Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
→ 187 76Os + (c) 187 Re 75 (d) 11 B+ 1 p 5 1 (e) 98 42 Mo 0 –1e → 3 42He 1 + 21H → 99 43Tc + 0n 7. The mass number drops by 28, and only alpha decay drops the mass number, and each alpha particle is 4, so 7 alpha particles are emitted. The atomic number drops by 10, which is 4 less than the drop of 14 caused by the alpha decays, so there must be 4 beta particles emitted, because each one raises the atomic number by 1. Making Connections 8. A typical report would include information on directing a particle beam through the body to cause maximum damage at a tumour location — and probably on the development of the “cobalt bomb” cobalt-60 cancer therapy machines by AECL, and the extensive worldwide use of this treatment program. GO TO www.science.nelson.com, Chemistry 11, Teacher Centre. 5.4 CALCULATING MASSES OF REACTANTS AND PRODUCTS PRACTICE (Page 227) Understanding Concepts 1. 2 Al2O3(s) → 3 O2(g) 125 g 101.96 g/mol 1 mol nAl O = 125 g 2 3 101.96 g nAl O = 1.23 mol 2 3 4 nAl = 1.23 mol 2 nAl = 2.45 mol 26.98 g mAl = 2.45 mol 1 mol mAl = 66.2 g or mAl mAl 4 Al(s) m 26.98 g/mol 1 mol A l2O3 l 26.98 g Al 4 mol A = 125 g Al 2O3 l2O3 1 mol A 101.96 g A l2O3 2 mol A l = 66.2 g The (maximum) mass of aluminum that can be produced is 66.2 g. 2. C3H8(g) 5 O2(g) → 3 CO2(g) 4 H2O(g) 10.0 g m 44.11 g/mol 32.00 g/mol 1 mol nC H = 10.0 g 3 8 44.11 g nC H = 0.227 mol 3 8 5 nO = 0.227 mol 2 1 nO = 1.13 mol 2 32.00 g mO = 1.13 mol 2 1 mol 120 Unit 2 Quantities in Chemical Reactions Copyright © 2002 Nelson Thomson Learning mO = 36.3 g mO 1 mol C 5 mol O2 32.00 g O 3H8 = 10.0 g C 2 3H8 44.11 g C H 1 m o l C H 1 mol O 2 3 8 3 8 = 36.3 g 2 or 2 mO 2 The (minimum) mass of oxygen required is 36.3 g. 3. 2 NaCl(aq) Pb(NO3)2(aq) → 2 NaNO3(aq) PbCl2(s) 2.57 g m 58.44 g/mol 278.10 g/mol 1 mol nNaCl = 2.57 g 58.44 g nNaCl = 0.0440 mol 1 nPbCl = 0.0440 mol 2 2 nPbCl = 0.0220 mol 2 278.10 g mPbCl = 0.0220 mol 2 1 mol mPbCl = 6.11 g 2 or 1 mol P bCl 278.10 g PbCl2 1m ol N aCl mPbCl = 2.57 g NaCl 2 2 58.44 g N aCl 2 mol N aCl 1 mol P bCl2 mPbCl = 6.11 g 2 The (maximum) mass of lead(II) chloride produced would be 6.11 g. 4. 2 Al(s) 3 H2SO4(aq) → 3 H2(g) 2 Al2(SO4)3(aq) 2.73 g m 26.98 g/mol 2.02 g/mol 1 mol nAl = 2.73 g 26.98 g nAl = 0.101 mol 3 nH = 0.101 mol 2 2 nH = 0.152 mol 2 2.02 g mH = 0.152 mol 2 1 mol mH = 0.307 g 2 or mH 2 mH 2 3 mol H 2.02 g H l 1 mol A = 2.73 g Al 2 2 2 mol A 1 mol 26.98 g A l l H2 = 0.307 g The (predicted) mass of hydrogen would be 0.307 g, or 307 mg. 5. 2 KOH(aq) Cu(NO3)2(aq) → 2 KNO3(aq) Cu(OH)2(s) 2.67 g m 56.11 g/mol 97.57 g/mol 1 mol nKOH = 2.67 g 56.11 g nKOH = 0.0476 mol 1 nCu(OH) = 0.0476 mol 2 2 nCu(OH) = 0.0238 mol 2 97.57 g mCu(OH) = 0.0238 mol 2 1 mol Copyright © 2002 Nelson Thomson Learning Chapter 5 Quantities in Chemical Equations 121 mCu(OH) = 2.32 g 2 or 1 mol Cu( OH) 1m ol K OH mCu(OH) = 2.67 g KOH 2 2 56.11 g K OH 2 mol K OH mCu(OH) = 2.32 g 97. 57 g Cu(OH) 2 1 mol C u(OH)2 2 The mass of copper(II) hydroxide produced would be 2.32 g. 6. 2 H2O(l) → 2 H2(g) O2(g) 25.0 g N 18.02 g/mol 6.02 × 1023 molecules/mol 1 mol nH O = 25.0 g 2 18.02 g nH O = 1.39 mol 2 1 nO = 1.39 mol 2 2 nO = 0.694 mol 6.02 × 1023 molecules 1 mol = 4.18 × 1023 molecules 2 = 0.694 mol NO 2 NO 2 or NO 6.02 × 1023 molecules O2 1 mol H 1m ol O2 2O = 25.0 g H2O 18.02 g H2O 2 mol H2O 1 mol O2 NO = 4.18 × 1023 molecules 2 2 In this reaction, 4.18 × 1023 molecules of oxygen would be produced. PRACTICE (Page 228) Understanding Concepts 7. The fundamental test for a scientific concept is its ability to predict. 8. The reaction equation coefficients show the ratio of substances in moles, so measurements must be changed to amounts before the ratio can be applied. 9. (a) The mole ratio CO2 : C8H18 (from the equation) is 16 : 2. (b) 2 C8H18(g) 25 O2(g) → 16 CO2(g) 18 H2O(g) 22.8 g m 114.26 g/mol 44.01 g/mol 1 mol nC H = 22.8 g 8 18 114.26 g nC H = 0.200 mol 8 18 16 nCO = 0.200 mol 2 2 nCO = 1.60 mol 2 44.01 g mCO = 1.60 mol 2 1 mol mCO = 70.3 g 2 or mCO 2 mCO 2 1 mol C 16 mol C O2 44.01 g CO2 8H18 = 22.8 g C 8H18 114.26 g C 2 mol C 1m ol C O2 8H18 8H18 = 70.3 g The mass of carbon dioxide produced will be 70.3 g. (c) The mole ratio O2 : C8H18 (from the equation) is 25 : 2. 122 Unit 2 Quantities in Chemical Reactions Copyright © 2002 Nelson Thomson Learning (d) 2 C8H18(g) 22.8 g 114.26 g/mol nC H 8 18 nC H 8 18 nO 2 nO 2 mO 2 mO 2 or mO 2 mO 2 25 O2(g) → 16 CO2(g) m 32.00 g/mol 1 mol = 22.8 g 114.26 g = 0.200 mol 25 = 0.200 mol 2 = 2.49 mol 32.00 g = 2.49 mol 1 mol = 79.8 g 18 H2O(g) 1 mol C 25 mol O2 32.00 g O 8H18 = 22.8 g C 2 8H18 114.26 g C 2 mol C 1 mol O 2 8H18 8H18 = 79.8 g The mass of oxygen consumed will be 79.8 g. (e) If oxygen is the limiting reagent, some carbon will not react completely to CO2(g), but instead will react incompletely to form CO(g), which is very toxic. 10. TiO2(s) 2 Cl2(g) 2 C(s) → TiCl4(l) 2 CO(g) 1.00 kg 79.88 g/mol TiCl4(l) 2 Mg(s) → Ti(s) 2 MgCl2(s) m 47.88 g/mol Note: This question requires the students to see that the mole ratio TiCl4 : TiO2 from the first equation can be combined with the mole ratio Ti : TiCl4 from the second equation. nTiO 2 nTiO 2 nTiCl 4 nTiCl 4 nTi nTi mTi mTi or 1 mol = 1.00 kg 79.88 g = 0.0125 kmol 1 = 0.125 kmol (mole ratio TiCl4 : TiO2 is 1 : 1 — step 1) 1 = 0.125 kmol 1 = 0.125 kmol (mole ratio Ti : TiCl4 is 1 : 1 — step 2) 1 = 0.125 kmol 47.88 g = 0.125 kmol 1 mol = 0.599 kg or 599 g mTi 1 mol T iO 2 1 mol T i Cl i 47.88 g Ti 1 mol T = 1.00 kg TiO2 4 1 mol T 79.88 g T iO2 1 mol T iO2 1 mol T iC l4 i mTi = 0.599 kg or 599 g The (maximum) mass of titanium produced would be 599 g. Copyright © 2002 Nelson Thomson Learning Chapter 5 Quantities in Chemical Equations 123 SECTION 5.4 QUESTIONS (Page 229) Understanding Concepts 1. Reaction equations read in amounts, not masses, so combining reagents by mass according to equation values will not work. 2. (a) Fe2O3(s) 3 H2(g) → 2 Fe(s) 3 H2O(g) 500 kg m 159.70 g/mol 55.85 g/mol 1 mol nFe O = 500 kg 2 3 159.70 g nFe O = 3.13 kmol 2 3 2 nFe = 3.13 kmol 1 nFe = 6.26 kmol 55.85 g mFe = 6.26 kmol 1 mol mFe = 350 kg or mFe mFe 1 mol Fe 2 mol Fe 55.85 g Fe 2O3 = 500 kg Fe 2O3 159.70 g F e2O3 1 mol F e2O3 1 mol F e = 350 kg The mass of iron produced would be 350 kg. (b) Fe2O3(s) 3 H2(g) → 2 Fe(s) 1.0 t = 1.0 Mg m 159.70 g/mol 2.02 g/mol 1 mol nFe O = 1.0 Mg 2 3 159.70 g nFe O = 0.0063 Mmol = 6.3 kmol 2 3 3 nH = 0.0063 Mmol 2 1 nH = 0.019 Mmol 2 2.02 g mH = 0.019 Mmol 2 1 mol mH = 0.038 Mg = 38 kg 3 H2O(g) 2 or mH 2 mH 2 1 mol Fe 3m ol H2 2.02 g H 2O3 = 1.0 Mg Fe 2 2O3 159.70 g F e2O3 1 mol F e2O3 1 mol H2 = 0.038 Mg = 38 kg The mass of hydrogen required would be 38 kg. (c) Fe2O3(s) 3 H2(g) → 2 Fe(s) m 159.70 g/mol 1 mol nH O = 220 kg 2 18.02 g nH O = 12.2 kmol 2 1 nFe O = 12.2 kmol 2 3 3 nFe O = 4.07 kmol 2 3 159.70 g mFe O = 4.07 kmol 2 3 1 mol 124 Unit 2 Quantities in Chemical Reactions 3 H2O(g) 220 kg 18.02 g/mol Copyright © 2002 Nelson Thomson Learning mFe O = 650 kg mFe O 1 mol H 1 mol F e2O3 159.70 g Fe2O3 2O = 220 kg H2O 18.02 g H2O 3m ol H2O 1 mol F e2O3 = 650 kg 2 3 or 2 3 mFe O 2 3 The mass of iron(III) oxide consumed would be 650 kg. 3. (a) 2 C6H12(l) 5 O2(g) → 2 C6H10O4(s) m 280 kg 84.18 g/mol 146.16 g/mol 1 mol nC H O = 280 kg 6 10 4 146.16 g nC H O = 1.92 kmol 6 10 4 2 nC H = 1.92 kmol 6 12 2 nC H = 1.92 kmol 6 12 84.18 g mC H = 1.92 kmol 6 12 1 mol mC H = 161 kg 2 H2O(g) 6 12 or mC H 6 12 mC H 6 12 1 mol C6 H10O4 2m ol C 84.18 g C6H12 6H12 = 280 kg C6 H10O4 146.16 g C6 H10O4 2 mol C6 H10O4 1 mol C 6H12 = 161 kg The mass of cyclohexane required is 161 kg. (b) 2 C6H12(l) 5 O2(g) → 2 C6H10O4(s) 125 kg m 84.18 g/mol 32.00 g/mol 1 mol nC H = 125 kg 6 12 84.18 g nC H = 1.48 kmol 6 12 5 nO = 1.48 kmol 2 2 nO = 3.71 kmol 2 32.00 g mO = 3.71 kmol 2 1 mol mO = 119 kg 2 H2O(g) 2 or mO 2 mO 2 1m ol C 5 mol O2 32.00 g O 6H12 = 125 kg C 2 6H12 84.18 g C H 2 m ol C H 1 mol O 2 6 12 6 12 = 119 kg The mass of oxygen required is 119 kg. (c) 2 C6H12(l) 5 O2(g) → 2 C6H10O4(s) m 200 g 84.18 g/mol 32.00 g/mol 1 mol nO = 200 g 2 32.00 g nO = 6.25 mol 2 2 nC H = 6.25 mol 6 12 5 nC H = 2.50 mol 2 H2O(g) 6 12 Copyright © 2002 Nelson Thomson Learning Chapter 5 Quantities in Chemical Equations 125 mC H 6 12 mC H 6 12 or mC H 6 12 mC H 6 12 84.18 g = 2.50 mol 1 mol = 210 g 1 mol O 2 2 mol C 84.18 g C6H12 6H12 = 200 g O2 32.00 g O2 5 mol O2 1 mol C 6H12 = 210 g The maximum mass of cyclohexane that can be reacted is 210 g. 4. 2 Fe(s) O2(g) → 2 FeO(s) 56.8 g m 55.85 g/mol 71.85 g/mol 1 mol nFe = 56.8 g 55.85 g nFe = 1.02 mol 2 nFeO = 1.02 mol 2 nFeO = 1.02 mol 71.85 g mFeO = 1.02 mol 1 mol mFeO = 73.1 g or mFeO mFeO e 2 mol FeO 71.85 g FeO 1 mol F = 56.8 g Fe 2 mol Fe 1 mol F eO 55.85 g Fe = 73.1 g The mass of iron(II) oxide that can be produced is 73.1 g. Applying Inquiry Skills 5. (a) Prediction From the balanced reaction equation, one mole of calcium ions produces one mole of solid calcium oxalate precipitate, so the amount of calcium ions in a sample will be the same as the amount of calcium oxalate precipitate that forms. (b) Experimental Design Excess sodium oxalate solution will be added to a measured sample (independent variable) of hard water. The precipitate formed will be dried and the mass measured (dependent variable). The stoichiometric method will be used to calculate an answer to the question. (c) The mass of calcium oxalate precipitate will be divided by its molar mass, 128.10 g/mol, to determine the amount of precipitate. The amount of calcium ions is the same, since the mole ratio is 1 : 1. (d) Materials • hard-water sample 126 • sodium oxalate solution • wash bottle with pure water • two 100-mL graduated cylinders • 250-mL beaker • filtration apparatus • filter paper • centigram balance Unit 2 Quantities in Chemical Reactions Copyright © 2002 Nelson Thomson Learning (e) Procedure 1. Use a graduated cylinder to add 10.0 mL of hard water to a 250-mL beaker. 2. Use another graduated cylinder to add about 10 mL of sodium oxalate solution to the beaker. 3. Allow the precipitate to settle. 4. Add another 10-mL aliquot of sodium oxalate solution to the beaker, observing whether or not more precipitate forms as the added solution mixes with the clear upper portion of the solution in the beaker. 5. Repeat step 4 until no more precipitate forms. 6. Measure and record the mass of a piece of filter paper. 7. Filter, wash, and dry the precipitate. 8. Measure and record the mass of the filter paper plus dry precipitate. 9. Dispose of all materials as instructed. Making Connections 6. (a) Raw materials, like aluminum oxide, are purchased according to the amount needed for reaction. (b) Products, like aluminum, are priced so that the amount produced will be profitable. (c) Pollutant amounts will be calculated and decisions on processes to control them will be made accordingly. (d) Amounts of all materials must be calculated to allow costing of things like packaging, disposal, and shipping. 5.5 CALCULATING LIMITING AND EXCESS REAGENTS PRACTICE (Page 231) Understanding Concepts 1. (a) 2 CH4(g) 3 O2(g) → m 16.05 g/mol 3.0 mol nCH 4 nCH 4 mCH 4 mCH 4 mCH 4 mCH 4 or 2 CO(g) 4 H2O(g) 2 = 3.0 mol 3 = 2.0 mol 16.05 g = 2.0 mol 1 mol = 32 g 2 mol CH4 16.05 g CH4 = 3 mol O2 3 mol O2 1 mol C H4 = 32 g The mass of methane that will react is 32 g. (b) CH4(g) 2 O2(g) → CO2(g) m 16.05 g/mol 3.0 mol 1 nCH = 3.0 mol 4 2 nCH = 1.5 mol 4 16.05 g mCH = 1.5 mol 4 1 mol mCH = 24 g 2 H2O(g) 4 or mCH 4 1 mol CH4 16.05 g CH4 = 3 mol O2 2 mol O2 1 mol C H4 Copyright © 2002 Nelson Thomson Learning Chapter 5 Quantities in Chemical Equations 127