn and 30.29 mol Li remaining 3.3 g Zn 30. 0. Cl2 reacts with P4 in a ten-to-one ratio.4 Mastering Concepts 83. Determine the limiting reactant if 25.312 mol LiBr 86. 25.84 g LiBr __ 27. Theoretical yield is calculated from a given reactant or the limiting reactant.345 mol MnO2 2 6. Write the balanced chemical equation for the reaction. The reaction of chlorine gas with solid 82.11 CHAPTER SOLUTIONS MANUAL 80. Li is in excess. 84. If 25.60 mol Li/0.1 g LiBr 81. When 16. Inc. What is the difference between actual yield and theoretical yield? Actual yield is the amount of product actually obtained experimentally.0 g Li 1 mol P __ 0.0 g Br2 1 mol Br __ 0.0 g of P4.90 g CI2 25.0 g of chlorine reacts with 23. How are actual yield and theoretical yield determined? Actual yield is determined through experimentation.0 g P4 a.0 g of lithium and 25. the limiting reactant.226 mol CI 23.156 Br2 or 23:1.92 g MnO2 0.80 g Br2 Actual ratio of mol Li to mol Br2 is 3.8 g Li remaining 1 mol Li Section 11. Thus.312 mol Li used 1 mol Br2 3.0 g of MnO2 are used.94 g Li 4 123.345/0.185 moles 1 mol Li _ 3. 0.0 g Cl2 2 70.156 mol Br 2 159. 2 16. Determine if the reaction went to completion. a division of The McGraw-Hill Companies.0 g _ 100 91.0 g of PbO is obtained? % yield 170. The unbalanced equation is: PbS(s) O2(g) 0 PbO(s) SO2(g) a.07 g C H OH 4 mol C H OH __ __ 2 5 2 5 1 mol C12H22O11 369 g C2H5OH % yield 1 mol C2H5OH actual yield __ 100 theoretical yield 349 g 100 94. CHAPTER . What experimental information do you need in order to calculate both the theoretical and the percent yield of any chemical reaction? The quantity of one reactant and the actual yield of the product 88. Balance the equation. 86. Inc. What is the percent yield if 170.27 g PbS 223. the initial products would have yielded four AB2 particles. lead(II) sulfide. actual yeild __ 100 percent yield theoretical yield 87. which is determined from the starting reactants. Lead(II) oxide is obtained by roasting galena.6 g PbO 2 mol PbS 1 mol PbO b.6 g Solutions Manual Copyright © Glencoe/McGraw-Hill. A metal oxide reacts with water to produce a metal hydroxide. Can the percent yield of a chemical reaction be more than 100%? Explain your answer.10% 186. Examine the reaction represented in Figure 11. and calculate the percent yield of the reaction. 2PbS 3O2 0 2PbO 2SO2 Theoretical yield 200.0 g of PbS is heated.19 g PbO 2 mol PbO __ __ 186. Ethanol (C2H5OH) is produced from the fermentation of sucrose (C12H22O11) in the presence of enzymes. Element A Element B theoretical yield: 684 g C12H22O11 1 mol C H O __ 12 22 11 342. in air. Using squares to represent Element A and circles to represent Element B.6% 369 g _ 91. but only three were produced.14. 226 Chemistry: Matter and Change • Chapter 11 1 mol PbS __ 239. What additional information would you need to determine the percent yield of metal hydroxide from this reaction? the mass of one substance in the reaction and the actual mass of metal hydroxide produced 89. There are enough remaining unreacted A and B particles to produce one more AB2 particle. No. C12H22O11(aq) H2O(g) 0 4C2H5OH(l) 4CO2(g) Determine the theoretical yield and the percent yield of ethanol if 684 g of sucrose undergoes fermentation and 349 g of ethanol is obtained. Explain your answer. What relationship is used to determine the percent yield of a chemical reaction? Mastering Problems 90.11 SOLUTIONS MANUAL 85.0 g PbS The reaction did not go to completion.23 g C12H22O11 46. you cannot produce more product than the theoretical yield. and determine the theoretical yield of PbO if 200. The percent yield is 75%. 3 2 2 6 144.333 mol SiO 2 2 6 mol HF 0.00 mol HF 20.11 g H SiF __ 48.CHAPTER 11 SOLUTIONS MANUAL 92.666 mol SiO2 limiting reactant. 6 d. a division of The McGraw-Hill Companies. In the second step.22 g Zr 1 mol Zr _ _ 6.5 g of actual yield __ 100 theoretical yield 97.09 g SiO2 obtained using the two-step Van Arkel process. Actual ratio of HF to SiO2 is 2.333 mol SiO2 used 0. In the first step. What is the mass of the excess reactant? 2.666 mol SiO 1 mol SiO2 2 60.0 g of CaCO3 is heated.0% % yield 1 mol HF 40.333 mol SiO2 60. What is the theoretical yield of H2SiF6? decomposes to produce calcium oxide (CaO) and carbon dioxide (CO2).0 g SiO2 and 40.0 g of ZrI4 is decomposed and 5. calcium carbonate (CaCO3) c. a.00 g Zr _ 100 73.0 g H2SiF6 is the theoretical yield.85 g of Zr 1 mol ZrI4 1 mol Zr actual yield __ 100 theoretical yield 5. What is the percent yield? CO2 is collected? 40.00 mol HF 0. theoretical yield 45.99 g CO 1 mol CO __ __ 103 g CO 1 mol CO2 2 b.333 mol H2SiF6 1 mol CaCO __ 100.0 g SiO2 in excess Solutions Manual Chemistry: Matter and Change • Chapter 11 227 .0 g SiO2 2 1 mol H2SiF6 48.00:1. Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produce hexafluorosilicic acid (H2SiF6).06 g CaCO 43. Thus. Van Arkel Process Pure zirconium is % yield 2 103 g CO2 93.00 mol HF or 3.00 mol HF 1 mol SiO __ 0. Determine the theoretical yield of CO2 if 235.666 mol SiO2 available 0.0 H2SiF6 94. Upon heating. a. HF is the 0.0 g HF __ 2.5 g CO __ 100 94.8 g H SiF __ 100 95. What is the percent yield of CO2 if 97.333 mol H2SiF6 CaCO3(s) 0 CaO(s) CO2(g) 235 g CaCO3 1 mol H SiF __ __ 0.09 g SiO __ 2 1 mol SiO2 20. Inc. SiO2(s) 6HF(aq) 0 H2SiF6(aq) 2H2O(l) Copyright © Glencoe/McGraw-Hill.8 g H2SiF6.7% % yield 45.4% yield 2 6 48. impure zirconium and iodine are heated to produce zirconium iodide (ZrI4). ZrI4 is decomposed to produce pure zirconium. 6.0 g H2SiF6 3 1 mol CaCO3 2 6 mol HF 0.85 g Zr __ b.0 g ZrI4 1 mol ZrI __ 4 598.00 g of pure Zr is obtained. ZrI4(s) 0 Zr(s) 2I2(g) Determine the percent yield of zirconium if 45.84 g ZrI4 91.0 g HF react to yield 45. 2. What is the limiting reactant? 40.333 mol SiO2 in excess 0.01 g HF Ratio of HF to SiO2 in the balanced equation is 6:1. CHAPTER .303 mol CO 250. and HCl is the limiting reactant. and coke (C) in an electric furnace.4.4030 mol P4O10 8.0 g of SiO2 are heated.05 g CH3OH/1 mol CH3OH) 9.4030 mol P4O10 0.71 g Molar mass 28. 228 4 2 mol Ca3(PO4)2 MnO2 4HCl 0 MnCl2 Cl2 2H2O 2Ca3(PO4)2(s) 6SiO2(s) 0 6CaSiO3(l) P4O10(g) 0. MnO2 (s) 4HCl (aq) 0 MnCl2 (aq) Cl2 (g) 2H2O (l) Step 2 is to determine which reactant is in excess.37 mol or 1:1.8060 mol Ca3(PO4)2 Step 3 is to determine amount of P4 produced in step 2 from 0.0 g of HCl react. the ratio is 0. In this reaction.92 g P4 97.1% 4 49.303 mol 0.71 g CH3OH 1 mol SiO __ 6. sand (SiO2).304 mol CH3OH (32.8060 mol 1 mol P O __ % yield 0. 86.38.303 mol CH3OH Percent yield (8. determine the percent yield of P4.94 g MnO2 1 mol HCI __ 1.01 g CO) 0. P4O10(g) 10C(s) 0 P4(g) 10CO(g) The P4O10 produced in the first reaction reacts with an excess of coke (C) in the second reaction. The actual yield of Cl2 is 20. CO 2H2 0 CH3OH When 8.88 g P __ __ 4 4 1 mol P4O10 1 mol P4 Step 4 is to determine the percent yield.08 g SiO2 1 mol Ca (PO ) __ 3 0.92 g.01 g/mol 32.46 g HCI According to the balanced equation.0 g of Ca3(PO4)2 and 400.50 g of carbon monoxide reacts with an excess of hydrogen.303 mol CO (1 mol CH3OH/1 mol CO) 0.05 g/mol Moles 0. actual yield __ 100 theoretical yield 45. 8. Determine the theoretical yield of P4 if 250.0 g.0 g SiO2 Step 2 is to determine amount of P4O10 produced. In this reaction.52 g of methanol is collected. Step 1: The balanced equation is given.657 mol of SiO 2 2 60. CO(g) CH3OH(l) Mass 8.0 g of MnO2 and 50. Ca3(PO4)2 reacts with SiO2 in a one-to-three ratio. Inc. 10 49. Complete Table 11.17 g Ca3(PO4)2 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. Step 1 is to determine the excess quantity in equation #1. given actual yield is 49.0 g P _ 100 90. If the actual yield of P4 is 45.0 g Ca3(PO4)2 0.37 mol HCI 36. a division of The McGraw-Hill Companies. Therefore MnO2 is in excess. Phosphorus (P4) is commercially prepared by heating a mixture of calcium phosphate (CaSiO3).8060 mol of Ca3(PO4)2 react.303 mol 0. The balanced equation is: Calculate the theoretical yield and the percent yield of chlorine if 86. The process involves two reactions. is produced when carbon monoxide reacts with hydrogen gas.50 g 9. SiO2 is in excess and 0.0 g. Chlorine forms from the reaction of hydrochloric acid with manganese(IV) oxide.11 SOLUTIONS MANUAL According to the balanced equation. wood alcohol.4030 mol P4O10.0 g HCl 1 mol MnO __ 0. Methanol.989 mol/1.50 g CO (1 mol CO/28. Methanol Reaction Data 400.71 g) 100 87. and calculate the percent yield for this reaction. 95.7% yield 96.0 g MnO2 50.92 g of P4 1 mol P 123.989 mol MnO 2 2 86.52 g/9. 4 2 310. MnO2 reacts with HCl in a one-to-four ratio. 15.50 mol of CaC2 reacts with 5.04 g H2 103. the mass of iron(II) oxide formed was plotted on the graph shown in Figure 11.0 g CuO 1 mol Cu 1 mol CuO __ __ 79. For each mass of iron burned. elemental copper and water are produced.CHAPTER 11 SOLUTIONS MANUAL Step 3 is to determine the grams of Cl2 produced. To obtain this compound. Why does the graph level off after 25.0 g of copper(II) oxide is used? CuO H2 0 Cu H2O 32.55 g CuO 1 mol CuO 63. Electrolysis Determine the theoretical and percent yield of hydrogen gas if 36.0 g of iron is burned? How many moles of oxygen are present in the fixed amount? Chemistry: Matter and Change • Chapter 11 229 .0 g of water undergoes electrolysis to produce hydrogen and oxygen and 3.1% 2 4.0 g H2O (1 mol H2O/18. What mole ratio would you use to determine the moles of NH4NO3 produced if the moles of CuS are known? 102. 1.90 g CI _ __ 2 4 mol HCI 24.3% 20.3 g CI2 2 mol NO __ 2 2 mol NO Mixed Review 98.0 g CI2 24. b.00 mol H2O (CaNCN) can be used as a nitrogen source for crops.80 g H _ 100 94. 2 1 mol CI2 2NO(g) O2(g) 0 2NO2(g) Step 4 is to determine percent yield. but it immediately converts to nitrogen dioxide as it reacts with oxygen. Inc.00 mol N2 80.3 g CI2. given that 20. Nitrogen oxide is present in urban pollution. Ammonium sulfide reacts with copper(II) nitrate in a double replacement reaction.80 g of hydrogen is collected.00 mol of N2? 5.02 g H2O) 2. (NH4)2S Cu(NO3)2 0 2NH4NO3 CuS 2H2O(l) 0 H2(g) O2(g) 2 mol NH NO __ 36.55 g Cu __ 25.02 g H _ 4. Fertilizer The compound calcium cyanamide 1 mol H __ 30 20 10 0 0 5 10 15 20 25 30 35 Mass of Fe (g) 4Fe(s) 1 3O2(g) 0 2Fe2O3(s) Different amounts of iron were burned in a fixed amount of oxygen. Write the balanced chemical equation for the formation of nitrogen dioxide from nitrogen oxide. Solutions Manual 2.37 mol HCl a. When copper(II) oxide is heated in the presence of hydrogen gas. a division of The McGraw-Hill Companies. What mole ratio would you use to convert from moles of nitrogen oxide to moles of nitrogen dioxide? _ 100 82. Iron reacts with oxygen as shown.0 g of Cl2 is actually formed. calcium carbide is reacted with nitrogen at high temperatures. 1 mol CI 70.6 g Cu 1 mol Cu 101. Mass of Fe2O3 Formed From Burning Fe Mass of Fe2O3 (g) Copyright © Glencoe/McGraw-Hill. 99.00 mol 4 3 1 mol CuS theoretical yield 2. What mass of copper can be obtained if 32.11 g CaNCN 1 mol CaNCN __ __ 1 mol N2 1 mol CaNCN 401 g CaNCN 100.04 g H 2 2 mol H2O 2 2 1 mol H2 % yield 3. CaC2(s) N2(g) 0 CaNCN(s) C(s) What mass of CaNCN can be produced if 7. 329 mol O2 1 mol Fe O 3 mole O __ __ 2 3 159. The graph levels off because the oxygen limits the reaction at that point. Heat the dish gently for 5 minutes. High results could mean the product was not completely dry. and allowed the beakers to sit overnight. divided it into two samples. and identify the limiting reactant. Apply Concepts When a campfire begins to die down and smolder. additional oxygen is added and the remaining coals can burn.0 mL 10. Reaction with Drop of Na3PO4 Reaction with Drop of Co(NO3)2 Trial Volume Na3PO4 Volume Co(NO3)2 ment that can be used to determine the percent yield of anhydrous copper(II) sulfate when copper(II) sulfate pentahydrate is heated to remove water. Cool the dish and remass.0 g Fe2O3 0. determine the theoretical yield of copper(II) 3 15.11 SOLUTIONS MANUAL sulfate. Is such a percent yield possible? Explain. each beaker had a purple precipitate formed at the bottom. then strongly for 5 minutes to drive off the water. Inc. Using the equation CuSO4·5H2O 0 CuSO4 5H2O and the initial mass of the copper(II) sulfate pentahydrate. Determine the mass of the anhydrous copper sulfate.0 mL 10. a division of The McGraw-Hill Companies. No. 108. b. you obtain a percent yield of product of 108%. Think Critically 104. Add 2. 105. 1 5. because there are two reactants. what reasons might explain such a result? When you fan the flame. or it was contaminated. CHAPTER . Yes. Apply Concepts Students conducted a lab to investigate limiting and excess reactants.0 mL no reaction purple precipitate Obtain and record the mass of an empty evaporating dish. Write a balanced chemical equation for the reaction.0 mL 10. 35. because there is only one reactant. the flame can be rekindled if you start to fan it. percent yields cannot be greater than 100%.0 mL no reaction purple precipitate 4 20. 2Na3PO4 (aq) 3Co(NO3)2 (aq) 0 Co3(PO4)2 (s) 6NaNO3 (aq) Solutions Manual Copyright © Glencoe/McGraw-Hill.5. and added one drop of sodium phosphate solution to one sample and one drop of cobalt(II) nitrate solution to the second sample. They then added a constant volume of cobalt(II) nitrate solution (Co(NO3)2). Their results are shown in Table 11. The students decanted the supernatant from each beaker. Silver nitrate and hydrochloric acid react to produce silver chloride and nitric acid. Explain why or why not. Assuming that your calculation is correct. stirred the contents. a. Divide the actual yield by the theoretical yield and multiply by 100 to determine percent yield of copper(II) sulfate. Analyze and Conclude In an experiment. Not enough information is given to identify which is the limiting reactant. Determine the actual yield of copper(II) sulfate. The next day. Record. Design an Experiment Design an experi- 230 Reaction Data for Co(NO3)2 and Na3PO4 Chemistry: Matter and Change • Chapter 11 a.0 mL purple no precipitate reaction 2 10. Observe and Infer Determine whether each reaction depends on a limiting reactant.0 mL no reaction purple precipitate 106. No.0 mL 10. Potassium chlorate decomposes to form potassium chloride and oxygen. Explain in terms of stoichiometry why the fire begins to flare up again. The students added different volumes of sodium phosphate solution (Na3PO4) to a beaker.7 g Fe2O3 2 2 mole Fe2O3 107.00 g of copper(II) sulfate pentahydrate and obtain and record the mass of the hydrate and evaporating dish. You observe that sugar dissolves more quickly in hot tea than in iced tea. VO2 2H2 0 V 2H2O Trials 2-4: Co(NO3)2 is limiting.59 g V2O5 1 mol H2 __ __ 181.38 g __ 0.94 g/mol 3.CHAPTER 11 b. 2 (1 mol V: 2.532 g H2 Cumulative Review 110. (Chapter 5) a.016 g H2 0.106 g 0.5 0.426 g 0. Determine the mass of hydrogen needed to complete the steps of this reaction. When 9.76 g.94 g/mol 1 mol VO2 2.21 g __ 0.999 g/mol Divided the smallest molar quantity.106 mol _ 1 0.99 艐 2 V2O5 H2 0 2VO2 H2O c.76 g VO2 __ _ 82. Copyright © Glencoe/McGraw-Hill. a division of The McGraw-Hill Companies. Challenge Problem 109. Co(NO3)2 is in excess because the addition of Na3PO4 caused an additional reaction. When the second vanadium oxide undergoes additional heating in the presence of hydrogen. aluminum [Ne]3s23p1 c. Based on the results.263 mol 50. because it is based only on observation. identify the limiting b.999 g/mol Divided the smallest molar quantity.106 mol 0.94 g/mol 15.106 mol 1 mol V: 2 mol O VO2 Solutions Manual Chemistry: Matter and Change • Chapter 11 231 .106 mol _ 1 0.106 mol 0. Write the electron configuration for each of the following atoms. Is this statement a hypothesis or a theory? Why? (Chapter 1) a hypothesis.016 g H2 __ 0. This new vanadium oxide has a mass of 8.38 g __ 0.106 g H2 1 mol H2 2 mol H2 8. Na3PO4 is in excess because the addition of Co(NO3)2 caused and additional reaction. SOLUTIONS MANUAL V: 5. 0.211 mol _ 1.263 mol _ 2.106 mol O: 4.59 g of a certain vanadium oxide is heated in the presence of hydrogen.38 g of vanadium metal forms.88 g/mol 1 mol V2O5 2. titanium [Ar]4s23d2 d. Trial 1: Na3PO4 is limiting. Inc. 0. radon [Xe]6s24f145d106p6 0. Write balanced equations for the steps of reactant and the excess reactant for each trial.426 g H2 1 mol H2 __ Total hydrogen mass 0. Determine the empirical formulas for the two vanadium oxides. a. 9.5 mol O) V2O5 V: O: 5. the reaction.38 g __ 0.211 mol 15.106 mol 50. fluorine [He]2s22p5 b.48 艐 2.106 mol Multiply by the mole ratio by 2.You state that higher temperatures increase the rate at which sugar dissolves in water. water and a new oxide of vanadium are formed. not on data 111. 5. NO2 and O3. Chemistry: Matter and Change • Chapter 11 ammonia produced when hydrogen and nitrogen are combined under ordinary conditions is extremely small.1 9. Haber Process The percent yield of 20 0 produced by combustion of gasoline in internal combustion engines. (Chapter 10) Percent Composition of Some Organic Compounds %C %H %O 40.4 g O (1 mol/16. N2(g) 3H2(g) 2NH3(g) 92 kJ.4 30 13. Make sure the equation. However. is included.02 g/mol 116.5 54. The process was of great importance because the Germans had to come up with a nitrogen compound that could be produced in large amounts. 117.00 g ) 2. What is the empirical formula of butanoic as diatomic molecules.28 mol O 4 2.16 gives percent composition data for several organic compounds.54 mol C/2. Solutions Manual Copyright © Glencoe/McGraw-Hill. Explain why the gaseous nonmetals exist b.7 Ethanol 9. Common pollutants are SO2.4 34.11 CHAPTER SOLUTIONS MANUAL 112. Air Pollution Research the air pollutants 115.03 6(18. through the use of stoichiometry. Check that stoichiometric calculations account for a decrease in the pollutant.5 36. a division of The McGraw-Hill Companies.28 mol 4.28 mol O 2 9. (Chapter 9) 54. 238. The monoatomic gases already have noble gas electron configurations.0 mol H/2.02 . The aim of the Haber process was to control a reaction so that a large amount of a useful product was yielded quickly.3 54.8 36. How are the molecular and empirical formulas of acetaldehyde and butanoic acid related? They are whole number multiples of one another. Research the conditions used in the Haber Process. Answers will vary. NO. Discuss the common pollutants and the reaction that produces them. 352. Figure 11. 232 Answers will vary. but other gaseous elements exist as single atoms.1 Formaldehyde Acetaldehyde Butanoic acid Compound name a.0 10 6.998) 352. and find out why the development of the process was of great importance. (Chapter 8) acid? Diatomic molecules achieve a noble gas electron configuration by forming covalent bonds. 9. how each pollutant could be reduced if more people used mass transit. What is the molecular mass of UF6? What is the molar mass of UF6? (Chapter 10) Additional Assessment 352.5 g C (1 mol/12. 114.28 mol O/2.1 g H (1 mol/1.0 mol 113.0 Percent by mass 52.28 mol O 1 4K(s) O2(g) 0 2K2O(s) The empirical formula is C2H4O.02 amu.01 g) 9.54 mol 36. the Haber Process combines the two gases under a set of conditions designed to maximize yield. Write a balanced equation for the reaction of potassium with oxygen.2 50 40 53. Inc. Show.01 g) 4. 9 mg 19.618 mol C6H4(OH)2 __ from reaction vs.08 104 mol 110. and how many milligrams are in excess? 2 mol C H (OH) __ 4 mol H O 110.1 mg of C6H4(OH)2 in excess 120. Supply of Various Chemicals in Dr.2 6 4 1.47 103 mol H2O2 0. Avogadro’s constant.0 mg H2O2 1 10 g 1 mg 34. 119.CHAPTER 11 SOLUTIONS MANUAL 2 mol C6H4(OH)2 __ from eqauation Document-Based Questions 4 mol H2O2 Bombardier Beetles Many insects secrete hydrogen peroxide (H2O2) and hydroquinone C6H4(OH)2. 24: no.47 103 mol H2O2 ___ 9. Bombardier beetles take this a step further by mixing these chemicals with a catalyst.0 g Ca(OH)2 300. d ? mg C6H4(OH)2 H2O2 0 C6H4O2 H2O O2 energy Use the graph below to answer Questions 2–5. constant mole ratios. Raitano’s Laboratory __ 3 100. conservation of mass. How many milligrams of benzoquinone will be produced? 1.0 g Na2CO3 500.02 g H2O2 1.0 g 1. d. Stoichiometry is based on the law of a. What is the excess reactant. 1 mol H2O2 Solutions Manual Chemistry: Matter and Change • Chapter 11 233 .0 g KClO3 200.08 104 mol C6H4(OH)2 AgNO3 100.0 mg C6H4(OH)2 1 10 g 1 mg 1 mol __ 9.9 mg C6H4(OH)2 used Figure 11. Bob.0 mg of hydroquinone (C6H4(OH)2) along with 50.47 103 mol H2O2 Standardized Test Practice pages 398–399 1.17 below shows the chemical reaction that results in the bombardier beetle’s defensive spray. conservation of energy.0 mg of hydrogen peroxide (H2O2). Data obtained from: Becker. H2O2 is the limiting reactant.00 g C6H4(OH)2 C6H4(OH)2 NaCl 700.0 mg 80. c.09 g C6H4O2 __ O C6H4(OH)2 Hydroquinone C6H4O2 Benzoquinone Copyright © Glencoe/McGraw-Hill.47 103 mol H2O2 H2O2 H2O O2 Energy 108. Researchers hope to use a similar method to reignite aircraft turbine engines. 118. The result is an exothermic chemical reaction and a spray of hot. b.0 g NaH2PO4 350. what is the limiting reactant? Balanced reaction: 2C6H4(OH)2 4H2O2 0 2C6H4O2 6H2O O2 100 mg 50 mg 2 mol C6H4O2 __ 4 mol H2O2 1 mg __ 1 mol C6H4O2 1 103 g 79.4 mg C6H4O2 produced Catalyst OH 2 2 O OH 2 1 mol C6H4(OH)2 1 103 g 80. Inc.12 g C H (OH) 1 mg __ __ 2 6 4 100. ChemMatters. irritating chemicals for any would-be predator. a division of The McGraw-Hill Companies.0 g __ __ 3 1 mol 50. April 2006. If the bombardier beetle stores 100. 98.94 g Na H P O 119.8 g __ 100 _ 39.00 g O 3 mol O __ __ 39.72 mol c. 9. 18. 37.0 g Na2CO3 4. Pure silver metal can be made using the reac- Cu(s) 2AgNO3(aq) 0 2Ag(s) Cu(NO3)2(aq) 2KClO3(s) 0 2KCl(s) 3O2(g) How many grams of copper metal will be needed to use up all of the AgNO3 in Dr. Na2CO3(aq) Ca(OH)2(aq) 0 2NaOH(aq) CaCO3(s) Using the amounts of chemicals available in Dr.43 mol c Find the limiting reactant.11 SOLUTIONS MANUAL 2.00 g Na2CO3 4. Sodium dihydrogen pyrophosphate (Na2H2P2O7).98 g NaH PO 2 mol NaH PO __ __ 2 2 2 444. 100 g a 100.097 mol NaOH 234 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. 8.17 g theoretical yield 100 32.0 g Na2H2P2O7 2 4 1 mol Na2H2P2O7 48.6% d. what is the maximum number of moles of NaOH that can be produced? a. 4. 130. Pure O2 gas can be generated from the decomposition of potassium chlorate (KClO3): tion shown below. The equation for this process is as follows.10 g Ca(OH)2 4.0 g NaH2PO4 7 2 2 2 2 7 4 1 mol Na2H2PO4 480. Raitano have to buy to make enough Na2H2P2O7? a. how much more NaH2PO4 will Dr.0 g Ca(OH)2 If half of the KClO3 in the lab is used and 12. 65. is manufactured by heating NaH2PO4 at high temperatures: 2NaH2PO4(s) 0 Na2H2P2O7(s) H2O(g) If 444.17 g O 2 2 2 mol KCIO3 percent 2 1 mol O2 actual yield 12. 0.0 g KCIO3 2 theoretical yield 100.05 mol b.717 mol Na2CO3 300.0 g of Na2H2P2O7 is needed.0 g 350.55w g Cu 18. Raitano’s lab.8 g of oxygen gas is produced.00 g b. what is the percent yield of this reaction? a.049 mol Ca(OH)2 b _1 (200. 12. Raitano’s laboratory? a.0 g AgNO3 __ 1 mol AgNO 1mol Cu __ __ 3 169. 1 mol Na CO __ 2 __ 1 mol Ca2 (OH)2 74. 32.0 g KCIO3 1 mol KCIO __ 3 122.88 g AgNO3 63.0 g KCIO3) 100. 74.7% c.8% b. a division of The McGraw-Hill Companies.0 g 130.0 g d.7 g d. 2 mol NaOH __ c 1 mol Na H P O __ 221.0 g 1 mol Ca(OH)2 8. The LeBlanc process is the traditional method 500.3 g c.5 g KCIO3 32. Inc.097 mol d. 4. 480 g 3 106.70 g b. 94.70 g Cu 1 mol Cu 2 mol AgNO3 of manufacturing sodium hydroxide.0 g c.0% 3. more commonly known as baking powder. 4.049 mol Ca(OH)2 Ca(OH)2 is the limiting reactant.7% 5. CHAPTER . Which elements tend to have the largest atomic radius in their periods? a.8% d __ 1 mol O2 theoretical yield O2 3.54 mol of O2 and 618 g of Hg. p c. d d.8% Copyright © Glencoe/McGraw-Hill.55 mol HgO 2 mol HgO 1. Red mercury(II) oxide decomposes at high temperatures to form mercury metal and oxygen gas: 2HgO(s) 0 2Hg(l) O2(g) If 3. 42. 13.CHAPTER 11 SOLUTIONS MANUAL 6. Dimethyl hydrazine (CH3)2N2H2 ignites sponta- W W Y Y 3 4 5 6 7 8 9 10 11 12 W W Y Z Z Z Z Z Z Z Z Z Z W W Y Z Z Z Z Z Z Z Z Z Z W W Y Z Z Z Z Z Z Z Z Z Z W W W W W W W W W W W W W W W W W Y W W W W W Y Y Z Z Z X X X X X X X X X X X X X X X X X X X X X X X X X X X X 7. First Ionization Energy of Period 3 Elements Element Atomic Number 1st Ionization Energy. how many moles of nitrogen gas will be released? Set up a mole ratio: 27 moles N2 18 1 13 14 15 16 17 2 neously upon contact with dinitrogen tetroxide (N2O4): 18 moles N2O4 1. If 18. (CH3)2N2H2(l) 2N2O4(l) 0 3N2(g) 4H2O(g) 2CO2(g) Because this reaction produces an enormous amount of energy from a small amount of reactants.55 mol of HgO decomposes to form 1. X c. f b Solutions Manual Chemistry: Matter and Change • Chapter 11 235 . what is the percent yield of this reaction? a. kJ/mol Sodium 11 496 Magnesium 12 736 Aluminum 13 578 Silicon 14 787 Phosphorus 15 1012 Selenium 16 1000 Chlorine 17 1251 Argon 18 1521 c 8.775 mol PERIODIC TABLE Y Y Y Y Y Y 9. Elements labeled W have their valence electrons in which sublevel? a.5% c.54 mol _ 100 86. a division of The McGraw-Hill Companies.2% b. Z 3 mole N __ 2 2 moles N2O4 3 moles N __ 2 2 moles N2O4 Use the table below to answer questions 10 and 11.775 mol O2 percent yield 1. Inc. W b. 56.0 mol of dinitrogen tetroxide is consumed in this reaction. Y d. it was used to drive the rockets on the Lunar Excursion Modules (LEMs) of the Apollo space program. s b.6% d. 86. Use the pictures below to answer Questions 13–17. First Ionization Energy b. 10 12 14 16 18 Atomic Number energy. First Ionization Energy (Kj/mol) Atomic Number vs.13 g Co2TiO4 3.13 g of the compound? a. Hydrogen sulfide displays this molecular shape.0310 mol 2 4 229. 3. Elements in the first few families have only 1 or 2 valence electrons. a 14. c 15.22 101 mol d. is in 7.39 101 mol b. 4. Inc. Ionization generally increases as you move across a period or row in the periodic table. Carbon dioxide displays this molecular shape. 1600 1400 1200 1000 800 600 400 200 c. 3. a division of The McGraw-Hill Companies. 12. in moles. Place atomic numbers on the x-axis. 2.10 102 mol c. This shape of molecule is referred to as trigonal planar. Molecules with this shape undergo sp2 hybridization.10 102 mol 1 mol Co TiO __ 0. similar to Figure 6.74 g Co2TiO4 16. b 236 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. d 17. Plot the data from this data table. Summarize the general trend in ionization . Molecules with this shape have four shared pairs of electrons and no lone pairs of electrons. e.11 CHAPTER SOLUTIONS MANUAL 10. b b 7. a.17 102 mol 13. therefore making it more likely for these elements to gain a few electrons rather than lose many. Elements on the right side of the periodic table have very high ionization energies because their outer shells are nearly filled. How much cobalt(III) titanate (Co2TiO4). which are relatively easy to remove since this will result in a complete outer shell. Data should form an approximately linear relationship with a few jagged edges.16 on page 191. 11. How does ionization energy relate to the number of valence electrons in an element? d.

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