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

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1 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

2 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

3 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) CH mol CO CO = mol CH 4 mol HO mol CO CO = 1.00 mol H O b. BaCl (aq) + AgNO (aq) AgCl(s) + Ba(NO ) (aq) mol AgCl mol BaCl = 1.00 mol AgCl BaCl Ba(NO ) mol BaCl BaCl c. C H 8 (g) + 5O (g) 4H O(l) + CO (g) 4 mol HO mol C H 8 C H =.00 mol H O 8 mol CO mol C H 8 C H = 1.50 mol CO 8 = mol Ba(NO ) d. H SO 4 (aq) + Fe(s) Fe (SO 4 ) (aq) + H (g) Fe (SO 4) mol H SO 4 = mol Fe (SO 4 ) mol H SO 4 mol H mol H SO 4 mol H SO = 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 = 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 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 = mol HNO 1. a. AgNO (aq) + LiOH(aq) AgOH(s) + LiNO (aq) molar masses: AgOH, g; LiNO, g AgOH 0.15 mol AgNO AgNO = 0.15 mol AgOH 0.15 mol AgOH g AgOH AgOH = 15.6 g AgOH LiNO 0.15 mol AgNO AgNO = 0.15 mol LiNO 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, 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 CaSO mol Al (SO 4 ) Al (SO ) 4 = 0.75 mol CaSO g CaSO mol CaSO 4 = 51.1 g CaSO 4 CaSO c. CaCO (s) + HCl(aq) CaCl (aq) + CO (g) + H O(l) molar masses: CaCl, g; CO, g; H O, 18.0 g CaCl 0.15 mol CaCO CaCO = 0.15 mol CaCl g CaCl 0.15 mol CaCl CaCl = 1.9 g CaCl CO 0.15 mol CaCO CaCO = 0.15 mol CO 166

5 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, g; H O, 18.0 g 8 mol CO 0.15 mol C 4 H 10 mol C H = mol CO g CO mol CO CO =.0 g CO 10 mol HO 0.15 mol C 4 H 10 mol C H = 0.65 mol H O 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, g; H O, 18.0 g 5 mol CO mol C 5 H 1 C H =.75 mol CO g CO.75 mol CO CO = 165 g CO 6 mol HO mol C 5 H 1 C H = 4.50 mol H O 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, g 4 mol HO 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 mol CH OH mol CH OH = mol CO 167

6 44.01 g CO 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 BaHPO mol Ba(OH) Ba(OH) = mol BaHPO 4. g BaHPO mol BaHPO 4 BaHPO = 175 g BaHPO 4 4 mol HO 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, g; CO, g mol CH5OH mol C 6 H 1 O 6 C H O = 1.50 mol C H 5 OH g CH5OH 1.50 mol C H 5 OH C H OH = 69.1 g C H 5 OH 5 mol CO mol C 6 H 1 O 6 C H O = 1.50 mol CO 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 = mol KI Cl b. 6Co(s) + P 4 (s) Co P (s) P mol Co 6 mol Co = mol P 4 c. Zn(s) + HNO (aq) Zn(NO ) (aq) + H (g) mol HNO 0.75 mol Zn = 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

7 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 = 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 = mol Si 8.09 g Si b. molar mass AuCl = 0.4 g 1 g AuCl.7 mg AuCl = 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 g FeCl 16. g FeCl = mol FeCl e. molar mass MgO = 40.1 g MgO g MgO = 19 mol MgO 40.1 g MgO 169

8 0. a. molar mass Ar = 9.95 g 1 g Ar 7.4 mg Ar = mol Ar 1000 mg 9.95 g Ar b. molar mass CS = g CS 5.7 g CS g CS = 0.69 mol CS c. molar mass Fe = g 1000 g Fe 784 kg Fe = mol Fe 1 kg g Fe d. molar mass CaCl = g CaCl g CaCl g CaCl = mol CaCl e. molar mass NiS = g NiS g NiS = 1.9 mol NiS 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 = mol 78.1 g PbCl mol PbCl = 1.18 g PbCl PbCl c. molar mass NH = 17.0 g 17.0 g NH NH NH = 1.65 g NH d. molar mass C 6 H 14 = g 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 = g g CH8. mol C H 8 C H = 98. g C H 8 8 = g C 6 H

9 b. molar mass of Ar = 9.95 g; 9.0 millimol = mol 9.95 g Ar mol Ar = 0.61 g Ar Ar c. molar mass of SiO = g g SiO mol SiO SiO = g SiO d. molar mass of CuCl = g g CuCl mol CuCl CuCl = g CuCl e. molar mass of CuCl = g g CuCl mol CuCl = 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 HSO 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 g Al = 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 HClO g Hg = mol HClO g Hg 171

10 c. K(s) + P(s) K P(s) molar mass K = 9.10 g K P g K = mol P 9.10 g K mol K d. CH 4 (g) + 4Cl (g) CCl 4 (l) + 4HCl(g) molar mass CH 4 = g CH4 4 mol Cl g CH 4 = 0.19 mol Cl g CH CH 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 = g; molar mass of HBr = 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 g Ti Ti = 148 g Ti 4 mol HBr 6.0 mol H = 1.4 mol HBr mol H g HBr 1.4 mol HBr = 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 SiN mol SiH 4 mol SiH = 0.10 mol Si N g SiN mol Si N 4 Si N = 18. g Si N 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

11 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 = mol Cu S mol Cu S mol Cu Cu S = mol Cu mol Cu 6.55 g Cu Cu = 9.98 g Cu S mol Cu S = mol S Cu S mol S.07 g S S =.5 g S 6. a. BCl (s) + H (g) B(s) + 6HCl(g) molar masses: BCl, g; B, g; HCl, 6.46 g BCl 15.0 g BCl g BCl = 0.18 mol BCl 0.18 mol BCl mol B 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, g; Cu O, 14.1 g; SO, g 17

12 CuS 15.0 g Cu S g Cu S = mol Cu S mol CuO 14.1 g CuO mol Cu S = 1.5 g Cu O mol Cu S Cu O mol SO g SO 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, g; Cu, 6.55 g; SO, g CuS 15.0 g Cu S g Cu S = mol Cu S mol Cu S 6 mol Cu 6.55 g Cu = 5.9 g Cu Cu S Cu SO g SO mol Cu S = 6.04 g SO Cu S SO d. CaCO (s) + SiO (s) CaSiO (s) + CO (g) molar masses: SiO, g; CaSiO, g; CO, g SiO 15.0 g SiO g SiO = mol SiO CaSiO g CaSiO mol SiO = 9.0 g CaSiO SiO CaSiO CO g CO 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, g; NH, 17.0 g 1.5 g (NH 4 ) CO ( ) ( ) NH CO g NH CO = mol (NH 4) CO 4 mol NH mol (NH 4 ) CO NH CO = mol NH ( ) g NH 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

13 molar masses: CaC, g; C H, 6.04 g CaC.75 g CaC g CaC = mol CaC CH mol CaC CaC = mol C H 6.04 g CH mol C H C H = 1.5 g C H 9. molar masses: C, 1.01 g; CO, 8.01 g; CO, g 5.00 g C C = mol C 1.01 g C carbon dioxide: C(s) + O (g) CO (g) CO mol C C = mol CO g CO mol CO CO = 18. g CO carbon monoxide: C(s) + O (g) CO(g) mol CO mol C = mol CO mol C 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, g; Na CO, g NaHCO 1.5 g NaHCO g NaHCO = mol NaHCO Na CO mol NaHCO mol NaHCO = mol Na CO g Na CO mol Na CO Na CO = g Na CO 1. Fe(s) + Cl (g) FeCl (s) millimolar masses: iron, mg; FeCl, 16. mg 1 mmol Fe 15.5 mg Fe = mmol Fe mg Fe 175

14 mmol FeCl mmol Fe = mmol FeCl mmol Fe 16. mg FeCl 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, g C6H1O6 5.5 g C 6 H 1 O g C H O = mol CH 6 1O mol CH5OH mol CH 6 1O 6 C H O = mol C H 5 OH g CH5OH 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, g HSO 4.5 g H SO 8.09 g H SO = mol H SO SO mol H SO H SO = mol SO g SO 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 = mol NH 4Cl 4 NH mol NH 4 Cl NH Cl = mol NH 17.0 g NH 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 P g P g P = mol P

15 5 mol O mol P 4 P = mol O.00 g O mol O O = 6.9 g O HgS(s) + 4CaO(s) 4Hg(l) + CaS(s) + CaSO 4 (s) molar masses: HgS,.7 g Hg, 00.6 g; 10.0 kg = g 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 = g Hg = 8.6 kg Hg 7. NH 4 NO (s) N (g) + O (g) + 4H O(g) molar masses: NH 4 NO, g; N, 8.0 g; 0,.00 g; H O, 18.0 g NH4NO 1.5 g NH 4 NO g NH NO = mol NH 4NO 4 mol N mol NH 4 NO mol NH NO = mol N g N mol N N = 0.47 g N O mol NH 4 NO mol NH NO = mol O 4.00 g O mol O O = 0.50 g O 4 mol HO mol NH 4 NO mol NH NO = 0.01 mol H O g HO 0.01 mol H O H O = 0.56 g H O As a check, note that 0.47 g g 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 C1HO g C 1 H O g C H O = mol C1HO

16 mol C1HO 11 1 mol C C H O 1 11 = mol C mol C 1.01 g C C = g C 9. SOCl (l) + H O(l) SO (g) + HCl(g) molar masses: SOCl, g; H O, 18.0 g SOCl 5.0 g SOCl = 0.94 mol SOCl 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 O 16CO + 18H O molar masses: C 8 H 18, 114. g; CO, g 1 lb of CO = g CO CO g CO 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 C8H CO 16 mol CO = 1.88 mol C 8H g C8H 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 ml C8H g C 8 H 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

17 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, g; H O, 18.0 g 5.00 g Na B 4 O g = mol Na B 4 O g H SO g H O g = mol H SO g = 0.77 mol H O Na B 4 O 7 is the limiting reactant. mol H SO 4 remaining unreacted = = 0.06 mol H O remaining unreacted = (0.049) = 0.15 mol mass of H SO 4 remaining = 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, g; H O, 18.0 g 5.00 g CaC g = 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 = g =.18 g H O c. NaCl(s) + H SO 4 (l) HCl(g) + Na SO 4 (s) molar masses: NaCl, g; H SO 4, g 5.00 g NaCl = mol NaCl g 179

18 5.00 g H SO g = 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) = mol H SO 4 mol H SO 4 remaining = = mol mass of H SO 4 remaining = mol g = 0.80 g H SO 4 d. SiO (s) + C(s) Si(l) + CO(g) molar masses: SiO, g; C, 1.01 g 5.00 g SiO g = 0.08 mol SiO 5.00 g C 1.01 g = mol C SiO is the limiting reactant; C is present in excess. mol C remaining = (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, g; SO, g; H O, 18.0 g 5.00 g S = mol S.07 g 5.00 g H SO g = 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 g SO mol H SO 4 = 4.90 g SO mol H SO SO 4 mol HO 18.0 g HO mol H SO 4 = 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, g; H SO g; Mn(SO 4 ), 47.1 g; H O, 18.0 g 5.00 g MnO g = MnO 180

19 5.00 g H SO g = 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) mol H SO 4 = 6.0 g Mn(SO 4 ) mol HSO4 Mn(SO ) mol HO 18.0 g HO mol H SO 4 = 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, g; H O, 18.0 g 5.00 g H S 4.09 g = mol H S 5.00 g O.00 g = 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 g SO mol O = 6.67 g SO mol O SO mol HO 18.0 g HO mol O = 1.88 g H O mol O H O d. AgNO (aq) + Al(s) Ag(s) + Al(NO ) (aq) Molar masses: AgNO, g; Al, 6.98 g; Ag, g; Al(NO ), 1.0 g 5.00 g AgNO g = mol AgNO g Al 6.98 g = 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 g Ag mol AgNO =.18 g Ag mol AgNO Ag Al(NO ) 1.0 g Al(NO ) mol AgNO =.09 g mol AgNO Al(NO ) 181

20 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, g; O,.00 g; CO, g; H O, 18.0 g 10.0 g C H g = 0.68 mol C H 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 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, g; AlCl, 1. g 10.0 g Al = mol Al 6.98 g 10.0 g Cl g = mol Cl For mol Cl (g), the amount of Al(s) required is mol Al mol Cl = 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 mol Cl = 1.5 g AlCl mol Cl AlCl c. NaOH(s) + CO (g) Na CO (s) + H O(l) molar masses: NaOH, g; CO, g; Na CO, g; H O, 18.0 g 10.0 g NaOH = mol NaOH g 10.0 g CO g = 0.7 mol CO 18

21 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 g Na CO mol NaOH = 1. g Na CO mol NaOH Na CO HO 18.0 g HO 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, g; HCl, 6.46 g; NaCl, g; H O, 18.0 g; CO, g 10.0 g NaHCO g = 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 mol NaHCO present is the limiting reactant. Because all the coefficients of the products are also each one, then if mol NaHCO reacts completely (with mol HCl), mol of each product will form mol NaCl g = 6.95 g NaCl mol H O 18.0 g mol CO g =.14 g H O = 5.4 g CO 48. a. CS (l) + O (g) CO (g) + SO (g) Molar masses: CS, g; O,.00 g; CO, g 1.00 g CS g = mol CS 1.00 g O.00 g = 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 g CO mol O = g CO mol O CO 18

22 b. NH (g) + CO (g) CN H 4 O(s) + H O(l) Molar masses: NH, 17.0 g; CO, g; H O, 18.0 g 1.00 g NH 17.0 g = mol NH 1.00 g CO 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 = g H O CO H O c. H (g) + MnO (s) MnO(s) + H O(l) Molar masses: H,.016 g; MnO, g; H O, 18.0 g 1.00 g H.016 g = mol H 1.00 g MnO g = 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 mol MnO = 0.07 g H O MnO H O d. I (s) + Cl (g) ICl(g) Molar masses: I, 5.8 g; Cl, g; ICl, 16.5 g 1.00 g I 5.8 g = mol I 1.00 g Cl g = 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 mol I = 1.8 g ICl I ICl 184

23 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 ; g HNO c. Zn(s) + HCl(aq) ZnCl (aq) + H (g) HCl is the limiting reactant; 1.87 g ZnCl ; 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, g; PbCO, 67.1 g 1.5 g Pb(C H O ) 5. g = mol Pb(C H O ) 5.95 g CO g = 0.15 mol CO Pb(C H O ) is the limiting reactant, which determines the yield of product. PbCO 67.1 g PbCO 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, g; H SO 4, g CuO.49 g CuO = 0.01 mol CuO g CuO HSO g H SO g H SO = 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

24 5. PbO(s) + C(s) Pb(l) + CO(g) molar masses: PbO,. g; C, 1.01 g; Pb, 07. g g PbO = 4.0 mol PbO. g g C 1.01 g = 416 mol C PbO is the limiting reactant. Pb 07. g Pb 4.0 mol PbO = g = 46.4 kg Pb PbO Pb 54. 4Fe(s) + O (g) Fe O (s) Molar masses: Fe, g; Fe O, g 1.5 g Fe = 0.04 mol Fe present 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 = mol O Because we have more O than this, Fe must be the limiting reactant. mol FeO 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, g; NaCl, 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 g AgNO = mol AgNO = 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 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 ( 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, g; Na SO 4, g 186

25 CaCl 5.1 g CaCl g CaCl = mol CaCl = mol Ca + ion Na SO g Na SO g Na SO = mol Na SO 4 = 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 ( 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 = mol BaO 5.0 ml solution g HCl 0.07 g HCl 1 ml solution = g HCl 6.46 g = mol HCl BaO is the limiting reactant 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, g; SiC, g; 1.0 kg = g g SiO g = mol SiO From the balanced chemical equation, if mol of SiO were to react completely (an excess of carbon is present), then mol of SiC should be produced (the coefficients of SiO and SiC are the same) mol SiC g = 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

26 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 = mol KClO mol O mol KClO mol KClO = mol O.00 g O 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 g = mol S g Na SO S 8 is the limiting reactant g = mol Na SO mol S 8. 8 mol Na SO5HO S 8 = mol Na S O 5H O 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, g LiOH CO 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

27 65. Xe(g) + F (g) XeF 4 (s) molar masses: Xe, 11. g; F, 8.00 g; XeF 4, 07. g 10. g Xe = Xe 11. g 100. g F 8.00 g =.6 mol F Xe is the limiting reactant. XeF4 07. g XeF 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, g; BaCl, 08. g; BaSO 4,.4 g 1.1 g SO g BaCl g = mol SO g = BaCl = Ba + SO 4 is the limiting reactant mol SO BaSO4.4 g BaSO4 4 SO BaSO =.7 g BaSO 4 Percent yield = actual yield theoretical yield =.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, mg mmol mg Ca(HCO ) 16.1 mg = mmol Ca(HCO ) mmol CaCO mmol Ca(HCO ) 1 mmol Ca(HCO ) = mmol CaCO mmol mg 1 mmol = mg = 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, g; NaHCO, g 189

28 10.0 g NH 15.0 g CO 17.0 g = mol NH g = mol CO CO is the limiting reactant. NaHCO mol CO CO = mol NaHCO mol NaHCO g = 8.6 g NaHCO 69. Fe(s) + S(s) FeS(s) molar masses: Fe, g; S,.07 g; FeS, 87.9 g 5.5 g Fe = mol Fe g 1.7 g S.07 g = 0.96 mol S Fe is the limiting reactant. FeS 87.9 g FeS 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, g 1.00 g glucose = mol glucose mol CO mol glucose glucose =. 10 mol CO. 10 mol CO 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 = mol Cu 6.55 g 50.0 g S.07 g = mol S Cu is the limiting reactant. CuS mol Cu = mol CuS Cu 190

29 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, mg; BaCl, 08. mg 150 mg SO 4 1 mmol 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 = g sample 10. g Cl g sample molar masses: Cl, 5.45 g; AgNO, g; AgCl, 14.4 g g Cl 5.45 g = mol Cl = g Cl mol Cl AgNO = mol AgNO Cl mol AgNO g mol Cl mol AgCl 14.4 g = 0.50 g AgNO required AgCl Cl = 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

30 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

31 78. a. NH (g) + HCl(g) NH 4 Cl(s) molar mass of NH = g 1.00 g NH g = mol NH NH4Cl mol NH NH = mol NH 4Cl b. CaO(s) + CO (g) CaCO (s) molar mass CaO = g 1.00 g CaO = mol CaO g CaCO mol CaO CaO = mol CaCO c. 4Na(s) + O (g) Na O(s) molar mass Na =.99 g 1.00 g Na = mol Na.99 g mol Na O mol Na = 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 = g 4.1 g CuSO g = mol CuSO 4 b. molar mass Ba(NO ) = 61. g 7.94 g Ba(NO ) 61. g = mol Ba(NO ) c. molar mass water = 18.0 g; 1.4 mg = g g 18.0 g = mol H O 19

32 d. molar mass W = 18.9 g 9.79 g W 18.9 g = 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 = g 4.65 g C H 5 OH 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 mol Hg 00.6 g = g Hg c. molar mass K CrO 4 = 194. g mol K CrO g = g K CrO 4 d. molar mass AlCl = 1. g 10.5 mol AlCl 1. g = g AlCl e. molar mass SF 6 = g mol SF g = g SF 6 f. molar mass NH = g 15 mol NH g =.1 10 g NH g. molar mass Na O = g mol Na O g = g Na O 194

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

34 O mol Na O mol Na O = mol O mol O.00 g = g O 85. Cu(s) + AgNO (aq) Cu(NO ) (aq) + Ag(s) millimolar masses: Cu, 6.55 mg; AgNO, mg 1 mmol 1.95 mg AgNO mg = mmol AgNO mmol AgNO 1 mmol Cu mmol AgNO = mmol Cu 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 = mol Zn 65.8 g H mol Zn Zn = mol H mol H.016 g = 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 = g g C H 6.04 g = mol C H 5 mol O mol C H mol C H = mol O mol O.00 g = g O 88. a. Na(s) + Br (l) NaBr(s) molar masses: Na,.99 g; Br, g; NaBr, 10.9 g 5.0 g Na = mol Na.99 g 196

35 5.0 g Br g = 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 mol Br = 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 mol Br = mol NaBr Br 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, g; CuSO 4, g 5.0 g Zn = mol Zn 65.8 g 5.0 g CuSO 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. ZnSO mol CuSO 4 CuSO = 0.01 mol ZnSO mol ZnSO 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, g; NH, 17.0 g; H O, 18.0 g; NaCl, g 5.0 g NH 4 Cl 5.49 g = mol NH 4Cl 197

36 5.0 g NaOH g = 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 mol of NH 4 Cl (the limiting reactant) reacts completely, then mol of each product will be formed mol NH 17.0 g = 1.6 g NH mol H O 18.0 g = 1.7 g H O mol NaCl g = 5.5 g NaCl d. Fe O (s) + CO(g) Fe(s) + CO (g) molar masses: Fe O, g; CO, 8.01 g; Fe, g; CO, g 5.0 g Fe O g = 0.01 Fe O 5.0 g CO 8.01 g = 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 = mol CO Fe O As there is mol of CO present, but we have determined that only 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 g Fe 0.01 Fe O =.5 g Fe Fe O Fe mol CO 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, g; O,.00 g; CO, g 5.0 g C H 5 OH g = mol C H 5 OH 5.0 g O.00 g = mol O 198

37 As there is less C H 5 OH present on a mole basis, see if this substance is the limiting reactant. mol O 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 g CO 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 = 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 mol O = 46.9 g NO O NO c. NaClO (aq) + Cl (g) ClO (g) + NaCl(aq) molar masses: NaClO, g; Cl, g; NaCl, g 5.0 g NaClO g = mol NaClO 5.0 g Cl g = 0.56 mol Cl See if NaClO is the limiting reactant. Cl mol NaClO mol NaClO = 0.18 mol Cl As 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 g NaCl mol NaClO = 16. g NaCl mol NaClO NaCl 199

38 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 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 g 91. Total quantity of H S = 50. L = g H S 1 L 8H S(aq) + 8Cl (aq) 16HCl(aq) + S 8 (s) molar masses: H S, 4.09 g; Cl, g; S 8, 56.6 g g H S 4.09 g = mol H S 1.0 g Cl g = 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

39 S g S8 mol H S = g S 8 removed 8 mol H S S g theory 40 g actual 100 g theory = 5.0 g 8 01

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