4. Balanced chemical equations tell us in what molar ratios substances combine to form products, not in what mass proportions they combine.

CHAPTER 9 1. The coefficients of the balanced chemical equation for a reaction give the relative numbers of molecules of reactants and products that are involved in the reaction.. The coefficients of the balanced chemical equation for a reaction indicate the relative numbers of moles of each reactant that combine during the process, as well as the number of moles of each product formed.. Although we define mass as the amount of matter in a substance, the units in which we measure mass are a human invention. Atoms and molecules react on an individual particle-by-particle basis, and we have to count individual particles when doing chemical calculations. 4. Balanced chemical equations tell us in what molar ratios substances combine to form products, not in what mass proportions they combine. 5. a. PCl (l) + H O(l) H PO (aq) + HCl(g) One molecule of liquid phosphorus trichloride reacts with three molecules of liquid water, producing one molecule of aqueous phosphorous acid and molecules of gaseous hydrogen chloride. One mole of phosphorus trichloride reacts with three moles of water to produce one mole of phosphorous acid and three moles of hydrogen chloride. b. XeF (g) + H O(l) Xe(g) + 4HF(g) + O (g) Two molecules of gaseous xenon difluoride react with two molecules of liquid water, producing two gaseous xenon atoms, four molecules of gaseous hydrogen fluoride, and one molecule of oxygen gas. Two moles of xenon difluoride reacts with two moles of water, to produce two moles of xenon, four moles of hydrogen fluoride, and one mole of oxygen. c. S(s) + 6HNO (aq) H SO 4 (aq) + H O(l) + 6NO (g) One sulfur atom reacts with six molecules of aqueous nitric acid, producing one molecule of aqueous sulfuric acid, two molecules of water, and six molecules of nitrogen dioxide gas. One mole of sulfur reacts with six moles of nitric acid, to produce one mole of sulfuric acid, two moles of water, and six moles of nitrogen dioxide. d. NaHSO (s) Na SO (s) + SO (g) + H O(l) Two formula units of solid sodium hydrogen sulfite react to produce one formula unit of solid sodium sulfite, one molecule of gaseous sulfur dioxide, and one molecule of liquid water. Two moles of sodium hydrogen sulfite react to produce one mole of sodium sulfite, one mole of sulfur dioxide, and one mole of water. 6. a. (NH 4 ) CO (s) NH (g) + CO (g) + H O(g) One formula unit of solid ammonium carbonate decomposes to produce two molecules of ammonia gas, one molecule of carbon dioxide gas, and one molecule of water vapor. 16

One mole of solid ammonium carbonate decomposes into two moles of gaseous ammonia, one mole of carbon dioxide gas, and one mole of water vapor. b. 6Mg(s) + P 4 (s) Mg P (s) Six atoms of magnesium metal react with one molecule of solid phosphorus (P 4 ) to make two formula units of solid magnesium phosphide. Six moles of magnesium metal react with one mole of phosphorus solid (P 4 ) to produce two moles of solid magnesium phosphide. c. 4Si(s) + S 8 (s) Si S 4 (l) Four atoms of solid silicon react with one molecule of solid sulfur (S 8 ) to form two molecules of liquid disilicon tetrasulfide. Four moles of solid silicon react with one mole of solid sulfur (S 8 ) to form two moles of liquid disilicon tetrasulfide. d. C H 5 OH(l) + O (g) CO (g) + H O(g) One molecule of liquid ethanol burns with three molecules of oxygen gas to produce two molecules of carbon dioxide gas and three molecules of water vapor. One mole of liquid ethanol burns with three moles of oxygen gas to produce two moles of gaseous carbon dioxide and three moles of water vapor. 7. False. The coefficients of the balanced chemical equation represent the ratios on a mole basis by which potassium hydroxide combines with sulfur dioxide. 8. Balanced chemical equations tell us in what molar ratios substances combine to form products, not in what mass proportions they combine. How could grams of reactant produce a total of grams of products? 9. 4Al(s) + O (g) Al O (s) For converting from a given number of moles of aluminum metal to the number of moles of oxygen needed for reaction, the correct mole ratio is mol O 4 mol Al For converting from a given number of moles of aluminum metal to the number of moles of product produced, the mole ratio is mol AlO 4 mol Al 10. Fe O (s) + H SO 4 (aq) Fe (SO 4 ) (s) + H O(l) For converting from a given number of moles of iron(iii) oxide to the number of moles of sulfuric acid required, the mole ratio is mol HSO 4 FeO For a given number of moles of iron(iii) oxide reacting completely, the mole ratios used to calculate the number of moles of each product are 164

Fe (SO 4) For Fe (SO 4 ) : FeO mol HO For H O: FeO 11. a. CO (g) + 4H (g) CH 4 (g) + H O(l) CH4 0.500 mol CO CO = 0.500 mol CH 4 mol HO 0.500 mol CO CO = 1.00 mol H O b. BaCl (aq) + AgNO (aq) AgCl(s) + Ba(NO ) (aq) mol AgCl 0.500 mol BaCl = 1.00 mol AgCl BaCl Ba(NO ) 0.500 mol BaCl BaCl c. C H 8 (g) + 5O (g) 4H O(l) + CO (g) 4 mol HO 0.500 mol C H 8 C H =.00 mol H O 8 mol CO 0.500 mol C H 8 C H = 1.50 mol CO 8 = 0.500 mol Ba(NO ) d. H SO 4 (aq) + Fe(s) Fe (SO 4 ) (aq) + H (g) Fe (SO 4) 0.500 mol H SO 4 = 0.167 mol Fe (SO 4 ) mol H SO 4 mol H 0.500 mol H SO 4 mol H SO = 0.500 mol H 4 1. a. 4Bi(s) + O (g) Bi O (s) mol BiO 0.50 mol Bi = 0.15 mol Bi O 4 mol Bi b. SnO (s) + H (g) Sn(s) + H O(g) Sn 0.50 mol SnO = 0.50 mol Sn SnO mol HO 0.50 mol SnO SnO = 0.500 mol H O c. SiCl 4 (l) + H O(l) SiO (s) + 4HCl(g) SiO 0.50 mol SiCl 4 = 0.50 mol SiO SiCl 4 165

4 mol HCl 0.50 mol SiCl 4 SiCl 4 = 1.00 mol HCl d. N (g) + 5O (g) + H O(l) 4HNO (aq) 4 mol HNO 0.50 mol N mol N = 0.500 mol HNO 1. a. AgNO (aq) + LiOH(aq) AgOH(s) + LiNO (aq) molar masses: AgOH, 14.91 g; LiNO, 68.95 g AgOH 0.15 mol AgNO AgNO = 0.15 mol AgOH 0.15 mol AgOH 14.91 g AgOH AgOH = 15.6 g AgOH LiNO 0.15 mol AgNO AgNO = 0.15 mol LiNO 68.95 g LiNO 0.15 mol LiNO LiNO = 8.6 g LiNO b. Al (SO 4 ) (aq) + CaCl (aq) AlCl (aq) + CaSO 4 (s) molar masses: AlCl, 1. g; CaSO 4, 16.15 g mol AlCl 0.15 mol Al (SO 4 ) Al (SO ) 4 = 0.50 mol AlCl 1. g AlCl 0.50 mol AlCl AlCl =. g AlCl mol CaSO4 0.15 mol Al (SO 4 ) Al (SO ) 4 = 0.75 mol CaSO 4 16.15 g CaSO4 0.75 mol CaSO 4 = 51.1 g CaSO 4 CaSO c. CaCO (s) + HCl(aq) CaCl (aq) + CO (g) + H O(l) molar masses: CaCl, 110.98 g; CO, 44.01 g; H O, 18.0 g CaCl 0.15 mol CaCO CaCO = 0.15 mol CaCl 4 110.98 g CaCl 0.15 mol CaCl CaCl = 1.9 g CaCl CO 0.15 mol CaCO CaCO = 0.15 mol CO 166

44.01 g CO 0.15 mol CO CO = 5.50 g CO HO 0.15 mol CaCO CaCO = 0.15 mol H O 18.0 g HO 0.15 mol H O H O =.5 g H O d. C 4 H 10 (g) + 1O (g) 8CO (g) + 10H O(g) molar masses: CO, 44.01 g; H O, 18.0 g 8 mol CO 0.15 mol C 4 H 10 mol C H = 0.500 mol CO 4 10 44.01 g CO 0.500 mol CO CO =.0 g CO 10 mol HO 0.15 mol C 4 H 10 mol C H = 0.65 mol H O 4 10 18.0 g HO 0.65 mol H O H O = 11. g H O 14. a. C 5 H 1 (l) + 8O (g) 5CO (g) + 6H O(l) molar masses: CO, 44.01 g; H O, 18.0 g 5 mol CO 0.750 mol C 5 H 1 C H =.75 mol CO 5 1 44.01 g CO.75 mol CO CO = 165 g CO 6 mol HO 0.750 mol C 5 H 1 C H = 4.50 mol H O 5 1 18.0 g HO 4.50 mol H O H O = 81.1 g H O b. CH OH(l) + O (g) 4H O(l) + CO (g) molar masses: H O, 18.0 g; CO, 44.01 g 4 mol HO 0.750 mol CH OH mol CH OH = 1.50 mol H O 18.0 g HO 1.50 mol H O H O = 7.0 g H O mol CO 0.750 mol CH OH mol CH OH = 0.750 mol CO 167

44.01 g CO 0.750 mol CO CO =.0 g CO c. Ba(OH) (aq) + H PO 4 (aq) BaHPO 4 (s) + H O(l) molar masses: BaHPO 4,. g; H O, 18.0 g BaHPO4 0.750 mol Ba(OH) Ba(OH) = 0.750 mol BaHPO 4. g BaHPO4 0.750 mol BaHPO 4 BaHPO = 175 g BaHPO 4 4 mol HO 0.750 mol Ba(OH) Ba(OH) = 1.50 mol H O 18.0 g HO 1.50 mol H O H O = 7.0 g H O d. C 6 H 1 O 6 (aq) C H 5 OH(aq) + CO (g) molar masses: C H 5 OH, 46.07 g; CO, 44.01 g mol CH5OH 0.750 mol C 6 H 1 O 6 C H O = 1.50 mol C H 5 OH 6 1 6 46.07 g CH5OH 1.50 mol C H 5 OH C H OH = 69.1 g C H 5 OH 5 mol CO 0.750 mol C 6 H 1 O 6 C H O = 1.50 mol CO 6 1 6 44.01 g CO 1.50 mol CO CO = 66.0 g CO 15. a. Cl (g) + KI(aq) KCl(aq) + I (s) mol KI 0.75 mol Cl = 0.550 mol KI Cl b. 6Co(s) + P 4 (s) Co P (s) P4 0.75 mol Co 6 mol Co = 0.0458 mol P 4 c. Zn(s) + HNO (aq) Zn(NO ) (aq) + H (g) mol HNO 0.75 mol Zn = 0.550 mol HNO Zn d. C 5 H 1 (l) + 8O (g) 5CO (g) + 6H O(g) 8 mol O 0.75 mol C 5 H 1 C H =.0 mol O 5 1 168

16. Before doing the calculations, the equations must be balanced. a. 4KO (s) + H O(l) O (g) + 4KOH(s) mol O 0.65 mol KOH 4 mol KOH = 0.469 mol O b. SeO (g) + H Se(g) Se(s) + H O(g) mol Se 0.65 mol H O = 0.98 mol Se mol H O c. CH CH OH(l) + O (g) CH CHO(aq) + H O(l) mol CHCHO 0.65 mol H O = 0.65 mol CH CHO mol H O d. Fe O (s) + Al(s) Fe(l) + Al O (s) mol Fe 0.65 mol Al O = 1.5 mol Fe Al O 17. the molar mass of the substance 18. Stoichiometry is the process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction. 19. a. molar mass Si = 8.09 g Si 4.15 g Si = 0.148 mol Si 8.09 g Si b. molar mass AuCl = 0.4 g 1 g AuCl.7 mg AuCl = 8.97 10 6 mol AuCl 1000 mg 0.4 g AuCl c. molar mass S =.07 g 1000 g S 1.05 kg S =.7 mol S 1 kg.07 g S d. molar mass FeCl = 16.0 g FeCl 0.000901 g FeCl 16. g FeCl = 5.55 10 6 mol FeCl e. molar mass MgO = 40.1 g 5.6 10 MgO g MgO = 19 mol MgO 40.1 g MgO 169

0. a. molar mass Ar = 9.95 g 1 g Ar 7.4 mg Ar = 1.81 10 mol Ar 1000 mg 9.95 g Ar b. molar mass CS = 76.15 g CS 5.7 g CS 76.15 g CS = 0.69 mol CS c. molar mass Fe = 55.85 g 1000 g Fe 784 kg Fe = 1.40 10 4 mol Fe 1 kg 55.85 g Fe d. molar mass CaCl = 110.98 g CaCl 0.00104 g CaCl 110.98 g CaCl = 9.7 10 6 mol CaCl e. molar mass NiS = 90.76 g 1.6 10 NiS g NiS = 1.9 mol NiS 90.76 g NiS 1. a. molar mass Ge = 7.59 g 7.59 g Ge.17 mol Ge = 158 g Ge Ge b. molar mass PbCl = 78.1 g; 4.4 millimol = 0.0044 mol 78.1 g PbCl 0.0044 mol PbCl = 1.18 g PbCl PbCl c. molar mass NH = 17.0 g 17.0 g NH 0.097 NH NH = 1.65 g NH d. molar mass C 6 H 14 = 86.17 g 4.6 10 86.17 g C6H14 mol C 6 H 14 C H 6 14 e. molar mass ICl = 16.5 g 16.5 g ICl 1.7 ICl = 78 g ICl ICl. a. molar mass of C H 8 = 44.09 g 44.09 g CH8. mol C H 8 C H = 98. g C H 8 8 =.67 10 5 g C 6 H 14 170

b. molar mass of Ar = 9.95 g; 9.0 millimol = 0.0090 mol 9.95 g Ar 0.0090 mol Ar = 0.61 g Ar Ar c. molar mass of SiO = 60.09 g 5.91 10 6 60.09 g SiO mol SiO SiO =.55 108 g SiO d. molar mass of CuCl = 14.45 g 14.45 g CuCl 0.000104 mol CuCl CuCl = 0.0140 g CuCl e. molar mass of CuCl = 99.00 g 99.00 g CuCl 0.000104 mol CuCl = 0.010 g CuCl CuCl. Before any calculations are done, the equations must be balanced. a. Co(s) + F (g) CoF (s) mol F 0.41 mol Co mol Co = 0.60 mol F b. Al(s) + H SO 4 (aq) Al (SO 4 ) (aq) + H (g) mol HSO4 0.41 mol Al = 0.60 mol H SO 4 mol Al c. K(s) + H O(l) KOH(aq) + H (g) mol HO 0.41 mol K mol K = 0.41 mol H O d. 4Cu(s) + O (g) Cu O(s) O 0.41 mol Cu 4 mol Cu = 0.10 mol O 4. Before any calculations are done, the equations must be balanced. a. Al(s) + Br (l) AlBr (s) molar mass Al = 6.98 g Al mol Br 0.557 g Al = 0.010 mol Br 6.98 g Al mol Al b. Hg(s) + HClO 4 (aq) Hg(ClO 4 ) (aq) + H (g) molar mass Hg = 00.6 g Hg mol HClO4 0.557 g Hg = 0.00555 mol HClO 4 00.6 g Hg 171

c. K(s) + P(s) K P(s) molar mass K = 9.10 g K P 0.557 g K = 0.00475 mol P 9.10 g K mol K d. CH 4 (g) + 4Cl (g) CCl 4 (l) + 4HCl(g) molar mass CH 4 = 16.04 g CH4 4 mol Cl 0.557 g CH 4 = 0.19 mol Cl 16.04 g CH CH 4 4 5. Before any calculations are done, the equations must be balanced. a. TiBr 4 (g) + H (g) Ti(s) + 4HBr(g) molar mass H =.016 g; molar mass Ti = 47.90 g; molar mass of HBr = 80.91 g H 1.5 g H.016 g H = 6.0 mol H 6.0 mol H Ti mol H =.10 mol Ti.10 mol Ti 47.90 g Ti Ti = 148 g Ti 4 mol HBr 6.0 mol H = 1.4 mol HBr mol H 80.91 g HBr 1.4 mol HBr = 1.00 10 g HBr HBr b. SiH 4 (g) + 4NH (g) Si N 4 (s) + 1H (g) molar mass SiH 4 =.1 g; molar mass Si N 4 = 140. g; molar mass H =.016 g SiH4 1.5 g SiH 4.1 g SiH = 0.89 mol SiH 4 SiN4 0.89 mol SiH 4 mol SiH = 0.10 mol Si N 4 140. g SiN4 0.10 mol Si N 4 Si N = 18. g Si N 4 4 4 4 1 mol H 0.89 mol SiH 4 mol SiH = 1.56 mol H.016 g H 1.56 mol H H =.14 g H 4 17

c. NO(g) + H (g) N (g) + H O(l) molar mass H =.016 g; molar mass N = 8.0 g; molar mass H O = 18.0 g H 1.5 g H.016 g H = 6.0 mol H N 6.0 mol H mol H =.10 mol N 8.0 g N.10 mol N N = 86.9 g N mol HO 6.0 mol H mol H = 6.0 mol H O 18.0 g HO 6.0 mol H O H O = 11 g H O d. Cu S(s) Cu(s) + S(g) molar mass Cu S = 159. g; molar mass Cu = 6.55 g; molar mass S =.07 g CuS 1.5 g Cu S 159. g Cu S = 0.0785 mol Cu S 0.0785 mol Cu S mol Cu Cu S = 0.157 mol Cu 0.157 mol Cu 6.55 g Cu Cu = 9.98 g Cu S 0.0785 mol Cu S = 0.0785 mol S Cu S 0.0785 mol S.07 g S S =.5 g S 6. a. BCl (s) + H (g) B(s) + 6HCl(g) molar masses: BCl, 117.16 g; B, 10.81 g; HCl, 6.46 g BCl 15.0 g BCl 117.16 g BCl = 0.18 mol BCl 0.18 mol BCl mol B 10.81 g B = 1.8 g B mol BCl B 0.18 mol BCl 6 mol HCl 6.46 g HCl = 14.0 g HCl mol BCl HCl b. Cu S(s) + O (g) Cu O(s) + SO (g) molar masses: Cu S, 159.17 g; Cu O, 14.1 g; SO, 64.07 g 17

CuS 15.0 g Cu S 159.17 g Cu S = 0.0944 mol Cu S mol CuO 14.1 g CuO 0.0944 mol Cu S = 1.5 g Cu O mol Cu S Cu O mol SO 64.07 g SO 0.0944 mol Cu S = 6.04 g SO mol Cu S SO c. Cu O(s) + Cu S(s) 6Cu(s) + SO (g) molar masses: Cu S, 159.17 g; Cu, 6.55 g; SO, 64.07 g CuS 15.0 g Cu S 159.17 g Cu S = 0.0944 mol Cu S 0.0944 mol Cu S 6 mol Cu 6.55 g Cu = 5.9 g Cu Cu S Cu SO 64.07 g SO 0.0944 mol Cu S = 6.04 g SO Cu S SO d. CaCO (s) + SiO (s) CaSiO (s) + CO (g) molar masses: SiO, 60.09 g; CaSiO, 116.17 g; CO, 44.01 g SiO 15.0 g SiO 60.09 g SiO = 0.496 mol SiO CaSiO 116.17 g CaSiO 0.496 mol SiO = 9.0 g CaSiO SiO CaSiO CO 44.01 g CO 0.496 mol SiO = 11.0 g CO SiO CO 7. The equation must first be balanced. (NH 4 ) CO (s) NH (g) + CO (g) + H O(g molar masses: (NH 4 ) CO, 96.09 g; NH, 17.0 g 1.5 g (NH 4 ) CO ( ) ( ) NH CO 4 96.09 g NH CO = 0.010 mol (NH 4) CO 4 mol NH 0.010 mol (NH 4 ) CO NH CO = 0.060 mol NH ( ) 4 17.0 g NH 0.060 mol NH NH = 0.44 g NH 8. The balanced equation for the reaction is: CaC (s) + H O(l) C H (g) + Ca(OH) (s) 174

molar masses: CaC, 64.10 g; C H, 6.04 g CaC.75 g CaC 64.10 g CaC = 0.0585 mol CaC CH 0.0585 mol CaC CaC = 0.0585 mol C H 6.04 g CH 0.0585 mol C H C H = 1.5 g C H 9. molar masses: C, 1.01 g; CO, 8.01 g; CO, 44.01 g 5.00 g C C = 0.416 mol C 1.01 g C carbon dioxide: C(s) + O (g) CO (g) CO 0.416 mol C C = 0.416 mol CO 44.01 g CO 0.416 mol CO CO = 18. g CO carbon monoxide: C(s) + O (g) CO(g) mol CO 0.416 mol C = 0.416 mol CO mol C 0.416 mol CO 8.01 g CO CO = 11.7 g CO 0. NaHCO (s) Na CO (s) + H O(g) + CO (g) molar masses: NaHCO, 84.01 g; Na CO, 106.0 g NaHCO 1.5 g NaHCO 84.01 g NaHCO = 0.01809 mol NaHCO Na CO 0.01809 mol NaHCO mol NaHCO = 0.009047 mol Na CO 106.0 g Na CO 0.009047 mol Na CO Na CO = 0.959 g Na CO 1. Fe(s) + Cl (g) FeCl (s) millimolar masses: iron, 55.85 mg; FeCl, 16. mg 1 mmol Fe 15.5 mg Fe = 0.775 mmol Fe 55.85 mg Fe 175

mmol FeCl 0.775 mmol Fe = 0.775 mmol FeCl mmol Fe 16. mg FeCl 0.775 mmol FeCl = 45.0 mg FeCl 1 mmol FeCl. C 6 H 1 O 6 (aq) C H 5 OH(aq) + CO (g) molar masses: C 6 H 1 O 6, 180. g; C H 5 OH, 46.07 g C6H1O6 5.5 g C 6 H 1 O 6 180. g C H O = 0.091 mol CH 6 1O 6 6 1 6 mol CH5OH 0.091 mol CH 6 1O 6 C H O = 0.586 mol C H 5 OH 6 1 6 46.07 g CH5OH 0.586 mol C H 5 OH C H OH 5 =.68 g ethyl alcohol. H SO (aq) H O(l) + SO (g) molar masses: H SO, 8.09 g; SO, 64.07 g HSO 4.5 g H SO 8.09 g H SO = 0.05177 mol H SO SO 0.05177 mol H SO H SO = 0.05177 mol SO 64.07 g SO 0.05177 mol SO SO =. g SO 4. NH 4 Cl(s) + NaOH(s) NH (g) + NaCl(s) + H O(g) molar masses: NH 4 Cl, 5.49 g; NH, 17.0 g NH4Cl 1.9 g NH 4 Cl 5.49 g NH Cl = 0.0599 mol NH 4Cl 4 NH 0.0599 mol NH 4 Cl NH Cl = 0.0599 mol NH 17.0 g NH 0.0599 mol NH NH = 0.44 g NH 5. P 4 (s) + 5O (g) P O 5 (s) molar masses: P 4, 1.88 g; O,.00 g P4 4.95 g P 4 1.88 g P = 0.0996 mol P 4 4 4 176

5 mol O 0.0996 mol P 4 P = 0.1998 mol O.00 g O 0.1998 mol O O = 6.9 g O 4 6. 4HgS(s) + 4CaO(s) 4Hg(l) + CaS(s) + CaSO 4 (s) molar masses: HgS,.7 g Hg, 00.6 g; 10.0 kg = 1.00 10 4 g 1.00 10 4 HgS g HgS = 4.97 mol HgS.7 g HgS 4.97 mol HgS 4 mol Hg 4 mol HgS = 4.97 mol Hg 4.97 mol Hg 00.6 g Hg Hg = 8.6 10 g Hg = 8.6 kg Hg 7. NH 4 NO (s) N (g) + O (g) + 4H O(g) molar masses: NH 4 NO, 80.05 g; N, 8.0 g; 0,.00 g; H O, 18.0 g NH4NO 1.5 g NH 4 NO 80.05 g NH NO = 0.0156 mol NH 4NO 4 mol N 0.0156 mol NH 4 NO mol NH NO = 0.0156 mol N 4 8.0 g N 0.0156 mol N N = 0.47 g N O 0.0156 mol NH 4 NO mol NH NO = 0.00780 mol O 4.00 g O 0.00780 mol O O = 0.50 g O 4 mol HO 0.0156 mol NH 4 NO mol NH NO = 0.01 mol H O 4 18.0 g HO 0.01 mol H O H O = 0.56 g H O As a check, note that 0.47 g + 0.50 g + 0.56 g = 1.49 g = 1.5 g. 8. C 1 H O 11 (s) 1C(s) + 11H O(g) molar masses: C 1 H O 11, 4. g; C, 1.01 C1HO11 1.19 g C 1 H O 11 4. g C H O =.476 10 mol C1HO 11 1 11 177

.476 10 mol C1HO 11 1 mol C C H O 1 11 = 0.0417 mol C 0.0417 mol C 1.01 g C C = 0.501 g C 9. SOCl (l) + H O(l) SO (g) + HCl(g) molar masses: SOCl, 119.0 g; H O, 18.0 g SOCl 5.0 g SOCl = 0.94 mol SOCl 119.0 g SOCl HO 0.94 mol SOCl SOCl = 0.94 mol H O 18.0 g HO 0.94 mol H O H O = 5.0 g H O 40. The balanced equation is: C 8 H 18 + 5O 16CO + 18H O molar masses: C 8 H 18, 114. g; CO, 44.01 g 1 lb of CO = 45.59 g CO CO 45.59 g CO 44.01 g CO = 10. CO From the balanced chemical equation, we can calculate the number of moles and number of grams of pure octane that would be required to produce 10. CO. mol C8H18 10. CO 16 mol CO = 1.88 mol C 8H 18 114. g C8H18 1.88 mol C 8 H 18 C H 8 18 = 147. g C 8 H 18 From the density of C 8 H 18 we can calculate the volume of 147. g C 8 H 18. 1 ml C8H18 147. g C 8 H 18 0.75 g C H = 196. ml C 8H 18 (.0 10 ml to two significant figures) 8 18 From the preceding, we know that to travel 1 mile, we need approximately 00 ml of octane 1 mi 1000 ml.7854 L = approximately 19 mi/gal 196. ml 1 L 1 gal 41. To determine the limiting reactant, first calculate the number of moles of each reactant present. Then determine how these numbers of moles correspond to the stoichiometric ratio indicated by the balanced chemical equation for the reaction. Specific answer depends on student response. 4. To determine the limiting reactant, first calculate the number of moles of each reactant present. Then determine how these numbers of moles correspond to the stoichiometric ratio indicated by the balanced chemical equation for the reaction. 178

4. The theoretical yield of a reaction represents the stoichiometric amount of product that should form if the limiting reactant for the process is completely consumed. 44. A reactant is present in excess if there is more of that reactant present than is needed to combine with the limiting reactant for the process. By definition, the limiting reactant cannot be present in excess. An excess of any reactant does not affect the theoretical yield for a process: the theoretical yield is determined by the limiting reactant. 45. a. Na B 4 O 7 (s) + H SO 4 (aq) + 5H O(l) 4H BO (s) + Na SO 4 (aq) molar masses: Na B 4 O 7, 01. g; H SO 4, 98.09 g; H O, 18.0 g 5.00 g Na B 4 O 7 01. g = 0.049 mol Na B 4 O 7 5.00 g H SO 4 5.00 g H O 98.09 g = 0.0510 mol H SO 4 18.0 g = 0.77 mol H O Na B 4 O 7 is the limiting reactant. mol H SO 4 remaining unreacted = 0.0510 0.049 = 0.06 mol H O remaining unreacted = 0.77 5(0.049) = 0.15 mol mass of H SO 4 remaining = 0.06 98.09 g =.56 g H SO 4 mass of H O remaining = 0.15 mol 18.0 g b. CaC (s) + H O(l) Ca(OH) (s) + C H (g) molar masses: CaC, 64.10 g; H O, 18.0 g 5.00 g CaC 64.10 g = 0.0780 mol CaC 5.00 g H O 18.0 g = 0.77 mol H O =.76 g H O CaC is the limiting reactant; water is present in excess. mol of H O remaining = 0.77 (0.0780) = 0.1 H O mass of H O remaining = 0.1 18.0 g =.18 g H O c. NaCl(s) + H SO 4 (l) HCl(g) + Na SO 4 (s) molar masses: NaCl, 58.44 g; H SO 4, 98.09 g 5.00 g NaCl = 0.0856 mol NaCl 58.44 g 179

5.00 g H SO 4 98.09 g = 0.0510 mol H SO 4 NaCl is the limiting reactant; H SO 4 is present in excess. mol H SO 4 that reacts = 0.5(0.0856) = 0.048 mol H SO 4 mol H SO 4 remaining = 0.0510 0.048 = 0.008 mol mass of H SO 4 remaining = 0.008 mol 98.09 g = 0.80 g H SO 4 d. SiO (s) + C(s) Si(l) + CO(g) molar masses: SiO, 60.09 g; C, 1.01 g 5.00 g SiO 60.09 g = 0.08 mol SiO 5.00 g C 1.01 g = 0.416 mol C SiO is the limiting reactant; C is present in excess. mol C remaining = 0.416 (0.08) = 0.50 mol mass of C remaining = 0.50 mol 1.01 g =.00 g C 46. a. S(s) + H SO 4 (aq) SO (g) + H O(l) Molar masses: S,.07 g; H SO 4, 98.09 g; SO, 64.07 g; H O, 18.0 g 5.00 g S = 0.1559 mol S.07 g 5.00 g H SO 4 98.09 g = 0.05097 mol H SO 4 According to the balanced chemical equation, we would need twice as much sulfuric acid as sulfur for complete reaction of both reactants. We clearly have much less sulfuric acid present than sulfur: sulfuric acid is the limiting reactant. The calculation of the masses of products produced is based on the number of moles of the sulfuric acid. mol SO 64.07 g SO 0.05097 mol H SO 4 = 4.90 g SO mol H SO SO 4 mol HO 18.0 g HO 0.05097 mol H SO 4 = 0.918 g H O mol H SO H O 4 b. MnO (s) + H SO 4 (aq) Mn(SO 4 ) + H O(l) molar masses: MnO, 86.94 g; H SO 4 98.09 g; Mn(SO 4 ), 47.1 g; H O, 18.0 g 5.00 g MnO 86.94 g = 0.0575 MnO 180

5.00 g H SO 4 98.09 g = 0.05097 mol H SO 4 According to the balanced chemical equation, we would need twice as much sulfuric acid as manganese(iv) oxide for complete reaction of both reactants. We do not have this much sulfuric acid, so sulfuric acid must be the limiting reactant. The amount of each product produced will be based on the sulfuric acid reacting completely. Mn(SO 4) 47.1 g Mn(SO 4) 0.05097 mol H SO 4 = 6.0 g Mn(SO 4 ) mol HSO4 Mn(SO ) mol HO 18.0 g HO 0.05097 mol H SO 4 = 0.918 g H O mol H SO H O 4 c. H S(g) + O (g) SO (g) + H O(l) Molar masses: H S, 4.09 g; O,.00 g; SO, 64.07 g; H O, 18.0 g 5.00 g H S 4.09 g = 0.1467 mol H S 5.00 g O.00 g = 0.156 mol O According to the balanced equation, we would need 1.5 times as much O as H S for complete reaction of both reactants. We don t have that much O, so O must be the limiting reactant that will control the masses of each product produced. mol SO 64.07 g SO 0.156 mol O = 6.67 g SO mol O SO mol HO 18.0 g HO 0.156 mol O = 1.88 g H O mol O H O d. AgNO (aq) + Al(s) Ag(s) + Al(NO ) (aq) Molar masses: AgNO, 169.9 g; Al, 6.98 g; Ag, 107.9 g; Al(NO ), 1.0 g 5.00 g AgNO 169.9 g = 0.094 mol AgNO 4 5.00 g Al 6.98 g = 0.185 mol Al According to the balanced chemical equation, we would need three moles of AgNO for every mole of Al for complete reaction of both reactants. We in fact have fewer moles of AgNO than aluminum, so AgNO must be the limiting reactant. The amount of product produced is calculated from the number of moles of the limiting reactant present: mol Ag 107.9 g Ag 0.094 mol AgNO =.18 g Ag mol AgNO Ag Al(NO ) 1.0 g Al(NO ) 0.094 mol AgNO =.09 g mol AgNO Al(NO ) 181

47. Before any calculations are attempted, the equations must be balanced. a. C H 8 (g) + 5O (g) CO (g) + 4H O(g) molar masses: C H 8, 44.09 g; O,.00 g; CO, 44.01 g; H O, 18.0 g 10.0 g C H 8 44.09 g = 0.68 mol C H 8 10.0 g O.00 g = 0.15 mol O For 0.68 mol C H 8, the amount of O that would be needed is 5 mol O 0.68 mol C H 8 C H = 1.14 mol O 8 Because we do not have this amount of O, then O is the limiting reactant. mol CO 44.01 g CO 0.15 mol O = 8.5 g CO 5 mol O CO 4 mol HO 18.0 g HO 0.15 mol O = 4.51 g H O 5 mol O H O b. Al(s) + Cl (g) AlCl (s) molar masses: Al, 6.98 g; Cl, 70.90 g; AlCl, 1. g 10.0 g Al = 0.706 mol Al 6.98 g 10.0 g Cl 70.90 g = 0.1410 mol Cl For 0.1410 mol Cl (g), the amount of Al(s) required is mol Al 0.1410 mol Cl = 0.09400 mol Al mol Cl We have far more than this amount of Al(s) present, so Cl (g) must be the limiting reactant that will control the amount of AlCl which forms. mol AlCl 1. g AlCl 0.1410 mol Cl = 1.5 g AlCl mol Cl AlCl c. NaOH(s) + CO (g) Na CO (s) + H O(l) molar masses: NaOH, 40.00 g; CO, 44.01 g; Na CO, 106.0 g; H O, 18.0 g 10.0 g NaOH = 0.500 mol NaOH 40.00 g 10.0 g CO 44.01 g = 0.7 mol CO 18

Without having to calculate, according to the balanced chemical equation, we would need twice as many moles of NaOH as CO for complete reaction. For the amounts calculated above, there is not nearly enough NaOH present for the amount of Cl used: NaOH is the limiting reactant. Na CO 106.0 g Na CO 0.500 mol NaOH = 1. g Na CO mol NaOH Na CO HO 18.0 g HO 0.500 mol NaOH =.5 g H O mol NaOH H O d. NaHCO (s) + HCl(aq) NaCl(aq) + H O(l) + CO (g) molar masses: NaHCO, 84.01 g; HCl, 6.46 g; NaCl, 58.44 g; H O, 18.0 g; CO, 44.01 g 10.0 g NaHCO 84.01 g = 0.1190 mol NaHCO 10.0 g HCl 6.46 g = 0.74 mol HCl Because the coefficients of NaHCO (s) and HCl(aq) are both one in the balanced chemical equation for the reaction, there is not enough NaHCO present to react with the amount of HCl present: the 0.1190 mol NaHCO present is the limiting reactant. Because all the coefficients of the products are also each one, then if 0.1190 mol NaHCO reacts completely (with 0.1190 mol HCl), 0.1190 mol of each product will form. 0.1190 mol NaCl 58.44 g = 6.95 g NaCl 0.1190 mol H O 18.0 g 0.1190 mol CO 44.01 g =.14 g H O = 5.4 g CO 48. a. CS (l) + O (g) CO (g) + SO (g) Molar masses: CS, 76.15 g; O,.00 g; CO, 44.01 g 1.00 g CS 76.15 g = 0.011 mol CS 1.00 g O.00 g = 0.015 mol O From the balanced chemical equation, we would need three times as much oxygen as carbon disulfide for complete reaction of both reactants. We do not have this much oxygen, and so oxygen must be the limiting reactant. CO 44.01 g CO 0.015 mol O = 0.458 g CO mol O CO 18

b. NH (g) + CO (g) CN H 4 O(s) + H O(l) Molar masses: NH, 17.0 g; CO, 44.01 g; H O, 18.0 g 1.00 g NH 17.0 g = 0.0587 mol NH 1.00 g CO 44.01 g = 0.07 mol CO The balanced chemical equation tells us that we would need twice as many moles of ammonia as carbon dioxide for complete reaction of both reactants. We have more than this amount of ammonia present, so the reaction will be limited by the amount of carbon dioxide present. HO 18.0 g HO 0.07 mol CO = 0.409 g H O CO H O c. H (g) + MnO (s) MnO(s) + H O(l) Molar masses: H,.016 g; MnO, 86.94 g; H O, 18.0 g 1.00 g H.016 g = 0.496 mol H 1.00 g MnO 86.94 g = 0.0115 mol MnO Because the coefficients of both reactants in the balanced chemical equation are the same, we would need equal amounts of both reactants for complete reaction. Therefore, manganese(iv) oxide must be the limiting reactant and controls the amount of product obtained. HO 18.0 g HO 0.0115 mol MnO = 0.07 g H O MnO H O d. I (s) + Cl (g) ICl(g) Molar masses: I, 5.8 g; Cl, 70.90 g; ICl, 16.5 g 1.00 g I 5.8 g = 0.0094 mol I 1.00 g Cl 70.90 g = 0.014 Cl From the balanced chemical equation, we would need equal amounts of I and Cl for complete reaction of both reactants. As we have much less iodine than chlorine, iodine must be the limiting reactant. mol ICl 16.5 g ICl 0.0094 mol I = 1.8 g ICl I ICl 184

49. a. UO (s) + 4HF(aq) UF 4 (aq) + H O(l) UO is the limiting reactant; 1.16 g UF 4, 0.1 g H O b. NaNO (aq) + H SO 4 (aq) Na SO 4 (aq) + HNO (aq) NaNO is the limiting reactant; 0.86 g Na SO 4 ; 0.741 g HNO c. Zn(s) + HCl(aq) ZnCl (aq) + H (g) HCl is the limiting reactant; 1.87 g ZnCl ; 0.076 g H d. B(OH) (s) + CH OH(l) B(OCH ) (s) + H O(l) CH OH is the limiting reactant; 1.08 g B(OCH ) ; 0.56 g H O 50. a. CO(g) + H (g) CH OH(l) CO is the limiting reactant; 11.4 mg CH OH b. Al(s) + I (s) AlI (s) I is the limiting reactant; 10.7 mg AlI c. Ca(OH) (aq) + HBr(aq) CaBr (aq) + H O(l) HBr is the limiting reactant; 1.4 mg CaBr ;. mg H O d. Cr(s) + H PO 4 (aq) CrPO 4 (s) + H (g) H PO 4 is the limiting reactant; 15.0 mg CrPO 4 ; 0.09 mg H 51. Pb(C H O ) (aq) + H O(l) + CO (g) PbCO (s) + HC H O (aq) molar masses: Pb(C H O ), 5. g; CO, 44.01 g; PbCO, 67.1 g 1.5 g Pb(C H O ) 5. g = 0.0084 mol Pb(C H O ) 5.95 g CO 44.01 g = 0.15 mol CO Pb(C H O ) is the limiting reactant, which determines the yield of product. PbCO 67.1 g PbCO 0.0084 mol Pb(C H O ) = 1.0 g PbCO Pb C H O PbCO ( ) 5. CuO(s) + H SO 4 (aq) CuSO 4 (aq) + H O(l) molar masses: CuO, 79.55 g; H SO 4, 98.09 g CuO.49 g CuO = 0.01 mol CuO 79.55 g CuO HSO4 5.05 g H SO 4 98.09 g H SO = 0.0515 mol H SO 4 4 Since the reaction is of 1:1 stoichiometry, CuO must be the limiting reactant since it is present in the lesser amount on a molar basis. 185

5. PbO(s) + C(s) Pb(l) + CO(g) molar masses: PbO,. g; C, 1.01 g; Pb, 07. g 50.0 10 g PbO = 4.0 mol PbO. g 50.0 10 g C 1.01 g = 416 mol C PbO is the limiting reactant. Pb 07. g Pb 4.0 mol PbO = 4.64 10 4 g = 46.4 kg Pb PbO Pb 54. 4Fe(s) + O (g) Fe O (s) Molar masses: Fe, 55.85 g; Fe O, 159.7 g 1.5 g Fe = 0.04 mol Fe present 55.85 g Calculate how many mol of O are required to react with this amount of Fe mol O 0.04 mol Fe 4 mol Fe = 0.0168 mol O Because we have more O than this, Fe must be the limiting reactant. mol FeO 159.7 g FeO 0.04 mol Fe = 1.79 g Fe O 4 mol Fe Fe O 55. Ag + (aq) + Cl (aq) AgCl(s) molar masses: AgNO, 169.91 g; NaCl, 58.44 g The number of moles of silver ion present will be the same as the number of moles of silver nitrate taken because each formula unit of silver nitrate contains one silver ion. AgNO 1.15 g AgNO 169.91 g AgNO = 0.00677 mol AgNO = 0.00677 mol Ag + ion The number of moles of chloride ion present will be the same as the number of moles of sodium chloride taken, because each formula unit of sodium chloride contains one sodium ion. NaCl 5.45 g NaCl 58.44 g NaCl = 0.09 mol NaCl = 0.09 mol Cl ion Because the balanced chemical equation indicates a 1:1 stoichiometry for the reaction, there is not nearly enough silver ion present (0.00677 mol) to precipitate the amount of chloride ion in the sample (0.09 mol). 56. CaCl (aq) + Na SO 4 (aq) CaSO 4 (s) + NaCl(aq) molar masses: CaCl, 110.98 g; Na SO 4, 14.05 g 186

CaCl 5.1 g CaCl 110.98 g CaCl = 0.0469 mol CaCl = 0.0469 mol Ca + ion Na SO4 4.95 g Na SO 4 14.05 g Na SO = 0.048 mol Na SO 4 = 0.048 mol SO 4 ion 4 Because the balanced chemical equation indicates a 1:1 stoichiometry for the reaction, there is not nearly enough sulfate ion present (0.048 mol) to precipitate the amount of calcium ion in the sample (0.0469 mol). Sodium sulfate (sulfate ion) is the limiting reactant. Calcium chloride (calcium ion) is present in excess. 57. BaO (s) + HCl(aq) H O (aq) + BaCl (aq) molar masses: BaO, 169. g; HCl, 6.46 g; H O, 4.0 g 1.50 g BaO 169. g = 8.860 10 mol BaO 5.0 ml solution 0.680 g HCl 0.07 g HCl 1 ml solution = 0.680 g HCl 6.46 g = 1.865 10 mol HCl BaO is the limiting reactant. 8.860 10 HO 4.0 g HO mol BaO = 0.01 g H O BaO H O 58. SiO (s) + C(s) CO(g) + SiC(s) molar masses: SiO, 60.09 g; SiC, 40.10 g; 1.0 kg = 1.0 10 g 1.0 10 g SiO 60.09 g = 16.64 mol SiO From the balanced chemical equation, if 16.64 mol of SiO were to react completely (an excess of carbon is present), then 16.64 mol of SiC should be produced (the coefficients of SiO and SiC are the same). 16.64 mol SiC 40.01 g = 6.7 10 g SiC = 0.67 kg SiC 59. The theoretical yield represents the yield we calculate from the stoichiometry of the reaction and the masses of reactants taken for the experiment. The actual yield is what is actually obtained in an experiment. The percent yield is the ratio of what is actually obtained to the theoretical amount that could be obtained, converted to a percent basis. 60. If the reaction is performed in a solvent, the product may have a substantial solubility in the solvent; the reaction may come to equilibrium before the full yield of product is achieved (see Chapter 16); loss of product may occur through operator error. 187

61. Percent yield = actual yield theoretical yield 100 = 1. g 1.44 g 100 = 85.4% 6. KClO (s) KCl(s) + O (g) molar mass: KClO, 1.55 g; O,.00 g KClO 4.74 g KClO 1.55 g KClO = 0.087 mol KClO mol O 0.087 mol KClO mol KClO = 0.0580 mol O.00 g O 0.0580 mol O O = 1.86 g O theoretical yield % yield = 1.51 g actual 1.86 g theoretical 100 = 81.% of theory 6. S 8 (s) + 8Na SO (aq) + 40H O(l) 8Na S O =5H O molar masses: S 8, 56.6 g; Na SO, 16.1 g; Na S O =5H O, 48. g.5 g S 8 56.6 g = 0.0167 mol S 8 1.1 g Na SO S 8 is the limiting reactant. 16.1 g = 0.109 mol Na SO 0.0167 mol S 8. 8 mol Na SO5HO S 8 = 0.1014 mol Na S O 5H O 0.1014 mol Na S O =5H O 48. g Na S O 5H O Na S O 5H O = 5. g Na S O 5H O.. Percent yield = actual yield theoretical yield 100 = 5.6 g 5. g 100 = 0.9% 64. LiOH(s) + CO (g) Li CO (s) + H O(g) molar masses: LiOH,.95 g; CO, 44.01 g LiOH CO 44.01 g CO 155 g LiOH = 14 g CO.95 g LiOH mol LiOH CO As the cartridge has only absorbed 10 g CO out of a total capacity of 14 g CO, the cartridge has absorbed 10 g 14 g 100 = 71.8% of its capacity. 188

65. Xe(g) + F (g) XeF 4 (s) molar masses: Xe, 11. g; F, 8.00 g; XeF 4, 07. g 10. g Xe = 0.990 Xe 11. g 100. g F 8.00 g =.6 mol F Xe is the limiting reactant. XeF4 07. g XeF4 0.990 Xe = 05 g XeF 4 Xe XeF 4 Percent yield = actual yield theoretical yield 100 = 145 g 05 g 100 = 70.7 % of theory 66. Ba + (aq) + SO 4 (aq) BaSO 4 (s) molar masses: SO 4, 96.07 g; BaCl, 08. g; BaSO 4,.4 g 1.1 g SO 4 5.0 g BaCl 96.07 g = 0.01166 mol SO 4 08. g = 0.041 BaCl = 0.041 Ba + SO 4 is the limiting reactant. 0.01166 mol SO BaSO4.4 g BaSO4 4 SO BaSO =.7 g BaSO 4 Percent yield = actual yield theoretical yield 4 4 100 =.0 g.7 g 100 = 74.% 67. Ca(HCO ) (aq) CaCO (s) + CO (g) + H O(l) millimolar masses: Ca(HCO ), 16.1 mg; CaCO, 100.1 mg.0 10 1 mmol mg Ca(HCO ) 16.1 mg = 1. 10 5 mmol Ca(HCO ) 1. 10 5 1 mmol CaCO mmol Ca(HCO ) 1 mmol Ca(HCO ) = 1. 10 5 mmol CaCO 1. 10 5 mmol 100.1 mg 1 mmol = 1. 10 mg = 1. 10 6 g CaCO 68. NaCl(aq) + NH (aq) + H O(l) + CO (s) NH 4 Cl(aq) + NaHCO (s) molar masses: NH, 17.0 g; CO, 44.01 g; NaHCO, 84.01 g 189

10.0 g NH 15.0 g CO 17.0 g = 0.587 mol NH 44.01 g = 0.408 mol CO CO is the limiting reactant. NaHCO 0.408 mol CO CO = 0.408 mol NaHCO 0.408 mol NaHCO 84.01 g = 8.6 g NaHCO 69. Fe(s) + S(s) FeS(s) molar masses: Fe, 55.85 g; S,.07 g; FeS, 87.9 g 5.5 g Fe = 0.0940 mol Fe 55.85 g 1.7 g S.07 g = 0.96 mol S Fe is the limiting reactant. FeS 87.9 g FeS 0.0940 mol Fe = 8.6 g FeS produced Fe FeS 70. C 6 H 1 O 6 (s) + 6O (g) 6CO (g) + 6H O(g) molar masses: glucose, 180. g; CO, 44.01 g 1.00 g glucose = 5.549 10 mol glucose 5.549 10 6 mol CO mol glucose glucose =. 10 mol CO. 10 mol CO 44.01 g = 1.47 g CO 71. Cu(s) + S(s) CuS(s) molar masses: Cu, 6.55 g; S,.07 g; CuS, 95.6 g 1.8 g Cu = 0.5004 mol Cu 6.55 g 50.0 g S.07 g = 1.559 mol S Cu is the limiting reactant. CuS 0.5004 mol Cu = 0.5004 mol CuS Cu 190

0.5004 mol CuS 95.6 g % yield = 40.0 g 47.8 g 100 = 8.7% = 47.8 g CuS 7. Ba + (aq) + SO 4 (aq) BaSO 4 (s) millimolar ionic masses: Ba +, 17. mg; SO 4, 96.07 mg; BaCl, 08. mg 150 mg SO 4 1 mmol 96.07 mg = 1.56 millimol SO 4 As barium ion and sulfate ion react on a 1:1 stoichiometric basis, then 1.56 millimol of barium ion is needed, which corresponds to 1.56 millimol of BaCl 08. mg 1.56 millimol BaCl = 5 milligrams BaCl needed 1 mmol 7. mass of Cl present = 1.054 g sample 10. g Cl 100.0 g sample molar masses: Cl, 5.45 g; AgNO, 169.9 g; AgCl, 14.4 g 0.1086 g Cl 5.45 g =.06 10 mol Cl = 0.1086 g Cl.06 10 mol Cl AgNO =.06 10 mol AgNO Cl.06 10 mol AgNO 169.9 g.06 10 mol Cl.06 10 mol AgCl 14.4 g = 0.50 g AgNO required AgCl Cl =.06 10 mol AgCl = 0.49 g AgCl produced 74. a. UO (s) + 4HF(aq) UF 4 (aq) + H O(l) One molecule (formula unit) of uranium(iv) oxide will combine with four molecules of hydrofluoric acid, producing one uranium(iv) fluoride molecule and two water molecules. One mole of uranium(iv) oxide will combine with four moles of hydrofluoric acid to produce one mole of uranium(iv) fluoride and two moles of water. b. NaC H O (aq) + H SO 4 (aq) Na SO 4 (aq) + HC H O (aq) Two molecules (formula units) of sodium acetate react exactly with one molecule of sulfuric acid, producing one molecule (formula unit) of sodium sulfate and two molecules of acetic acid. Two moles of sodium acetate will combine with one mole of sulfuric acid, producing one mole of sodium sulfate and two moles of acetic acid. 191

c. Mg(s) + HCl(aq) MgCl (aq) + H (g) One magnesium atom will react with two hydrochloric acid molecules (formula units) to produce one molecule (formula unit) of magnesium chloride and one molecule of hydrogen gas. One mole of magnesium will combine with two moles of hydrochloric acid, producing one mole of magnesium chloride and one mole of gaseous hydrogen. d. B O (s) + H O(l) B(OH) (aq) One molecule of diboron trioxide will react exactly with three molecules of water, producing two molecules of boron trihydroxide (boric acid). One mole of diboron trioxide will combine with three moles of water to produce two moles of boron trihydroxide (boric acid). 75. False. For 0.40 mol of Mg(OH) to react, 0.80 mol of HCl will be needed. According to the balanced equation, for a given amount of Mg(OH), twice as many moles of HCl is needed. 5 mol O 76. For O : C H 8 mol CO For CO : CH8 4 mol HO For H O: CH8 77. a. H O (l) H O(l) + O (g) mol HO 0.50 mol H O mol H O = 0.50 mol H O O 0.50 mol H O mol H O = 0.5 mol O b. KClO (s) KCl(s) + O (g) mol KCl 0.50 mol KClO mol KClO = 0.50 mol KCl mol O 0.50 mol KClO mol KClO = 0.75 mol O c. Al(s) + 6HCl(aq) AlCl (aq) + H (g) mol AlCl 0.50 mol Al = 0.50 mol AlCl mol Al mol H 0.50 mol Al = 0.75 mol H mol Al d. C H 8 (g) + 5O (g) CO (g) + 4H O(l) mol CO 0.50 mol C H 8 C H = 1.5 mol CO 8 4 mol HO 0.50 mol C H 8 C H =.0 mol H O 8 19

78. a. NH (g) + HCl(g) NH 4 Cl(s) molar mass of NH = 17.01 g 1.00 g NH 17.01 g = 0.0588 mol NH NH4Cl 0.0588 mol NH NH = 0.0588 mol NH 4Cl b. CaO(s) + CO (g) CaCO (s) molar mass CaO = 56.08 g 1.00 g CaO = 0.0178 mol CaO 56.08 g CaCO 0.0178 mol CaO CaO = 0.0178 mol CaCO c. 4Na(s) + O (g) Na O(s) molar mass Na =.99 g 1.00 g Na = 0.045 mol Na.99 g mol Na O 0.045 mol Na = 0.017 mol Na O 4 mol Na d. P(s) + Cl (g) PCl (l) molar mass P = 0.97 g 1.00 g P = 0.0 mol P 0.97 g mol PCl 0.0 mol P mol P = 0.0 mol PCl 79. a. molar mass CuSO 4 = 159.6 g 4.1 g CuSO 4 159.6 g = 0.064 mol CuSO 4 b. molar mass Ba(NO ) = 61. g 7.94 g Ba(NO ) 61. g = 0.004 mol Ba(NO ) c. molar mass water = 18.0 g; 1.4 mg = 0.0014 g 0.0014 g 18.0 g = 6.88 10 5 mol H O 19

d. molar mass W = 18.9 g 9.79 g W 18.9 g = 5. 10 mol W e. molar mass S =.07 g; 1.45 lb = 1.45(454) = 658 g 658 g S = 0.5 mol S.07 g f. molar mass C H 5 OH = 46.07 g 4.65 g C H 5 OH 46.07 g = 0.10 C H 5 OH g. molar mass C = 1.01 g 1.01 g C = 1.00 mol C 1.01 g 80. a. molar mass HNO = 6.0 g 5.0 mol HNO 6.0 g =. 10 g HNO b. molar mass Hg = 00.6 g 0.00005 mol Hg 00.6 g = 0.061 g Hg c. molar mass K CrO 4 = 194. g.1 10 5 mol K CrO 4 194. g = 4.49 10 g K CrO 4 d. molar mass AlCl = 1. g 10.5 mol AlCl 1. g = 1.40 10 g AlCl e. molar mass SF 6 = 146.1 g 4.9 10 4 mol SF 6 146.1 g = 7. 10 6 g SF 6 f. molar mass NH = 17.01 g 15 mol NH 17.01 g =.1 10 g NH g. molar mass Na O = 77.98 g 0.0105 mol Na O 77.98 g = 0.997 g Na O 194

81. Before any calculations are done, the equations must be balanced. a. BaCl (aq) + H SO 4 (aq) BaSO 4 (s) + HCl(aq) HSO4 0.145 mol BaCl = 0.145 mol H SO 4 BaCl b. AgNO (aq) + NaCl(aq) AgCl(s) + NaNO (aq) NaCl 0.145 mol AgNO = 0.145 mol NaCl AgNO c. Pb(NO ) (aq) + Na CO (aq) PbCO (s) + NaNO (aq) Na CO 0.145 mol Pb(NO ) = 0.145 mol Na CO Pb(NO ) d. C H 8 (g) + 5O (g) CO (g) + 4H O(g) 5 mol O 0.145 mol C H 8 C H = 0.75 mol O 8. SO (g) + O (g) SO (g) 8 molar masses: SO, 64.07 g; SO, 80.07 g; 150 kg = 1.5 10 5 g 1.5 10 5 g SO 64.07 g =.4 10 mol SO.4 10 mol SO mol SO mol SO =.4 10 mol SO.4 10 mol SO 80.07 g = 1.9 10 5 g SO = 1.9 10 kg SO 8. ZnS(s) + O (g) ZnO(s) + SO (g) molar masses: ZnS, 97.45 g; SO, 64.07 g; 1.0 10 kg = 1.0 10 5 g 1.0 10 5 g ZnS 97.45 g = 1.06 10 mol ZnS 1.06 10 mol SO mol ZnS mol ZnS = 1.06 10 mol SO 1.06 10 mol SO 64.07 g 84. Na O (s) + H O(l) 4NaOH(aq) + O (g) molar masses: Na O, 77.98 g; O,.00 g.5 g Na O 77.98 g = 0.0417 mol Na O = 6.6 10 4 g SO = 66 kg SO 195

O 0.0417 mol Na O mol Na O = 0.009 mol O 0.009 mol O.00 g = 0.667 g O 85. Cu(s) + AgNO (aq) Cu(NO ) (aq) + Ag(s) millimolar masses: Cu, 6.55 mg; AgNO, 169.9 mg 1 mmol 1.95 mg AgNO 169.9 mg = 0.01148 mmol AgNO 0.01148 mmol AgNO 1 mmol Cu mmol AgNO = 0.005740 mmol Cu 0.005740 mmol Cu 6.55 g = 0.65 mg Cu 86. Zn(s) + HCl(aq) ZnCl (aq) + H (g) molar masses: Zn, 65.8 g; H,.016 g.50 g Zn = 0.084 mol Zn 65.8 g H 0.084 mol Zn Zn = 0.084 mol H 0.084 mol H.016 g = 0.0771 g H 87. C H (g) + 5O (g) 4CO (g) + H O(g) molar masses: C H, 6.04 g; O,.00 g; 150 g = 1.5 10 g 1.5 10 g C H 6.04 g = 5.760 mol C H 5 mol O 5.760 mol C H mol C H = 14.40 mol O 14.40 mol O.00 g = 4.6 10 g O 88. a. Na(s) + Br (l) NaBr(s) molar masses: Na,.99 g; Br, 159.8 g; NaBr, 10.9 g 5.0 g Na = 0.175 mol Na.99 g 196

5.0 g Br 159.8 g = 0.019 mol Br Intuitively, we would suspect that Br is the limiting reactant, because there is much less Br than Na on a mole basis. To prove that Br is the limiting reactant, the following calculation is needed: mol Na 0.019 mol Br = 0.0658 mol Na. Br Clearly, there is more Na than this present, so Br limits the reaction extent and the amount of NaBr formed. mol NaBr 0.019 mol Br = 0.0658 mol NaBr Br 0.0658 mol NaBr 10.9 g = 6.4 g NaBr b. Zn(s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu(s) molar masses: Zn, 65.8 g; Cu, 6.55 g; ZnSO 4, 161.5 g; CuSO 4, 159.6 g 5.0 g Zn = 0.07648 mol Zn 65.8 g 5.0 g CuSO 4 159.6 g = 0.01 mol CuSO 4 As the coefficients of Zn and CuSO 4 are the same in the balanced chemical equation, an equal number of moles of Zn and CuSO 4 would be needed for complete reaction. There is less CuSO 4 present, so CuSO 4 must be the limiting reactant. ZnSO4 0.01 mol CuSO 4 CuSO = 0.01 mol ZnSO 4 0.01 mol ZnSO 4 161.5 g Cu 0.01 mol CuSO 4 CuSO 4 = 5.1 g ZnSO 4 4 = 0.01 mol Cu 0.01 mol Cu 6.55 g =.0 g Cu c. NH 4 Cl(aq) + NaOH(aq) NH (g) + H O(l) + NaCl(aq) molar masses: NH 4 Cl, 5.49 g; NaOH, 40.00 g; NH, 17.0 g; H O, 18.0 g; NaCl, 58.44 g 5.0 g NH 4 Cl 5.49 g = 0.0948 mol NH 4Cl 197

5.0 g NaOH 40.00 g = 0.150 mol NaOH As the coefficients of NH 4 Cl and NaOH are both one in the balanced chemical equation for the reaction, an equal number of moles of NH 4 Cl and NaOH would be needed for complete reaction. There is less NH 4 Cl present, so NH 4 Cl must be the limiting reactant. As the coefficients of the products in the balanced chemical equation are also all one, if 0.0948 mol of NH 4 Cl (the limiting reactant) reacts completely, then 0.0948 mol of each product will be formed. 0.0948 mol NH 17.0 g = 1.6 g NH 0.0948 mol H O 18.0 g = 1.7 g H O 0.0948 mol NaCl 58.44 g = 5.5 g NaCl d. Fe O (s) + CO(g) Fe(s) + CO (g) molar masses: Fe O, 159.7 g; CO, 8.01 g; Fe, 55.85 g; CO, 44.01 g 5.0 g Fe O 159.7 g = 0.01 Fe O 5.0 g CO 8.01 g = 0.1785 mol CO Because there is considerably less Fe O than CO on a mole basis, let s see if Fe O is the limiting reactant. mol CO 0.01 Fe O = 0.099 mol CO Fe O As there is 0.1785 mol of CO present, but we have determined that only 0.099 mol CO would be needed to react with all the Fe O present, then Fe O must be the limiting reactant. CO is present in excess. mol Fe 55.85 g Fe 0.01 Fe O =.5 g Fe Fe O Fe mol CO 44.01 g CO 0.01 Fe O = 4.1 g CO Fe O CO 89. a. C H 5 OH(l) + O (g) CO (g) + H O(l) molar masses: C H 5 OH, 46.07 g; O,.00 g; CO, 44.01 g 5.0 g C H 5 OH 46.07 g = 0.547 mol C H 5 OH 5.0 g O.00 g = 0.781 mol O 198

As there is less C H 5 OH present on a mole basis, see if this substance is the limiting reactant. mol O 0.547 mol C H 5 OH C H OH = 1.68 O. 5 From the above calculation, C H 5 OH must not be the limiting reactant (even though there is a smaller number of moles of C H 5 OH present) because more oxygen than is present would be required to react completely with the C H 5 OH present. Oxygen is the limiting reactant. mol CO 44.01 g CO 0.781 mol O =.9 g CO mol O CO b. N (g) + O (g) NO(g) molar masses: N, 8.0 g; O,.00 g; NO, 0.01 g 5.0 g N 8.0 g = 0.89 mol N 5.0 g O.00 g = 0.781 mol O As the coefficients of N and O are the same in the balanced chemical equation for the reaction, an equal number of moles of each substance would be necessary for complete reaction. There is less O present on a mole basis, so O must be the limiting reactant. mol NO 0.01 g NO 0.781 mol O = 46.9 g NO O NO c. NaClO (aq) + Cl (g) ClO (g) + NaCl(aq) molar masses: NaClO, 90.44 g; Cl, 70.90 g; NaCl, 58.44 g 5.0 g NaClO 90.44 g = 0.764 mol NaClO 5.0 g Cl 70.90 g = 0.56 mol Cl See if NaClO is the limiting reactant. Cl 0.764 mol NaClO mol NaClO = 0.18 mol Cl As 0.764 mol of NaClO would require only 0.18 mol Cl to react completely (and since we have more than this amount of Cl ), then NaClO must indeed be the limiting reactant. mol NaCl 58.44 g NaCl 0.764 mol NaClO = 16. g NaCl mol NaClO NaCl 199

d. H (g) + N (g) NH (g) molar masses: H,.016 g; N, 8.0 g; NH, 17.0 g 5.0 g H.016 g = 1.40 mol H 5.0 g N 8.0 g = 0.89 mol N See if N is the limiting reactant. mol H 0.89 mol N N =.677 mol H N is clearly the limiting reactant, because there is 1.40 mol H present (a large excess). mol NH 17.0 g NH 0.89 mol N = 0.4 g NH N NH 90. N H 4 (l) + O (g) N (g) + H O(g) molar masses: N H 4,.05 g; O,.00 g; N, 8.0 g; H O, 18.0 g 0.0 g N H 4.05 g = 0.64 mol N H 4 0.0 g O.00 g = 0.65 mol O The two reactants are present in nearly the required ratio for complete reaction (due to the 1:1 stoichiometry of the reaction and the very similar molar masses of the substances). We will consider N H 4 as the limiting reactant in the following calculations. N 8.0 g N 0.64 mol N H 4 = 17.5 g N N H N 4 mol HO 18.0 g HO 0.64 mol N H 4 =.5 g H O N H H O 4-5 1.5 10 g 91. Total quantity of H S = 50. L = 7.5 10 4 g H S 1 L 8H S(aq) + 8Cl (aq) 16HCl(aq) + S 8 (s) molar masses: H S, 4.09 g; Cl, 70.90 g; S 8, 56.6 g 7.5 10 4 g H S 4.09 g =.0 10 5 mol H S 1.0 g Cl 70.90 g = 1.41 10 mol Cl There is a large excess of chlorine present compared to the amount of Cl that would be needed to react with all the H S present in the water sample: H S is the limiting reactant for the process. 00

.0 10 5 S8 56.6 g S8 mol H S = 7.1 10 4 g S 8 removed 8 mol H S S 9. 1.5 g theory 40 g actual 100 g theory = 5.0 g 8 01