Answers and Solutions to Text Problems

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1 Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x items. For example, one mole of water contains molecules of water. 7.2 Avogadro s number is , which equals the number of items in one mole. 7.3 a. One mole of carbon contains atoms of carbon. b. One mole of iron contains atoms of iron. 2.0 moles Fe atoms Fe = atoms Fe 1 mole Fe c moles CO molecules CO 2 = molecules CO 2 1 mole CO a. 2.5 moles S 6.02 x atoms S = atoms S 1 mole S b mole Ag 6.02 x atoms Ag = atoms Ag 1 mole Ag c. 8.0 moles H 2 O 6.02 x molecules H 2 O = molecules H 2 O 1 mole H 2 O 7.5 The subscripts indicate the moles of each element in one mole of that compound. a. 24 moles of H b. 5.0 moles quinine 20 moles C atoms 6.02 x atoms C = atoms C 1 mole quinine 1 mole C c moles quinine 2 moles N atoms atoms N = atoms N 1 mole quinine 1 mole N atoms 7.6 The subscripts indicate the moles of each element in one mole of that compound. a. 3.0 moles Al 2 (SO 4 ) 3 3 moles S atoms = 9.0 moles S 1 mole Al 2 (SO 4 ) 3 b moles Al 2 (SO 4 ) 3 2 moles Al ions Al ions = Al ions 1 mole q Al 2 (SO 4 ) 3 1 mole Al ions c. 1.5 moles Al 2 (SO 4 ) 3 3 moles SO 4 ions SO 4 ions = SO 4 2- ions 1 mole Al 2 (SO 4 ) 3 1 mole SO 4 ions 7.7 a. 2.5 moles of C atoms 6.02 x atoms of C = 1.5 x atoms C 1 mole of C atoms b. 2.0 moles PCl 3 3 moles Cl atoms 6.02 x atoms Cl = 3.6 x atoms Cl 1 mole PCl 3 1 mole Cl atoms c. 5.0 x atoms O 1 moles O atoms 1 mole SO 3 = 2.8 moles SO x atoms O 3 mole O atoms

2 Chemical Quantities 7.8 a. 1.5 moles of Au atoms atoms of Au = atoms Au 1 mole of Au atoms b moles NBr 3 3 moles Br atoms 6.02 x atoms Br = atoms Br 1 mole NBr 3 1 mole Br atoms c H atoms 1 mole H atoms 1 mole C 2 H 6 O = 1.4 moles C 2 H 6 O 6.02 x H atoms 6 moles H atoms 7.9 a. 1 mole of Na and 1 mole of Cl: 23.0 g g = 58.5 g/mole NaCl b. 2 mole of Fe and 3 moles of O: g g = g/mole Fe 2 O 3 c. 2 mole of Li and 1 mole of C and 3 moles of O: 13.8 g g g = 73.8 g/mole Li 2 CO 3 d. 2 mole of Al and 3 moles of S and 12 moles of O: 54.0 g g g = g/mole Al 2 (SO 4 ) 3 e. 1 mole of Mg and 2 moles of O and 2 moles of H: 24.3 g g g = 58.3 g/mole Mg(OH) 2 f. 16 moles of C and 19 moles of H and 3 moles of N and 5 moles of O and 1 mole of S: g g g g g = g/mole C 16 H 19 N 3 O 5 S 7.10 a. 1 mole of Fe and 1 mole of S and 4 moles of O: 55.9 g g g = g/mole FeSO 4 b. 2 moles of Al and 3 moles of O: 54.0 g g = g/mole Al 2 O 3 c. 7 moles of C, 5 moles of H, 1 mole of N, 3 moles of O, and 1 mole S: 84.0 g g g g and 32.1 = g/mole C 7 H 5 NO 3 S d. 3 moles of C and 8 moles of H and 1 mole of O: 36.0 g g g = 60.0 g/mole C 3 H 8 O e. 2 moles of N and 8 moles of H and 1 mole of C and 3 moles of O: 28.0 g g g g = 96.0 g/mole (NH 4 ) 2 CO 3 f. 1 mole of Zn and 4 moles of C and 6 moles of H and 4 moles of O: 65.4 g g g g = g/mole Zn(C 2 H 3 O 2 ) a mole Na 23.0 g Na = 46.0 g Na 1 mole Na b mole Ca 40.1 g Ca = 112 g Ca 1 mole Ca c mole Sn g Sn = 14.8 g Sn 1 mole Sn

3 Chapter 7 Answers and Solutions 7.12 a mole K 39.1 g K = 58.7 g K 1 mole K b. 2.5 mole C 12.0 g C = 30. g C 1 mole C c mole P 31.0 g P = 7.8 g P 1 mole P 7.13 a mole NaCl 58.5 g NaCl = 29.3 g NaCl 1 mole NaCl b mole Na 2 O 62.0 g Na 2 O = 109 g Na 2 O 1 mole Na 2 O c mole H 2 O 18.0 g H 2 O = 4.05 g H 2 O 1 mole H 2 O 7.14 a. 2.0 mole MgCl g MgCl 2 = 190 g MgCl 2 1 mole MgCl 2 b. 3.5 mole C 3 H g C 3 H 8 = 150 g C 3 H 8 1 mole C 3 H 8 c mole C 2 H 6 O 46.0 g C 2 H 6 O = 230. g C 2 H 6 O ( g) 1 mole C 2 H 6 O 7.15 a mole K 39.1 g K = 19.6 g K 1 mole K b mole Cl g Cl 2 = 35.5 g Cl 2 1 mole Cl 2 c mole Na 2 CO g Na 2 CO 3 = 53.0 g Na 2 CO 3 1 mole Na 2 CO a mole N g N 2 = 63.0 g N 2 1 mole N 2 b mole NaBr g NaBr = 232 g NaBr 1 mole NaBr c mole C 6 H g C 6 H 14 = 194 g C 6 H 14 1 mole C 6 H a mole MgSO g MgSO 4 = 602 g MgSO 4 1 mole MgSO 4 b mole CO g CO 2 = 11 g CO 2 1 mole CO a mole C 3 H g C 3 H 6 = 11 g C 3 H 6 1 mole C 3 H 6 b mole C 15 H 22 ClNO g C 15 H 22 ClNO 2 = 7.1 g C 15 H 22 ClNO 2 1 mole C 15 H 22 ClNO 2

4 7.19 a g Ag 1 mole Ag = mole Ag g Ag b g C 1 mole C = mole C 12.0 g C Chemical Quantities c g NH 3 1 mole NH 3 = mole NH g NH 3 d g SO 2 1 mole SO 2 = 1.17 mole SO g SO a g Ca 1 mole Ca = mole Ca 40.1 g Ca b g S 1 mole S = mole S 32.1 g S c g H 2 O 1 mole H 2 O = 2.22 mole H 2 O 18.0 g H 2 O d g O 2 1 mole O 2 = mole O 2 (Use atomic mass to 4 sig figs) g O a g CO 2 1 mole CO 2 = mole CO g CO 2 b g Al(OH) 3 1 mole Al(OH) 3 = mole Al(OH) g Al(OH) 3 c g MgCl 2 1 mole MgCl 2 = mole MgCl g MgCl a g NH 3 1 mole NH 3 = mole NH g NH 3 b g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2 = mole Ca(NO 3 ) g Ca(NO 3 ) 2 c g SO 3 1 mole SO 3 = mole SO g SO g NaOH 1 mole NaOH = 12 mole NaOH 40.0 g NaOH g Au 1 mole Au = mole Au g Au 7.25 a. 25 g S 1 mole S = 0.78 moles of S 32.1 g S b. 125 g SO 2 1 mole SO 2 1 mole S = 1.95 moles of S 64.1 g SO 2 1 mole SO 2 c. 275 g Al 2 S 3 1 mole Al 2 S 3 3 moles S = 5.50 moles of S g Al 2 S 3 1 mole Al 2 S 3

5 Chapter 7 Answers and Solutions 7.26 a. 75 g C 1 mole C = 6.3 moles C 12.0 g C b mole C 2 H 6 2 mole C = 0.50 mole of C 1 mole C 2 H 6 c. 5.0 x CO 2 molecules 1 mole CO 2 molecules 1 mole C atoms 6.02 x CO 2 molecules 1 mole CO 2 molecules = 8.3 mole C atoms 7.27 a g N 1 mole N 6.02 x atoms of N = atoms of N 14.0 g N 1 mole N b. 1.5 moles N 2 O 4 2 mole N 6.02 x atoms of N = atoms of N 1 mole N 2 O 4 1 mole N c. 2.0 moles N 2 2 mole N 6.02 x atoms of N = atoms of N 1 mole N 2 1 mole N 7.28 a. 5.0 g Ag atoms 1 mole Ag atoms 6.02 x Ag atoms = Ag Atoms g Ag atoms 1 mole Ag atoms b mole Ag 2 S 6.02 x Ag 2 S 2 Ag atoms = Ag Atoms 1 mole Ag 2 S 1 Ag 2 S c g AgCl 1 mole AgCl 6.02 x AgCl units 1 Ag ion = ions g AgCl 1 mole AgCl units 1 AgCl unit 7.29 a g Mg 100 = 39.0% Mg 38.0 g F 100 = 61.0% F 62.3 g MgF g MgF 2 b g Ca 100 = 54.1% Ca 32.0 g O 100 = 43.2%O 74.1 g Ca(OH) g Ca(OH) g H 100 = 2.7% H 74.1 g Ca(OH) 2 c g C 100 = 40.0% C 8.0 g H 100 = 6.7%H g C 4 H 8 O g C 4 H 8 O g O 100 = 53.3% O g C 4 H 8 O a g S 100 = 22.6% S g Na 2 SO 4 b g S 100 = 64.1% S g Al 2 S 3 c g S 100 = 40.1% S 80.1 g SO a. 3.2 g N 1 mole N = 0.23 mole N 14.0 g N 1.8 g O 1 mole O = 0.11 mole O 0.23 mole N 0.11 mole = 2.1 moles N 0.11 mole O 0.11 mole = 1.0 mole O formula = N 2 O

6 b. 8.0 g C 1 mole C = 0.67 mole C 12.0 g C 2.0 g H 1 mole H = 2.0 moles H 1.0 g H Chemical Quantities 2.0 moles H 0.67 mole = 3.0 moles H 0.67 mole C 0.67 mole = 1.0 mole C formula = CH 3 c. 1.6 g H x 1 mole H = 1.6 moles H 1.0 g H 22. g N x 1 mole N = 1.6 moles N 14.0 g N 76 g O x 1 mole O = 4.8 moles O 1.6 moles H 1.6 moles = 1.0 mole H 1.6 moles N 1.6 moles = 1.0 mole N 4.8 moles O 1.6 moles = 3.0 mole O formula = HNO a. 2.9 g Ag 1 mole Ag = mole Ag g Ag 0.43 g S 1 mole S = mole S 32.1 g S mole Ag mole = 2.1 moles Ag mole S mole = 1.0 mole S formula = Ag 2 S b. 22 g Na 1 mole Na = 0.96 mole Na 23.0 g Na 7.7 g O 1 mole O = 0.48 moles O 0.96 moles Na 0.48 mole = 2.0 moles Na 0.48 mole O 0.48 mole = 1.0 mole O formula = Na 2 O c. 19 g Na x 1 mole Na = 0.83 mole Na 23.0 g Na 0.83 g H x 1 mole H = 0.83 mole H 1.0 g H 27 g S x 1 mole S = 0.84 mole S 32.1 g S 53 g O x 1 mole O = 3.3 moles O 0.83 mole Na 0.83 moles = 1.0 mole Na 0.83 mole H 0.83 moles = 1.0 mole H 0.83 mole S 0.83 moles = 1.0 mole S 3.3 moles O 0.83 moles = 4.0 mole O formula = NaHSO 4

7 Chapter 7 Answers and Solutions 7.33 a. Converting from % by mass to grams in a100 g sample gives 71 g K and 29 g S 71 g K 1 mole K = 1.8 moles K 39.1 g K 29 g S 1 mole S = 0.90 moles S 32.1 g S 1.8 moles K 0.90 mole = 2.0 moles K 0.90 mole S 0.90 mole = 1.0 mole S formula = K 2 S b. Converting from % by mass to grams in a100 g sample gives 55 g Ga and 45 g F 55 g Ga 1 mole Ga = 0.79 mole Ga 69.7 g Ga 45 g F 1 mole F = 2.4 moles F 19.0 g F 0.79 mole Ga 0.79 mole = 1.0 mole Ga 2.4 moles F 0.79 mole = 3.0 moles F formula = GaF 3 c. Converting from % by mass to grams in a100 g sample gives 31 g B and 69 g O 31 g B 1 mole B = 2.9 moles B 10.8 g B 69 g O 1 mole O = 4.3 moles O 2.9 mole B 2.9 mole = 1.0 mole B x 2 = 2.0 moles B 4.3 moles O 2.9 mole = 1.5 moles O x 2 = 3.0 moles O formula = B 2 O a. Converting from % by mass to grams in a100 g sample gives 56 g Ca and 44 g S 56 g Ca 1 mole Ca = 1.4 moles Ca 40.1 g Ca 44 g S 1 mole S = 1.4 moles S 32.1 g S 1.4 moles Ca 1.4 mole = 1.0 moles Ca 1.4 mole S 1.4 mole = 1.0 mole S formula = CaS b. Converting from % by mass to grams in a100 g sample gives 78 g Ba and 22 g F 78 g Ba x 1 mole Ba = 0.57 mole Ba g Ba 22 g F x 1 mole F = 1.2 moles F 19.0 g F 0.57 mole Ba 0.57 mole = 1.0 mole Ba 1.2 moles F 0.57 mole = 2.1 moles F formula = BaF 2

8 Chemical Quantities c. Converting from % by mass to grams in a100 g sample gives 76 g Zn and 24 g P 76 g Zn 1 mole Zn = 1.2 moles Zn 65.4 g Zn 24 g P 1 mole P = 0.77 mole P 31.0 g P 1.2 moles Zn 0.77 mole = 1.55 moles Zn 2 = 3.1 moles Zn 0.77 mole P 0.77 mole = 1.0 moles P 2 = 2.0 moles P formula = Zn 3 P a. Two SO 2 molecules react with one O 2 molecule to produce two SO 3 molecules. Two moles of SO 2 react with one mole of O 2 to produce two moles of SO 3. b. Four P atoms react with five O 2 molecules to produce two P 2 O 5 molecules. Four moles of P react with five moles of O 2 to produce two moles of P 2 O a. Two aluminum atoms react with three chlorine molecules to produce two formula units of AlCl 3. Two moles of Aluminum react with three moles of chlorine to produce two moles of AlCl 3. b. Four HCl molecules react with one O 2 molecule to produce two Cl 2 molecules and two H 2 O molecules. Four moles of HCl react with one mole of O 2 to form two moles of Cl 2 and two moles of H 2 O a. Reactants: 2 moles of SO 2 plus 1 mole of O 2 = 2 moles (64.1 g/mole) + 1 mole (32.0 g/mole) = g Products: 2 moles of SO 3 = 2 moles (80.1 g/mole) = g b. Reactants: 4 moles of P and 5 moles of O 2 = 4 moles (31.0 g/mole) + 5 moles (32.0 g/mole) = 284 g Products: 2 moles of P 2 O 5 = 2 moles (142 g/mole) = 284 g 7.38 a. Reactants: 2 moles of Al and 6 moles of Cl = 54.0 g g = 267 g Products: 2 moles of Al and 6 moles of Cl = 54.0 g g = 267 g b. Reactant: 4 moles of H and 4 moles of Cl plus 2 moles of O = 4.0 g g g = g Products: 4 moles of Cl plus 4 moles of H and 2 moles of O = g g g = g 7.39 a. 2 mole SO 2 and 1 mole O 2 2 mole SO 2 and 2 mole SO 3 1 mole O 2 2 mole SO 2 2 mole SO 3 2 mole SO 2 1 mole O 2 and 2 mole SO 3 2 mole SO 3 1 mole O 2 b 4 mole P and 5 mole O 2 4 mole P and 2 mole P 2 O 5 5 mole O 2 4 mole P 2 mole P 2 O 5 4 mole P 5 mole O 2 and 2 mole P 2 O 5 2 mole P 2 O 5 5 mole O 2

9 Chapter 7 Answers and Solutions 7.40 a. 2 mole Al and 3 mole Cl 2 2 mole Al and 2 mole AlCl 3 3 mole Cl 2 2 mole Al 2 mole AlCl 3 2 mole Al 3 mole Cl 2 and 2 mole AlCl 3 2 mole AlCl 3 3 mole Cl 2 b. 4 mole HCl and 1 mole O 2 4 mole HCl and 2 mole Cl 2 1 mole O 2 4 mole HCl 2 mole Cl 2 4 mole HCl 4 mole HCl and 2 mole H 2 O 1 mole O 2 and 2 mole Cl 2 2 mole H 2 O 4 mole HCl 2 mole Cl 2 1 mole O 2 1 mole O 2 and 2 mole H 2 O 2 mole Cl 2 and 2 mole H 2 O 2 mole H 2 O 1 mole O 2 2 mole H 2 O 2 mole Cl a. 2.0 mole H 2 1 mole O 2 = 1.0 mole O 2 2 mole H 2 b. 5.0 mole O 2 2 mole H 2 = 10. mole H 2 1 mole O 2 c. 2.5 mole O 2 2 mole H 2 O = 5.0 mole H 2 O 1 mole O a. 1.0 mole N 2 3 moles H 2 = 3.0 moles H 2 1 mole N 2 b mole NH 3 1 mole N 2 = 0.30 mole N 2 2 moles NH 3 c. 1.4 moles H 2 2 moles NH 3 = 0.93 mole NH 3 3 mole H a mole SO 2 5 mole C = 1.25 mole C 2 mole SO 2 b. 1.2 mole C 4 mole CO = 0.96 mole CO 5 mole C c mole CS 2 2 mole SO 2 = 1.0 mole SO 2 1 mole CS 2 d. 2.5 mole C 1 mole CS 2 = 0.50 mole CS 2 5 mole C 7.44 a mole C 2 H 2 5 mole O 2 = 5.00 mole O 2 2 mole C 2 H 2 b. 3.5 mole C 2 H 2 4 mole CO 2 = 7.0 mole CO 2 2 mole C 2 H 2 c mole H 2 O 2 mole C 2 H 2 = 0.50 mole C 2 H 2 2 mole H 2 O d mole O 2 4 mole CO 2 = mole CO 2 5 mole O 2

10 7.45 a. 2.5 mole Na 2 mole Na 2 O 62.0 g Na 2 O = 78 g Na 2 O 4 mole Na 1 mole Na 2 O b g Na 1 mole Na 1 mole O g O 2 = 6.26 g O g Na 4 mole Na 1 mole O 2 c g Na 2 O 1 mole Na 2 O 1 mole O g O 2 = 19.4 g O g Na 2 O 2 mole Na 2 O 1 mole O 2 Chemical Quantities 7.46 a. 1.8 mole H 2 2 mole NH g NH 3 = 20. g NH 3 3 mole H 2 1 mole NH 3 b g N 2 1 mole N 2 3 mole H g H 2 = 0.60 g H g N 2 1 mole N 2 1 mole H 2 c. 12 g H 2 1 mole H 2 2 mole NH g NH 3 = 68 g NH g H 2 3 mole H 2 1 mole NH a. 8.0 mole NH 3 3 mole O g O 2 = 190 g O 2 4 mole NH 3 1 mole O 2 b g O 2 1 mole O 2 2 mole N g N 2 = 3.79 g N g O 2 3 mole O 2 1 mole N 2 c. 34 g NH 3 1 mole NH 3 6 mole H 2 O 18.0 g H 2 O = 54 g H 2 O 17.0 g NH 3 4 mole NH 3 1 mole H 2 O 7.48 a mole Fe 2 O 3 3 mole C 12.0 g C = 90.0 g C 1 mole Fe 2 O 3 1 mole C b g C 1 mole C 3 mole CO 28.0 g CO = 84.0 g CO 12.0 g C 3 mole C 1 mole CO c g Fe 2 O 3 1 mole Fe 2 O 3 2 mole Fe 55.9 g Fe = 4.20 g Fe g Fe 2 O 3 1 mole Fe 2 O 3 1 mole Fe 7.49 a g C 1 mole C 1mole CS g CS 2 = 50.8 g CS g C 5 moles C 1 mole CS g CS 2 (actual) 100 = 70.9% 50.8 g CS 2 (theoretical) b g SO 2 1 mole SO 2 1mole CS g CS 2 = 19.0 g CS g SO 2 2 moles SO 2 1 mole CS g CS 2 (actual) 100 = 63.1% 19.0 g CS 2 (theoretical) 7.50 a. Theorectical yield of Fe: 65.0 g Fe 2 O 3 1 mole Fe 2 O 3 2 mole Fe 55.9 g Fe = 45.5 Fe g Fe 2 O 3= 1 mole Fe 2 O 3 1 mole Fe Percent yield: 15.0 g Fe (actual) 100 = 33.0 % Fe actually produced 45.5 g Fe(theoretical)

11 Chapter 7 Answers and Solutions b. Theorectical yield of CO 2 : 75.0 g CO 1 mole CO 3 mole CO g CO 2 = 118 g CO g CO 3 moles CO 1 mole CO 2 Percent yield: 15.0 g CO 2 (actual) 100 = 12.7 % CO 2 actually produced g CO 2 (theoretical) g Al 1 mole Al 2 moles Al 2 O g Al 2 O 3 = 94.4 g Al 2 O g Al 4 moles Al 1 mole Al 2 O 3 Use the percent yield to convert theoretical to actual: 94.4 g Al 2 O 3 x 75.0 g Al 2 O 3 = 70.8g Al 2 O 3 (actual) 100 g Al 2 O Theorectical yield of CO 2 : 45.0 g C 3 H 8 1 mole C 3 H 8 3 moles CO g CO 3 = 135 g CO g C 3 H 8 1 moles C 3 H 8 1 mole CO 3 Use the percent yield to convert theoretical to actual: 135 g CO 2 x 60.0 g CO 2 = 81.0 g CO 2 (actual) 100 g CO In the expression for K eq the products are divided by the reactants with each concentration raised to a power that equals the coefficient in the equation: a. K eq = [CS 2 ][H 2 ] 4 b. K eq =[N 2 ][O 2 ] [CH 4 ][H 2 S] 2 [NO] 2 c. K eq = [CS 2 ][O 2 ] 4 [SO 3 ] 2 [CO 2 ] 7.54 In the expression for K eq the products are divided by the reactants with each concentration raised to a power that equals the coefficient in the equation: a. K eq = [H 2 ][Br 2 ] b. K eq = [CH 3 OH] [HBr] 2 [CO][H 2 ] 2 c. K eq = [CS 2 ][H 2 ] 4 [H 2 S] 2 [CH 4 ] 7.55 a. A large K eq means that the products are favored. b. A small Keq means that the reactants are favored a. A small K eq means that the reactants are favored. b. A large Keq means that the products are favored a g K 100 = 40.3% K 52.0g Cr x 100 = 26.8 %Cr g K 2 CrO g K 2 CrO g O 100 = 33.7% O g K 2 CrO 4 b g Al 100 = 12.9% Al 3.0g H 100 = 1.4 %H g Al(HCO 3 ) g Al(HCO 3 ) g C 100 = 17.1% C g O 100 = 68.6% O g Al(HCO 3 ) g Al(HCO 3 ) 3

12 Chemical Quantities c g C 100 = 40.0% C 12.0g H 100 = 6.7 %H g C 6 H 12 O g C 6 H 12 O g O 100 = 53.3% O g C 6 H 12 O a g P 100 = 43.7 % P 284 g P 4 O 10 b g P 100 = 46.0 % P g Mg 3 P 2 c g P 100 = 20.0 % P g Ca 3 (PO 4 ) a g S 1 mole S = mole S 32.1 g S 7.81 g F 1 mole F = mole F 19.0 g F mole S mole = 1.00 mole S moles F mole = 6.00 moles F formula = SF 6 b g Ag 1 mole Ag = mole Ag g Ag g N 1 mole N = mole N 14.0 g N 2.83 g O 1 mole O = mole O mole Ag mole = 1.00 mole Ag mole N mole = 1.00 mole N moles O mole = 3.00 moles O formula = AgNO 3 c g Au 1 mole Au = mole Au g Au 10.9 g O 1 mole O = mole O mole Au mole = 1.00 mole Au x 2 = 2.00 moles Au moles O mole = 1.50 moles O x 2 = 3.00 mole O formula = Au 2 O a. 61 g Sn 1 mole Sn = 0.51 mole Sn g Sn 39 g F 1 mole F = 2.1 moles F 19.0 g F 0.51 mole Sn 0.51 mole = 1.0 mole Sn 2.1 moles F 0.51 mole = 4.1 moles F formula = SnF 4

13 Chapter 7 Answers and Solutions b g N 1 mole N = 1.85 moles N 14.0 g N 74.1 g O 1 mole O = 4.63 moles O 1.85 moles N 1.85 mole = 1.00 mole N 4.63 moles O 1.85 mole = 2.50 moles O Calculation of whole numbers 1.00 mole N 2 = 2.00 mole N 2.50 moles O 2 = 5.00 moles O Formula N 2 O 5 c g Al 1 mole Al = mole Al 27.0 g Al 25.4 g P 1 mole P = mole P 31.0 g P 52.5 g O 1 mole O = 3.28 moles O mole Al mole = 1.00 mole Al moles P mole = 1.00 moles O 4.00 moles O mole = 4.00 mole O formula = AlPO kg body wt 60.0 kg H 2 O 1000 g 1 mole H 2 O = 2.33 x 10 3 moles H 2 O 100 kg body wt 1 kg 18.0 g H 2 O ml blood 400. mg C 2 H 6 O 1 g 1 mole C 2 H 6 O = mole C 2 H 6 O 100 ml blood 1000 mg 46.0 g C 2 H 6 O 7.63 a. 124 g C 2 H 6 O 1 mole C 2 H 6 O 1 mole C 6 H 12 O 6 = 1.35 mole C 6 H 12 O g C 2 H 6 O 2 mole C 2 H 6 O b kg C 6 H 12 O g 1 mole C 6 H 12 O 6 2 mole C 2 H 6 O 46.0 g C 2 H 6 O 1 kg g C 6 H 12 O 6 1 mole C 6 H 12 O 6 1 mole C 2 H 6 O 7.64 a. C 2 H 6 O + 3O 2 2CO 2 + 3H 2 O b. 4.0 mole C 2 H 6 O 3 mole O 2 = 12 mole O 2 1 mole C 2 H 6 O = 123 g C 2 H 6 O c. 88 g CO 2 1 mole CO 2 3 mole O g O 2 = 96 g O g CO 2 2 mole CO 2 1 mole O 2 d. 125 g C 2 H 6 O 1 mole C 2 H 6 O 2 mole CO g CO 2 = 239 g CO g C 2 H 6 O 1 mole C 2 H 6 O 1 mole CO g C 2 H 6 O 1 mole C 2 H 6 O 3 mole H 2 O 18.0 g H 2 O = 147 g 46.0 g C 2 H 6 O 1 mole C 2 H 6 O 1 mole H 2 O

14 Chemical Quantities NH 3 + 5F 2 N 2 F 4 + 6HF a mole HF 2 mole NH 3 = 1.33 mole NH 3 6 mole HF 4.00 mole HF 5 mole F 2 = 3.33 mole F 2 6 mole HF b mole NH 3 5 mole F g F 2 = 143 g F 2 2 mole NH 3 1 mole F 2 c g NH 3 1 mole NH 3 1 mole N 2 F g N 2 F 4 = 10.4 g N 2 F g NH 3 2 mole NH 3 1 mole N 2 F Theoretical yield 50.0 g Fe 2 O 3 1 mole Fe 2 O 3 2 moles Fe 55.9 g Fe = 35.0 g Fe g Fe 2 O 3 1 mole Fe 2 O 3 1 mole Fe Percent yield 32.8 g Fe x 100 = 93.7 % Fe (actual yield) C 2 H 2 (g) + 3O 2 (g) 2CO 2 (g) + H 2 O(g) 22.5 g C 2 H 2 1 mole C 2 H 2 2 moles CO g CO 2 = 76.2 g CO 2 (theoretical) 26.0 g C 2 H 2 1 mole C 2 H 2 1 mole CO g (actual) 100 = 81.4% (percent yield) 76.2 g (theoretical) 7.68 Theoretical yield would be 30.0 g NH g NH 3 = 46.2 g NH % 46.2 g NH 3 1 mole NH 3 1 mole N g N 2 = 38.0 g N 2 reacted 17.0 g NH 3 2 moles NH 3 1 mole N 2