Chapter 4 The Major Classes of Chemical Reactions

CHAPTER 4 THE MAJOR CLASSES OF CHEMICAL REACTIONS 4.1 The distribution of its bonding electrons and the shape of the molecule are both unsymmetrical. 4.2 Ionic and polar covalent compounds are most likely to be soluble in water. Because water is polar, the partial charges in water molecules are able to interact with the charges, either ionic or dipole-induced, in other substances. 4.3 Ions must be present in an aqueous solution for it to conduct an electric current. Ions come from ionic compounds or from other electrolytes such as acids and bases. 4.4 The ions on the surface of the solid attract the water molecules (cations attract the “negative” end and anions attract the “positive” end of the water molecules). The interaction of the solvent with the ions overcomes the attraction of the oppositely charged ions for one another, and they are released into the solution. 4.5 a) 2 — This is the only solution containing a divalent cation (Ca2+). b) 3 — This is the only solution containing a divalent anion (SO42–). c) 1 — This is the only solution where both the cation (NH4+) and the anion (Br–) are univalent. 4.6 The box in (2) best represents a volume of magnesium nitrate solution. Upon dissolving the salt in water, magnesium nitrate, Mg(NO3)2, would dissociate to form one Mg2+ ion for every two NO3– ions, thus forming twice as many nitrate ions. Only box (2) has twice as many nitrate ions (red circles) as magnesium ions (blue circles). 4.7 In some cases, the force of the attraction between the ions is so strong that it cannot be overcome by the interaction of the ions with the water molecules. These materials will be insoluble in water. 4.8 The interaction with water depends on the structure of the molecule. If the interaction is good, the substance will be soluble; otherwise, the substance will not be very soluble. For example, if a covalent molecule contains polar groups, they will interact well with the polar solvent water. A few covalent molecules, such as HCl, produce ions when dissolving in water. 4.9 In general, covalent compounds which produce ions in aqueous solution interact with the water molecules to form either H+(aq) or OH–(aq), producing an acidic or basic (alkaline) solution, respectively. These compounds contain combined hydrogen atoms. Examples include hydrochloric acid (HCl), acetic acid (HC2H3O2), and ammonia (NH3). 4.10 a) Benzene is likely to be insoluble in water because it is non-polar and water is polar. b) Sodium hydroxide, an ionic compound, is likely to be soluble in water since the ions from sodium hydroxide will be held in solution through ion-dipole attractions with water. c) Ethanol (CH3CH2OH) will likely be soluble in water because the alcohol group (-OH) will hydrogen bond with the water. d) Potassium acetate, an ionic compound, will likely dissolve in water to form sodium ions and acetate ions that are held in solution through ion-dipole attractions to water. 4.11 a) Lithium nitrate, an ionic compound, would be expected to be soluble in water, and the solubility rules confirm this. b) Glycine (H2NCH2COOH) is a covalent compound, but it contains polar N-H and O-H bonds. This would make the molecule interact well with polar water molecules, and make it likely that it would be soluble. 4-1 c) Pentane (C5H12) has no bonds of significant polarity, so it would not be expected to be soluble in the polar solvent water. d) Ethylene glycol (HOCH2CH2OH) molecules contain polar O-H bonds, similar to water, so it would be expected to be soluble. 4.12 a) An aqueous solution that contains ions conducts electricity. CsI is a soluble ionic compound, and a solution of this salt in water contains Cs+ and I– ions. Its solution conducts electricity. b) HBr is a strong acid that dissociates completely in water. Its aqueous solution contains H+ and Br– ions, so it conducts electricity. 4.13 a) Yes; KOH is a strong base. b) No; glucose is neither a salt, acid, nor base, so it would be a nonelectrolyte (even though it’s soluble in water). 4.14 Plan: To determine the total moles of ions released, write a dissolution equation showing the correct molar ratios, and convert the given amounts to moles if necessary. Solution: a) Each mole of NH4Cl dissolves in water to form 1 mole of NH4+ ions and 1 mole of Cl– ions, or a total of 2 moles of ions.  2 mol ions   = 0.74 mol of ions  1 mol NH 4 Cl  ( 0.37 mol NH 4 Cl )  b) Each mole of Ba(OH)2•8H2O forms 1 mole of barium ions (Ba2+) and 2 mol of hydroxide ions (OH–), or a total of 3 mol of ions.  1 mol Ba(OH) 2 • 8H 2 O    3 mol ions ( 35.4 g Ba(OH)2 • 8H 2 O )    315.4 g Ba(OH) • 8H O 1 mol Ba(OH) • 8H O 2 2  2 2   = 0.336715 = 0.337 mol of ions c) Each mole of LiCl produces 2 moles of ions (1 mole of lithium ions and 1 mole of chloride ions).    2 mol ions  1 mol LiCl FU LiCl   6.022 x 1023 FU LiCl   1 mol LiCl     –5 –5 = 1.17901 x 10 = 1.18 x 10 mol of ions (3.55 x 10 18 4.15 ) a) Each mole of Rb2SO4 dissolves in water to form 2 moles of Rb+ ions and 1 mole of SO42– ions, or a total of 3 moles of ions.  3 mol ions   = 2.415 = 2.42 mol of ions  1 mol Rb 2SO 4  b) Each mole of Ca(NO3)2 forms 1 mole of calcium ions (Ca2+) and 2 moles of nitrate ions (NO3–), or a total of 3 moles of ions.  1 mol Ca(NO3 ) 2  3 mol ions  3.85 x 10−3 g Ca(NO3 ) 2     164.10 g Ca(NO3 ) 2  1 mol Ca(NO3 ) 2  ( 0.805 mol Rb 2SO4 )  ( ) = 7.03839 x 10–5 = 7.04 x 10–5 mol of ions c) Each moles of Sr(HCO3)2 produces 3 moles of ions (1 mole of strontium ions and two moles of bicarbonate ions).    1 mol Sr(HCO3 ) 2 3 mol ions 4.03 x 1019 F.U.Sr(HCO3 ) 2   23  6.022 x 10 FU Sr(HCO )   1 mol Sr(HCO )  3 2  3 2   –4 –4 = 2.0076 x 10 = 2.01 x 10 mol of ions ( 4.16 ) Plan: To determine the total moles of ions released, write a dissolution equation showing the correct molar ratios, and convert the given amounts to moles if necessary. 4-2 Solution: a) Recall that phosphate, PO43–, is a polyatomic anion and does not dissociate further in water. K3PO4(s) → 3 K+(aq) + PO43–(aq) 4 moles of ions are released when one mole of K3PO4 dissolves, so the total number of moles released is (0.83 mol K3PO4) (4 mol ions/mol K3PO4) = 3.32 = 3.3 mol of ions b) NiBr2•3H2O(s) → Ni2+(aq) + 2 Br –(aq) Three moles of ions are released when 1 mole of NiBr2•3H2O dissolves. The waters of hydration become part of the larger bulk of water. Convert the grams of NiBr2•3H2O to moles using the molar mass (be sure to include the mass of the water):  1 mol NiBr2 • 3H 2 O    3 mol ions 8.11 x 10−3 g NiBr2 • 3H 2 O     272.54 g NiBr2 • 3H 2 O   1 mol NiBr2 • 3H 2 O  = 8.9271 x 10–5 = 8.93 x 10–5 mol of ions c) FeCl3(s) → Fe3+(aq) + 3 Cl–(aq) Recall that a mole contains 6.022 x 1023 entities, so a mole of FeCl3 contains 6.022 x 1023 units of FeCl3, (more easily expressed as formula units). Since the problem specifies only 1.23 x 1021 formula units, we know that the amount is some fraction of a mole.    4 mol ions  1 mol FeCl3 1.23 x 1021 FU FeCl3   6.022 x 1023 FU FeCl   1 mol FeCl  3  3   = 8.17004 x 10–3 = 8.17 x 10–3 mol of ions ( ) ( 4.17 ) 3 ions produced a) Na2HPO4(s) → 2 Na+(aq) + HPO42– (0.734 mol Na2HPO4) (3 mol ions/mol Na2HPO4) = 2.202 = 2.20 mol of ions b) CuSO4•5H2O(s) → Cu+2(aq) + SO42–(aq) 2 ions produced  1 mol CuSO 4 • 5H 2 O   2 mol ions ( 3.86 g CuSO4 • 5H 2 O )    217.63 g CuSO • 5H O 1 mol CuSO • 5H O 4 2  4 2   = 3.5473 x 10–2 = 3.55 x 10–2 mol of ions c) NiCl2(s) → Ni (aq) + 2 Cl–(aq) 3 ions produced    3 mol ions  1 mol NiCl2 8.66 x 1020 FU NiCl2   6.022 x 1023 FU NiCl   1 mol NiCl  2  2   = 4.31418 x 10–3 = 4.31 x 10–3 mol of ions 2+ ( 4.18 ) a) AlCl3(s) → Al3+(aq) + 3 Cl–(aq)  10−3 L   2.45 mol AlCl3   1 mol Al3 + Moles Al3+ = (100. mL )     1 mL   L   1 mol AlCl3    10−3 L   2.45 mol AlCl3   3 mol Cl− Moles Cl– = (100. mL )     1 mL   L   1 mol AlCl3    6.022 x 1023 Al3 + Al3+ ions = 0.245 mol Al3 +   1 mol Al3 +  ( )  6.022 x 1023 Cl− Cl– ions = 0.735 mol Cl−   1 mol Cl−  b) Li2SO4(s) → 2 Li+(aq) + SO42–(aq) ( )  –  = 0.735 mol Cl   23 23 3+  = 1.47539 x 10 = 1.48 x 10 Al ions   23 23 –  = 4.42617 x 10 = 4.43 x 10 Cl ions   2.59 g Li 2SO 4   1 mol Li 2SO 4 Moles Li+ = (1.80 L )   1L    109.95 g Li 2SO 4 4-3  3+  = 0.245 mol Al    2 mol Li +     1 mol Li 2SO 4  +  = 0.08480 = 0.0848 mol Li   2.59 g Li 2SO4   1 mol Li 2SO4 Moles SO42– = (1.80 L )   1L   109.95 g Li 2SO 4  6.022 x 1023 Li + Li+ ions = 0.084802 mol Li +   1 mol Li +  ( )   1 mol SO42 −     1 mol Li 2SO 4  = 0.04240 = 0.0424 mol SO42–    22 22 +  = 5.106787 x 10 = 5.11 x 10 Li ions   6.022 x 1023 SO 24 −  22 22 2– SO42– ions = 0.042401 mol SO 24 −   = 2.55339 x 10 = 2.55 x 10 SO4 ions 2−  1 mol SO 4   c) KBr(s) → K+(aq) + Br–(aq)  10−3 L   1.68 x 1022 FU KBr   1 K +  21 + K+ ions = ( 225 mL )     = 3.78 x 10 K ions  1 mL   L 1 FU KBr     ( )  10−3 L   1.68 x 1022 FU KBr   1 Br −  21 – Br– ions = ( 225 mL )     = 3.78 x 10 Br ions  1 mL   L     1 FU KBr   1 mol K + Moles K+ = 3.78 x 1021 K +   6.022 x 1023 K +  ( )  –3 –3 +  = 6.27698 x 10 = 6.28 x 10 mol K   1 mol Br − Moles Br– = 3.78 x 1021 Br −   6.022 x 1023 Br −  ( 4.19 )  –3 –3 –  = 6.27698 x 10 = 6.28 x 10 mol Br  a) MgCl2(s) → Mg2+(aq) + 2 Cl–(aq)  10−3 L   1.75 mol MgCl2 Moles Mg2+ = ( 88.mL )   1 mL   L    10−3 L   1.75 mol MgCl2 Moles Cl– = ( 88. mL )   1 mL   L     1 mol Mg 2 +   1 mol MgCl  2   2 mol Cl−   1 mol MgCl  2  6.022 x 1023 Mg 2 + Mg2+ ions = 0.154 mol Mg 2 +   1 mol Mg 2 +  ( )  2+  = 0.154 = 0.15 mol Mg   –  = 0.308 = 0.31 mol Cl   22 23 2+  = 9.27388 x 10 = 9.3 x 10 Mg ions   6.022 x 1023 Cl−  23 23 – Cl– ions = 0.308 mol Cl−   = 1.854776 x 10 = 1.9 x 10 Cl ions −  1 mol Cl   b) Al2(SO4)3(s) → 2 Al3+(aq) + 3 SO42–(aq)  10−3 L   0.22 g Al2 (SO 4 )3   1 mol Al2 (SO 4 )3   2 mol Al3 +  Moles Al3+ = ( 321 mL )       1 mL   1L   342.17 g Al2 (SO 4 )3   1 mol Al2 (SO 4 )3    = 4.12777 x 10–4 = 4.1 x 10–4 mol Al3+  10−3 L   0.22 g Al2 (SO4 )3   1 mol Al2 (SO4 )3   3 mol SO24 −  Moles SO42– = ( 321 mL )         1L  342.17 g Al2 (SO4 )3   1 mol Al2 (SO4 )3   1 mL   = 6.191659 = 6.2 x 10–4 mol SO42–  6.022 x 1023 Al3 +  20 20 3+ Al3+ ions = 4.12777 x 10−4 mol Al3 +   = 2.4857 x 10 = 2.5 x 10 Al ions  1 mol Al3 +   ( ) ( )  6.022 x 1023 SO24 − SO42– ions = 6.191659 x 10−4 mol SO24 −   1 mol SO2 − 4  ( ) 4-4  20 20 2–  = 3.7286 x 10 = 3.7 x 10 SO4 ions  c) CsNO3(s) → Cs+(aq) + NO3–(aq)  8.83 x 1021 FU CsNO3    1 Cs + Cs+ ions = (1.65 L )        L    1 FU CsNO3  = 1.45695 x 1022 = 1.46 x 1022 Cs+ ions  8.83 x 1021 FU CsNO3   1 NO3−  NO3– ions = (1.65 L )       1 FU CsNO  L 3    = 1.45695 x 1022 = 1.46 x 1022 NO3– ions   1 mol Cs + = 0.0241938 = 0.0242 mol Cs+ Moles Cs+ = 1.45695 x 1022 Cs +   6.022 x 1023 Cs +    ( )  1 mol NO3− Moles NO3– = 1.45695 x 1022 NO3−   6.022 x 1023 NO − 3  ( )  –  = 0.0241938 = 0.0242 mol NO3  4.20 Plan: The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely to yield H+ ions and associated anions. One mole of HClO4, HNO3 and HCl each produce one mole of H+ upon dissociation, so moles H+ = moles acid. Molarity is expressed as moles/L instead of as M. Solution: a) Moles H+ = mol HClO4 = (1.40 L) (0.25 mol/L) = 0.35 mol H+ b) Moles H+ = mol HNO3 = (1.8 mL) (10–3 L/1 mL) (0.72 mol/L) = 1.296 x 10–3 = 1.3 x 10–3 mol H+ c) Moles H+ = mol HCl = (7.6 L) (0.056 mol/L) = 0.4256 = 0.43 mol H+ 4.21 The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely to yield H+ ions and associated anions. One mole of HBr, HI and HNO3 each produce one mole of H+ upon dissociation, so moles H+ = moles acid. Molarity is expressed as moles/L instead of as M. a) Moles H+ = mol HBr = (1.4 mL) (10–3 L/1mL) (0.75 mol/L) = 1.05 x 10–3 = 1.0 x 10–3 mol H+ b) Moles H+ = mol HI = (2.47 mL) (10–3L/1 mL) (1.98 mol/L) = 4.8906 x 10–3 = 4.89 x 10–3 mol H+ c) Moles H+ = mol HNO3 = (395 mL) (10–3 L/1 mL) (0.270 mol/L) = 0.10665 = 0.107 mol H+ 4.22 In this problem, there is no intermediate rounding. The ions are not calculated separately for those compounds where the number of cations equals the number of anions. The significant figures in each intermediate answer are underlined. a) The volume of the seawater is needed.  103 g  cm3   1 mL   10−3 L  = 0.961538 L Volume (seawater) = (1.00 kg )   1 kg   1.04 g   1 cm3   1 mL       The moles of each ion are needed. If an ion comes from more than one source, the total moles are needed. NaCl  1 mol NaCl   1 mol ion  Moles Na+ = mol Cl– = ( 26.5 g NaCl )    = 0.4534565 mol  58.44 g NaCl   1 mol NaCl  MgCl2  1 mol MgCl2 Moles Mg2+ = ( 2.40 g MgCl2 )   95.21 g MgCl2  1 mol MgCl2 Moles Cl– = ( 2.40 g MgCl2 )   95.21 g MgCl2   1 mol Mg 2 +     1 mol MgCl2   2 mol Cl−     1 mol MgCl2  2+  = 0.025207 mol Mg   –  = 0.050415 mol Cl  MgSO4  1 mol MgSO 4   1 mol ion  Moles Mg2+ = mol SO42– = ( 3.35 g MgSO 4 )    = 0.0278285 mol  120.38 g MgSO 4   1 mol MgSO 4  4-5 CaCl2  1 mol CaCl2 Moles Ca2+ = (1.20 g CaCl 2 )   110.98 g CaCl2  1 mol CaCl2 Moles Cl– = (1.20 g CaCl 2 )   110.98 g CaCl2   1 mol Ca 2 +     1 mol CaCl 2   2 mol Cl−     1 mol CaCl 2  2+  = 0.0108128 mol Ca   –  = 0.0216255 mol Cl  KCl  1 mol KCl  1 mol ion  Moles K+ = mol Cl– = (1.05 g KCl )    = 0.0140845 mol  74.55 g KCl  1 mol KCl  NaHCO3  1 mol NaHCO3   1 mol ion  Moles Na+ = mol HCO3– = ( 0.315 g NaHCO3 )    = 0.00374955 mol  84.01 g NaHCO3   1 mol NaHCO3  NaBr  1 mol NaBr   1 mol ion  Moles Na+ = mol Br– = ( 0.098 g NaBr )    = 0.0009524735 mol  102.89 g NaBr   1 mol NaBr  Total moles of each ion: Cl–: 0.4534565 + 0.050415 + 0.0216255 + 0.0140845 = 0.5395815 mol Cl– + 0.4534565 + 0.00374955 + 0.0009524735 = 0.458158523 mol Na+ Na : 2+ Mg : 0.025207 + 0.0278285 = 0.0530355 mol Mg2+ SO42–: 0.0278285 mol SO42– 0.0108128 mol Ca2+ Ca2+: + 0.0140845 mol K+ K: – HCO3 : 0.00374955 mol HCO3– 0.0009524735 mol Br– Br–: Dividing each of the numbers of moles by the volume (0.961538 L) and rounding to the proper number of significant figures gives the molarities. M = (mol ion) / L solution M Cl– = (0.5395815 mol) / (0.961538 L) = 0.561165 = 0.561 M Cl– M Na+ = (0.458158523 mol) / (0.961538 L) = 0.476485 = 0.476 M Na+ M Mg2+ = (0.0530355 mol) / (0.961538 L) = 0.0551569 = 0.0552 M Mg2+ M SO42– = (0.0278285 mol) / (0.961538 L) = 0.02894165 = 0.0289 M SO42– M Ca2+ = (0.0108128 mol) / (0.961538 L) = 0.0112453 = 0.0112 M Ca2+ M K+ = (0.0140845 mol) / (0.961538 L) = 0.0146479 = 0.0146 M K+ M HCO3– = (0.00374955 mol) / (0.961538 L) = 0.00389953 = 0.00390 M HCO3– M Br– = (0.0009524735 mol) / (0.961538 L) = 0.00099057 = 0.00099 M Br– b) The alkali metal cations are Na+ and K+. Simply sum the molarities of the individual ions. 0.476485 M Na+ + 0.0146479 M K+ = 0.4911329 = 0.491 M total for alkali metal cations c) The alkaline earth metal cations are Mg2+ and Ca2+. Simply sum the molarities of the individual ions. 0.0551569 M Mg2+ + 0.0112453 M Ca2+ = 0.0664022 = 0.0664 M total for alkaline-earth cations d) The anions are Cl–, SO42–, HCO3–, and Br–. Simply sum the molarities of the individual ions. 0.561165 M Cl– + 0.02894165 M SO42– + 0.00389953 M HCO3– + 0.00099057 M Br– = 0.59499675 = 0.595 M total for anions 4.23 The moles of the calcium ions and the iron ions are needed. The moles of each of the ions to be replaced must be multiplied by the charge to get the total moles of charge. Since sodium has a +1 charge the total moles of charge equals the moles of sodium ions.  0.015 mol Ca 2 +   2 mol charge  = 30. mol charge from Ca2+ 1.0 x103 L    2+    L 1 mol Ca    ( )  Fe (1.0 x 10 L )  0.0010 mol L 3  3+   3 mol charge  = 3.0 mol charge from Fe3+   3+      1 mol Fe 4-6 Moles Na+ =  +  mol Na  ( ( 30. + 3.0 ) mol charge )  11mol charge    = 33 mol Na+ 4.24 Ions in solution that do not participate in the reaction do not appear in a net ionic equation. These spectator ions remain as dissolved ions throughout the reaction. These ions are only present to balance charge. 4.25 The reaction given has the following total ionic and net ionic equations: Total ionic Ba2+(aq) + 2 NO3–(aq) + 2 Na+(aq) + CO32–(aq) → BaCO3(s) + 2 Na+(aq) + 2 NO3–(aq) Net ionic Ba2+(aq) + CO32–(aq) → BaCO3(s) The “new” equations are: Molecular: BaCl2(aq) + K2CO3(aq) → BaCO3(s) + 2 KCl(aq) Total ionic: Ba2+(aq) + 2 Cl–(aq) + 2 K+(aq) + CO32–(aq) → BaCO3(s) + 2 K+(aq) + 2 Cl–(aq) Molecular: BaBr2(aq) + (NH4)2CO3(aq) → BaCO3(s) + 2 NH4Br(aq) Total ionic: Ba2+(aq) + 2 Br–(aq) + s NH4+(aq) + CO32–(aq) → BaCO3(s) + 2 NH4+(aq) + 2 Br–(aq) 4.26 If the electrostatic attraction between the ions is greater than the attraction of the ions for water molecules, the ions will form a precipitate. This is the basis for the solubility rules. 4.27 a) Since the possible products (CaCl2 and KNO3) are both soluble, no reaction would take place. b) According to the table, KNO3 is soluble but PbCl2 is insoluble. The K+ and NO3– would be spectator ions, because their salt is soluble. 4.28 Assuming that the left beaker is AgNO3 (because it has gray Ag+ ion) and the right must be NaCl, then the NO3– is blue, the Na+ is brown, and the Cl– is green. Molecular equation: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) Total ionic equation: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → AgCl(s) + Na+(aq) + NO3–(aq) Net ionic equation: Ag+(aq) + Cl–(aq) → AgCl(s) 4.29 Plan: Check to see if any of the ion pairs are not soluble according to the solubility rules. a) Molecular: Hg2(NO3)2(aq) + 2 KI(aq) → Hg2I2(s) + 2 KNO3(aq) Total ionic: Hg22+(aq) + 2 NO3–(aq) + 2 K+(aq) + 2 I–(aq) → Hg2I2(s) + 2 K+(aq) + 2 NO3–(aq) Net ionic: Hg22+(aq) + 2 I–(aq) → Hg2I2(s) Spectator ions are K+ and NO3–. b) Molecular: FeSO4(aq) + Ba(OH)2(aq) → Fe(OH)2(s) + BaSO4(s) Total ionic: Fe2+(aq) + SO42–(aq) + Ba2+(aq) + 2 OH–(aq) → Fe(OH)2(s) + BaSO4(s) Net ionic: This is the same as the total ionic equation, because there are no spectator ions. 4.30 Check to see if any of the ion pairs are not soluble according to the solubility rules. a) Molecular: 3 CaCl2(aq) + 2 Cs3PO4(aq) → Ca3(PO4)2(s) + 6 CsCl(aq) Total ionic: 3 Ca2+(aq) + 6 Cl–(aq) + 6 Cs+(aq) + 2 PO43–(aq) → Ca3(PO4)2(s) + 6 Cs+(aq) + 6 Cl–(aq) Net ionic: 3Ca2+(aq) + 2PO43–(aq) → Ca3(PO4)2(s) Spectator ions are Cs+ and Cl–. b) Molecular: Na2S(aq) + ZnSO4(aq) → ZnS(s) + Na2SO4(aq) Total ionic: 2 Na+(aq) + S2–(aq) + Zn2+(aq) + SO42–(aq) → ZnS(s) + 2 Na+(aq) + SO42–(aq) Net ionic: S2–(aq) + Zn2+(aq) → ZnS(s) Spectator ions are Na+ and SO42-. 4.31 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Solution: a) No precipitate will form. The ions Na+ and SO42– will not form an insoluble salt according to solubility rule #1: All common compounds of Group 1A ions are soluble. The ions Cu2+ and NO3– will not form an insoluble salt according to the solubility rule #2: All common nitrates are soluble. 4-7 4.32 b) A precipitate will form because silver ions, Ag+, and iodide ions, I–, will combine to form a solid salt, silver iodide, AgI. The ammonium and nitrate ions do not form a precipitate. Molecular: NH4I(aq) + AgNO3(aq) → AgI(s) + NH4NO3(aq) Total ionic: NH4+(aq) + I–(aq) + Ag+(aq) + NO3–(aq) → AgI(s) + NH4+(aq) + NO3–(aq) Net ionic: Ag+(aq) + I–(aq) → AgI(s) a) Barium carbonate (BaCO3) precipitates; the potassium hydroxide is a strong base. Molecular: K2CO3(aq) + Ba(OH)2(aq) → BaCO3(s) + 2 KOH(aq) Total ionic: 2 K+(aq) + CO32–(aq) + Ba2+(aq) + 2 OH–(aq) → BaCO3(s) + 2 K+(aq) + 2 OH–(aq) Net ionic: Ba2+(aq) + CO32–(aq) → BaCO3(s) b) Aluminum phosphate (AlPO4) precipitates; the sodium nitrate is soluble. Molecular: Al(NO3)3(aq) + Na3PO4(aq) → AlPO4(s) + 3 NaNO3(aq) Total ionic: Al3+(aq) + 3 NO3–(aq) + 3 Na+(aq) + PO43–(aq) → AlPO4(s) + 3 NO3–(aq) + 3 Na+(aq) Net ionic: Al3+(aq) + PO43–(aq) → AlPO4(s) 4.33 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Solution: a) New cation-anion combinations are potassium nitrate and iron(II) chloride. The rules state that all common nitrates and chlorides (with some exceptions) are soluble, so no precipitate forms. b) New cation-anion combinations are ammonium chloride and barium sulfate. Again, the rules state that most chlorides are soluble; however, another rule states that sulfate compounds containing barium are insoluble. Barium sulfate is a precipitate and its formula is BaSO4. Molecular: (NH4)2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2 NH4Cl(aq) Total ionic: 2 NH4+(aq) + SO42–(aq) + Ba2+(aq) + 2 Cl–(aq) → BaSO4(s) + 2 NH4+(aq) + Cl–(aq) Net ionic: Ba2+(aq) + SO42–(aq) → BaSO4(s) 4.34 a) Nickel(II) sulfide (NiS) precipitates, sodium sulfate is soluble. Molecular: Na2S(aq) + NiSO4(aq) → NiS(s) + Na2SO4(aq) Total ionic: 2 Na+(aq) + S–2(aq) + Ni2+(aq) + SO42–(aq) → NiS(s) + 2 Na+(aq) + SO42–(aq) Net ionic: Ni2+(aq) + S2–(aq) → NiS(s) b) Lead(II) bromide (PbBr2) precipitates, potassium nitrate is soluble. Molecular: Pb(NO3)2(aq) + 2 KBr(aq) → PbBr2(s) + 2 KNO3(aq) Total ionic: Pb2+(aq) + 2 NO3–(aq) + 2 K+(aq) + 2 Br–(aq) → PbBr2(s) + 2 K+(aq) + 2 NO3–(aq) Net ionic: Pb2+(aq) + 2 Br–(aq) → PbBr2(s) 4.35 Plan: Write a balanced equation for the chemical reaction described in the problem. By applying the solubility rules to the two possible products (NaNO3 and PbI2), determine that PbI2 is the precipitate. By using molar relationships, determine how many moles of Pb(NO3)2 (and thus Pb2+ ion) are required to produce 0.628 g of PbI2. Solution: The reaction is: Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq).  1 mol PbI 2   1 mol Pb 2 +     1 mL  1 M Pb2+ = ( 0.628 g PbI 2 )        −3   461.0 g PbI2   1 mol PbI2   35.0 mL   10 L  = 0.0389216 = 0.0389 M Pb2+ 4.36 The reaction is AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq).  1 mol AgCl   1 mol Ag +     1 mL  1 M Ag+ = ( 0.842 g AgCl )        −3   143.4 g AgCl   1 mol AgCl   25.0 mL   10 L  = 0.2348675 = 0.235 M Ag+ 4-8 4.37 Plan: The balanced equation for this reaction is AgNO3(aq) + Cl–(aq) → AgCl(s) + NO3–(aq). First, determine the moles of Cl– present in the 25.00 mL sample. Second, convert moles of Cl– into grams, and convert the sample volume into grams using the given density. The mass percent of Cl– is found by dividing one result by the other. Solution:  10−3 L   0.3020 mol AgNO3   1 mol Cl−   35.45 g Cl  = 0.467098 g Cl (unrounded) ( 43.63 mL )       −  L   1 mol AgNO3   1 mol Cl   1 mL    1.04 g  Mass of sample = ( 25.00 mL )   = 26.0 g sample  mL  Mass Cl 0.467098 g Cl x 100% = x 100% = 1.79653 = 1.80% Cl Mass % Cl = Mass Sample 26.0 g Sample 4.38 a) Total: Al2(SO4)3(aq) + 6 NaOH(aq) → 3 Na2SO4(aq) + 2 Al(OH)3(s) Net ionic: Al3+(aq) + 3 OH–(aq) → Al(OH)3(s) b) Determine the limiting reactant:  10−3 L   15.8 g Al2 (SO 4 )3   1 mol Al2 (SO 4 )3  Moles Al2(SO4)3 = ( 627 mL )     1 mL   L   342.17 g Al2 (SO 4 )3    = 0.028952 mol Al2(SO4)3 (unrounded)  10−3 L   0.533 mol NaOH  Moles NaOH = (185.5 mL )   = 0.0988715 mol NaOH (unrounded)  1 mL   L    NaOH is the limiting reactant. Finishing with limiting reactant  2 mol Al(OH)3   78.00 g Al(OH)3  Grams Al(OH)3 = ( 0.0988715 mol NaOH )   = 2.570659 = 2.57 g Al(OH)3   6 mol NaOH   1 mol Al(OH)3  4.39 The molecular equation is: H2SO4(aq) + Sr(OH)2(aq) → SrSO4(s) + 2 H2O(l) The total ionic equation is: 2 H+(aq) + SO42–(aq) + Sr2+(aq) + 2 OH–(aq) → SrSO4(s) + 2 H2O(l) Since there are no spectator ions, the total and net ionic equations are the same. 4.40 H+(aq) + OH–(aq) → H2O(l) 4.41 a) Any three of HCl, HBr, HI, HNO3, H2SO4, or HClO4 b) Any three of NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 c) They dissociate 100% into ions in aqueous solution. 4.42 a) There are many possibilities including: acetic acid, chlorous acid, nitrous acid, .... b) NH3 c) Strong acids or bases dissociate 100%; weak acids or bases dissociate much less than this (typically less than 10%) in aqueous solution. The electrical conductivity of a solution of a strong acid or base would be much higher than that of a weak acid or base of equal concentration. 4.43 a) The formation of a gas, SO2(g), and formation of water drive this reaction to completion, because both products remove reactants from solution. b) The formation of a precipitate, Ba3(PO4)2(s), will cause this reaction to go to completion. This reaction is one between an acid and a base, so the formation of water molecules through the combination of H+ and OH– ions also drives the reaction. 4-9 4.44 Molecular: HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(l) Net ionic: HC2H3O2(aq) + OH–(aq) → C2H3O2–(aq) + H2O(l) Molecular: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Net ionic: H+(aq) + OH–(aq) → H2O(l) The difference is because CH3COOH is a weak acid and HCl is a strong acid. 4.45 a) Molecular equation: KOH(aq) + HI(aq) → KI(aq) + H2O(l) Total ionic equation: K+(aq) + OH–(aq) + H+(aq) + I–(aq) → K+(aq) + I–(aq) + H2O(l) Net ionic equation: OH–(aq) + H+(aq) → H2O(l) The spectator ions are K+(aq) and I–(aq) b) Molecular equation: NH3(aq) + HCl(aq) → NH4Cl(aq) Total ionic equation: NH3(aq) + H+(aq) + Cl–(aq) → NH4+(aq) + Cl–(aq) NH3 is a weak base and is written in the molecular form. HCl is a strong acid and is written in the dissociated form. NH4Cl is a soluble compound, because all ammonium compounds are soluble. Net ionic equation: NH3(aq) + H+(aq) → NH4+(aq) Cl– is the only spectator ion. 4.46 a) Molecular: CsOH(aq) + HNO3(aq) → CsNO3(aq) + H2O(l) Total ionic: Cs+(aq) + OH–(aq) + H+(aq) + NO3–(aq) → Cs+(aq) + NO3–(aq) + H2O(l) Net ionic: OH–(aq) + H+(aq) → H2O(l) Spectator ions are Cs+ and NO3–. b) Molecular: Ca(OH)2(aq) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + 2 H2O(l) Total ionic: Ca2+(aq) + 2 OH–(aq) + 2 HC2H3O2 (aq) → Ca2+(aq) + 2 C2H3O2–(aq) + 2 H2O(l) Net ionic: OH–(aq) + HC2H3O2(aq) → C2H3O2–(aq) + H2O(l) Spectator ion is Ca2+. 4.47 Calcium carbonate dissolves in HCl(aq) because the carbonate ion, a base, reacts with the acid to form CO2(g). Total ionic equation: CaCO3(s) + H+(aq) + 2 Cl–(aq) → Ca2+(aq) + 2 Cl–(aq) + H2O(l) + CO2(g) Net ionic equation: CaCO3(s) + H+(aq) → Ca2+(aq) + H2O(l) + CO2(g) 4.48 The HNO3 reacts with the bound OH-, producing H2O: Total ionic equation: Zn(OH)2(s) + 2 H+(aq) + 2 NO3–(aq) → Zn2+(aq) + 2 NO3–(aq) + 2 H2O(l) Net ionic equation: Zn(OH)2(s) + 2 H+(aq) → Zn2+(aq) + 2 H2O(l) 4.49 The reaction is: KOH(aq) + CH3COOH(aq) → KCH3COO(aq) + H2O(l) −3  1mol CH3COOH     1 mL  1  0.1180 mol KOH   10 L  M =    ( 25.98 mL )     −3    L    1 mL   1 mol KOH   52.50 mL   10 L  = 0.05839314 = 0.05839 M CH3COOH 4.50 The reaction is: 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) −3  1 mol H 2SO 4     1 mL  1  0.1750 mol NaOH   10 L  M=    ( 36.25 mL )     −3     L    1 mL   2 mol NaOH   25.00 mL   10 L  = 0.126875 = 0.1269 M H2SO4 4-10 4.51 The reaction is: 2 NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)   1 mL   2 mol NaHCO3   1L  2.6 mol H 2SO 4   10−3 L  V=    ( 78 mL )    −3    L    1 mL   1 mol H 2SO 4   1.5 mol NaHCO3   10 L  = 270.4 = 2.7 x 102 mL NaHCO3 4.52 The reaction is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) −3   1 mL   1 mol NaOH   1  0.1528 mol HCl   10 L  M=    −3    ( 25.00 mL )     L    1 mL   1mol HCl   ( 39.21 − 2.24 ) mL   10 L  = 0.103327 = 0.1033 M NaOH 4.53 The reaction is: UO2(s) + 4 HF(aq) → UF4(s) + 2 H2O(l)  103 g   1 mol UO 2  4 mol HF   1L V = ( 2.25 kg UO 2 )  = 13.3333 = 13.3 L HF  1 kg   270.0 g UO  1 mol UO  2.50 mol HF   2  2    4.54 a) Since H is bonded to a nonmetal, O.N. (H) = +1. Then O.N. (S) = -2. b) Assume O.N. (O) = -2. Then O.N. (S) = +4. 4.55 No, O.N. (N) = -3, and O.N. (H) = +1, O.N. (Cl) = -1 throughout. Since no elements changes oxidation number, this cannot be a redox reaction. 4.56 An oxidizing agent causes something else to be oxidized; i.e., to lose electrons. The oxidizing agent accepts these electrons and is reduced. 4.57 Electrons cannot appear and disappear as if by magic; they must come from somewhere and go to somewhere. Therefore, something must be oxidized and something else be reduced. 4.58 a) The S in SO42– (i.e., H2SO4) has O.N. = +6, and in SO2, O.N. (s) = +4, so the S has been reduced (and the I– oxidized), so the H2SO4 is an oxidizing agent. b) The oxidation numbers remain constant throughout; H2SO4 transfers a proton to F– to produce HF, so it acts as an acid. 4.59 Oxidizing agent = NO2; Reducing agent = NH3; the NO2 is the oxidizing agent (O.N. (N) goes from +4 in NO2 to 0 in N2), and the NH3 is the reducing agent (O.N.(N) goes from -3 to 0). 4.60 Refer to the rules in Table 4.3. Remember that oxidation number (O.N.) is not the same thing as ionic charge. The O.N. is the charge an atom would have if electrons were transferred completely. The compounds in this problem contain covalent bonds, or a combination of ionic and covalent bonds (Na2C2O4). Covalent bonds do not involve the complete transfer of electrons. a) CF2Cl2. The rules dictate that F and Cl each have an O.N. of -1; two F and two Cl yield a sum of -4, so C must have a +4 O.N. C = +4 b) Na2C2O4. The rule dictates that Na has a +1 O.N.; rule 5 dictates that O has a -2 O.N.; two Na and four O yield a sum of -6 [(+2) + (-8)]. Therefore, the total of the O.N.’s on the two C atoms is +6 and each C is +3. C = +3 c) HCO3–. H is combined with nonmetals and has an O.N. of +1; O has an O.N. of -2. To have an overall oxidation state equal to -1, C must be +4 because (+1) + (+4) + (-6) = -1. C = +4 d) C2H6. Each H has an O.N. of +1; six H gives +6. The sum of O.N.’s for the two C atoms must be -6, so each C is -3. C = -3 4.61 a) -1 b) +3 c) +5 d) -1 4-11 4.62 a) NH2OH: b) N2H4: c) NH4+: d) HNO2: (O.N. for N) + 3(+1 for H) + 1(-2 for O) = 0 O.N. for N = -1 2(O.N. for N) + 4(+1 for H) = 0 O.N. for N = -2 (O.N. for N) + 4(+1 for H) = +1 O.N. for N = -3 (O.N. for N) + 1(+1 for H) + 2(-2 for O) = 0 O.N. for N = +3 4.63 a) +4 c) +4 4.64 a) AsH3. H is combined with a nonmetal, so its O.N. is +1 (Rule 3). The O.N. for As is -3. b) H3AsO4. The H’s in this formula are acidic hydrogens. When an acid dissociates in water, it forms H3O+ or H+ ion, therefore, the O.N. of H in this compound is +1, or +3 for 3 H’s. Oxygen’s O.N. is -2, with total O.N. of -8 (4 times -2), so As needs to have an O.N. of +5. c) AsCl3. Cl has an O.N. of -1, total of -3, so As must have an O.N. of +3. a) As = -3 b) As = +5 c) As = +3 4.65 a) +5 4.66 a) MnO42–: b) Mn2O3: c) KMnO4: (O.N. for Mn) + 4(-2 for O) = -2 {2(O.N. for Mn)} + 3(-2 for O) = 0 1(+1 for K) + (O.N. for Mn) + 4(-2 for O) = 0 4.67 a) +6 c) +3 4.68 Oxidizing agent: substance that accepts the electrons released by the substance that is oxidized. Reducing agent: substance that provides the electrons accepted by the substance that is reduced. First, assign oxidation numbers to all atoms. Second, recognize that the agent is the compound that contains the atom that is gaining or losing electrons, not just the atom itself. a) 5 H2C2O4(aq) + 2 MnO4–(aq) + 6 H+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) H = +1 Mn = +7 H = +1 Mn = +2 C = +4 H = +1 C = +3 O = -2 O = -2 O = -2 O = -2 Hydrogen and oxygen do not change oxidation state. The Mn changes from +7 to +2 (reduction). Therefore, MnO4– is the oxidizing agent. C changes from +3 to +4 (oxidation), so H2C2O4 is the reducing agent. b) 3 Cu(s) + 8 H+(aq) + 2 NO3–(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l) Cu = 0 H = +1 N = +5 Cu = +2 N = +2 H = +1 O = -2 O = -2 O = -2 Cu changes from 0 to +2 (is oxidized) and Cu is the reducing agent. N changes from +5 (in NO3–) to +2 (in NO) and is reduced, so NO3– is the oxidizing agent. 4.69 a) Oxidizing agent = H+ b) Oxidizing agent = H2O2 4.70 a) Oxidizing agent is NO3– because nitrogen changes from +5 O.N. in NO3– to +4 O.N. in NO2. Reducing agent is Sn because its O.N. changes from 0 as the element to +4 in SnCl62–. b) Oxidizing agent is MnO4– because manganese changes from +7 O.N. in MnO4– to +2 O.N. in Mn2+. Reducing agent is Cl– because its O.N. changes from -1 in Cl– to 0 as the element to Cl2. 4.71 a) Oxidizing agent = Cr2O72– b) Oxidizing agent = NO3– 4.72 S is in Group 6A(16), so its highest possible O.N. is +6 and its lowest possible O.N. is 6 - 8 = -2. a) The lowest O.N. for S [(Group 6A(16)] is 6 - 8 = -2, which occurs in S2–. Therefore, when S2– reacts in an oxidation-reduction reaction, S can only increase its O.N. (oxidize), so S2– can only function as a reducing agent. b) The highest O.N. for S is +6, which occurs in SO42–. Therefore, when SO42– reacts in an oxidation-reduction reaction, the S can only decrease its O.N. (reduce), so SO42– can only function as an oxidizing agent. c) The O.N. of S in SO2 is +4, so it can increase or decrease its O.N. Therefore, SO2 can function as either an oxidizing or reducing agent. b) -1 b) -3 b) +6 d) -2 c) +5 O.N. for Mn = +6 O.N. for Mn = +3 O.N. for Mn = +7 Reducing agent = Sn Reducing agent = Fe2+ Reducing agent = SO32– Reducing agent = Zn 4-12 4.73 N is Group 5A (15), so its highest possible O.N. is +5 and its lowest possible O.N. is 5 - 8 = -3. a) In N3–, O.N. (N) = -3, so it can only be oxidized (i.e., act as a reducing agent). b) In NO3–, O.N. (N) = +5, so it can only be reduced (i.e., act as an oxidizing agent). c) In NO2–, O.N. (N) = +3, so it can be either oxidized or reduced (i.e., can act as a reducing or oxidizing agent). 4.74 a) Step: 1: Assign oxidation numbers to all elements. Step: 2: Identify oxidized and reduced species. a) The O.N. of Cr decreases from +6 in K2CrO4 to +3 in Cr(NO3)3, so chromium is gaining electrons and is reduced. The O.N. of Fe increases from +2 in Fe(NO3)2 to +3 in Fe(NO3)3, so iron is losing electrons and is oxidized. Step: 3: Compute number of electrons lost in oxidation and gained in reduction. Step: 4: Multiply by factors to make e– lost equal e– gained. Cr gained 3 electrons so Fe should be multiplied by 3 to lose 3 electrons Step: 5: Finish balancing by inspection. 8 HNO3(aq) + K2CrO4(aq) + 3 Fe(NO3)2(aq) → 2 KNO3(aq) + 3 Fe(NO3)3(aq) + Cr(NO3)3(aq) + 4 H2O(l) Reducing agent: Fe(NO3)2 Oxidizing agent: K2CrO4 b) 8 HNO3(aq) + 3 C2H6O(l) + K2Cr2O7(aq) → 2 KNO3(aq) + 3 C2H4O(l) + 7 H2O(l) + 2 Cr(NO3)3(aq) Reducing agent: C2H6O Oxidizing agent: K2Cr2O7 c) 6 HCl(aq) + 2 NH4Cl(aq) + K2Cr2O7(aq) → 2 KCl(aq) + 2 CrCl3(aq) + N2(g) + 7 H2O(l) Reducing agent: NH4Cl Oxidizing agent: K2Cr2O7 d) KClO3(aq) + 6 HBr(aq) → 3 Br2(l) + 3 H2O(l) + KCl(aq) Oxidizing agent: KClO3 Reducing agent: HBr 4.75 a) 2 HCl(aq) + 2 FeCl2(aq) + H2O2(aq) → 2 FeCl3(aq) + 2 H2O(l) H2O2 is the oxidizing agent (O.N.(O) goes from -1 to -2) and FeCl2 is the reducing agent (O.N.(Fe) goes from +2 to +3). b) I2(s) + 2 Na2S2O3(aq) → Na2S4O6(aq) + 2 NaI(aq) I2 is the oxidizing agent (O.N.(I) goes from 0 to -1) and Na2S2O3 is the reducing agent (O.N.(S) goes from +2 to +2.5). c) 8 HNO3(aq) + 6 KI(aq) → 2 NO(g) + 3 I2(s) + 4 H2O(l) + 6 KNO3(aq) HNO3 is the oxidizing agent (O.N.(N) goes from +5 to +2) and KI is the reducing agent (O.N.(I) goes from -1 to 0). d) 3 PbO(s) + 2 NH3(aq) → N2(g) + 3 H2O(l) + 3 Pb(s) PbO is the oxidizing agent (O.N.(Pb) goes from +2 to 0) and NH3 is the reducing agent (O.N.(N) goes from -3 to 0). 4.76 Plan: Determine the moles of permanganate ion present. The moles of permanganate ion along with the balanced chemical equation give the moles of hydrogen peroxide. The molar mass of hydrogen peroxide is required to finish the problem. The unrounded numbers from each step are carried to the next step to minimize rounding errors. Solution:  0.105 mol MnO 4−  10−3 L  a) Moles MnO4– =  43.2 mL ) = 4.536 x 10–3 = 4.54 x 10–3 mol MnO4–     1 mL  ( L     5 mol H 2 O 2 b) Moles H2O2 = 4.536 x 10−3 MnO −4   2 mol MnO − 4  ( )   = 0.01134 = 0.0113 mol H2O2   34.02 g H 2 O 2  c) Mass H2O2 = ( 0.01134 mol H 2 O 2 )   = 0.3857868 = 0.386 g H2O2  1 mol H 2 O 2   0.3857868 g H 2 O 2  d) Mass percent H2O2 =   x 100% = 2.79556 = 2.80% H2O2  13.8 g Sample  e) H2O2 (The hydrogen peroxide reduces the manganese from +7 to +2.) 4-13 4.77  0.05961 mol Cr2 O72 −   10−3 L   1 mol C2 H 5 OH   46.07 g C 2 H5 OH  Mass C2H5OH =  35.46 mL )   (      1 mL   2 mol Cr O 2 −   1 mol C H OH  1L 2 5  2 7      = 0.0486907 g C2H5OH (unrounded)  0.0486907 g C 2 H5 OH  Mass percent C2H5OH =   x 100% = 0.173895 = 0.1739% C2H5OH 28.00 g Sample   4.78 a) decomposition 4.79 By definition, elements cannot decompose into anything simpler, so they could not be reactants in a decomposition reaction. 4.80 Combination and displacement; possibly decomposition 4.81 A common example of a combination/redox reaction is the combination of a metal and nonmetal to form an ionic salt, such as 2 Mg(s) + O2(g) → 2 MgO(s). A common example of a combination/non-redox reaction is the combination of a metal oxide and water to form an acid, such as: CaO(s) + H2O(l) → Ca(OH)2(aq). 4.82 Yes, since the oxidation number of O2 will change from zero (in the element) to some negative value (typically -2) in the product(s) during the reaction. 4.83 a) Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g) Displacement: one Ca atom displaces 2 H atoms. b) 2 NaNO3(s) → 2 NaNO2(s) + O2(g) Decomposition: one reactant breaks into two products. c) C2H2(g) + 2 H2(g) → C2H6(g) Combination: reactants combine to form one product. 4.84 a) 2 HI(g) → H2(g) + I2(g) b) Zn(s) + 2 AgNO3(aq) → Zn(NO3)2(aq) + 2 Ag(s) c) 2 NO(g) + O2(g) → N2O4(l) 4.85 Recall the definitions of each type of reaction: Combination: X + Y → Z; decomposition: Z → X + Y Single displacement: X + YZ → XZ + Y double displacement: WX + YZ → WZ + YX combination a) 2 Sb(s) + 3 Cl2(g) → 2 SbCl3(s) b) 2 AsH3(g) → 2 As(s) + 3 H2(g) decomposition c) 3 Mn(s) + 2 Fe(NO3)3(aq) → 3 Mn(NO3)2(aq) + 2 Fe(s) displacement 4.86 a) Mg(s) + 2 H2O(g) → Mg(OH)2(s) + H2(g) b) Cr(NO3)3(aq) + Al(s) → Al(NO3)3(aq) + Cr(s) c) PF3(g) + F2(g) → PF5(g) 4.87 a) The combination between a metal and a nonmetal gives a binary ionic compound. Ca(s) + Br2(l) → CaBr2(s) b) Many metal oxides release oxygen gas upon thermal decomposition. ∆ 2 Ag2O(s) → 4 Ag(s) + O2(g) c) This is a single displacement reaction. Mn is a more reactive metal and displaces Cu2+ from solution. Mn(s) + Cu(NO3)2(aq) → Mn(NO3)2(aq) + Cu(s) b) combination c) displacement decomposition displacement combination displacement displacement combination 4-14 4.88 a) Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) elect b) 2 LiCl(l)  → 2 Li(l) + Cl2(g) c) SnCl2(aq) + Co(s) → CoCl2(aq) + Sn(s) 4.89 a) N2(g) + 3 H2(g) → 2 NH3(g) ∆ b) 2 NaClO3(s) → 2 NaCl(s) + 3 O2(g) c) Ba(s) + 2 H2O(l) → Ba(OH)2(aq) + H2(g) 4.90 a) Fe(s) + 2 HClO4(aq) → Fe(ClO4)2(aq) + H2(g) ∆ b) S8(s) + 8 O2(g) → 8 SO2(g) elect → Ba(s) + Cl2(g) c) BaCl2(l)  4.91 a) Cs, a metal, and I2, a nonmetal, react to form the binary ionic compound, CsI. 2 Cs(s) + I2(s) → 2 CsI(s) b) Al is a stronger reducing agent than Mn and is able to displace Mn from solution, i.e., cause the reduction from Mn2+(aq) to Mn0(s). 2 Al(s) + 3 MnSO4(aq) → Al2(SO4)3(aq) + 3 Mn(s) c) Sulfur dioxide, SO2, is a nonmetal oxide that reacts with oxygen, O2, to form the higher oxide, SO3. ∆ 2 SO2(g) + O2(g) → 2 SO3(g) It is not clear from the problem, but energy must be added to force this reaction to proceed. d) Propane is a three-carbon hydrocarbon with the formula C3H8. It burns in the presence of oxygen, O2, to form carbon dioxide gas and water vapor. Although this is a redox reaction that could be balanced using the oxidation number method, it is easier to balance by considering only atoms on either side of the equation. First, balance carbon and hydrogen (because they only appear in one species on each side of the equation), and then balance oxygen. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) e) Total ionic equation: 2 Al(s) + 3 Mn2+(aq) + 3 SO42–(aq) → 2 Al3+(aq) + 3 SO42–(aq) + 3 Mn(s) Net ionic equation: 2 Al(s) + 3 Mn2+(aq) → 2 Al3+(aq) + 3 Mn(s) Note that the molar coefficients are not simplified because the number of electrons lost (6 e–) must equal the electrons gained (6 e–). 4.92 a) C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) b) PCl3(l) + Cl2(g) → PCl5(s) c) Zn(s) + 2 HBr(aq) → ZnBr2(aq) + H2(g) d) 2 KI(aq) + Br2(l) → 2 KBr(aq) + I2(s) e) 2 I–(aq) + Br2(l) → I2(s) + 2 Br–(aq) 4.93 ∆ The balanced chemical equation is 2 HgO(s) → 2 Hg(l) + O2(g)  103 g   1 mol HgO   1 mol O 2   32.00 g O 2  Mass O2 = ( 4.27 kg HgO )  = 315.420 = 315 g O2  1 kg   216.6 g HgO   2 mol HgO   1 mol O    2    The other product is mercury.  103 g   1 mol HgO  2 mol Hg   200.6 g Hg   1 kg  Mass Hg = ( 4.27 kg HgO )   1 kg   216.6 g HgO  2 mol HgO   1 mol Hg   103 g        = 3.95458 = 3.95 kg Hg 4-15 4.94 elect The balanced chemical equation is CaCl2(l)  → Ca(s) + Cl2(g) Note, the reaction cannot be done in the presence of water as elemental calcium would displace the hydrogen from the water.  1 mol CaCl2   1 mol Cl2   70.90 g Cl2  Mass Cl2 = ( 874 g CaCl2 )     = 558.358 = 558 g Cl2  110.98 g CaCl 2   1 mol CaCl2   1 mol Cl2  The other product is calcium.  1 mol CaCl2   1 mol Ca   40.08 g Ca  Mass Ca = ( 874 g CaCl2 )     = 315.64 = 316 g Ca  110.98 g CaCl 2   1 mol CaCl2   1 mol Ca  4.95 To determine the reactant in excess, write the balanced equation (metal + O2 → metal oxide), convert reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. 4 Li(s) + O2(g) → 2 Li2O(s)  1 mol Li   2 mol Li 2 O  a) Moles Li2O if Li limiting = (1.62 g Li )    = 0.1166979 mol Li2O (unrounded)  6.941 g Li   4 mol Li   1 mol O 2  2 mol Li 2 O  Moles Li2O if O2 limiting = ( 6.00 g O 2 )    = 0.375 mol Li2O (unrounded)  32.00 g O 2  1 mol O 2  Li is the limiting reactant; O2 is in excess. b) 0.1166979 = 0.117 mol Li2O c) Li is limiting, thus there will be none remaining (0 g Li).  1 mol Li   2 mol Li 2 O   29.88 g Li 2 O  Grams Li2O = (1.62 g Li )   = 3.4869 = 3.49 g Li2O    6.941 g Li   4 mol Li   1 mol Li 2 O   1 mol Li  1 mol O 2   32.00 g O 2  Grams O2 used = (1.62 g Li )   = 1.867166 g O2 (unrounded)    6.941 g Li  4 mol Li  1 mol O 2  The beginning part of these calculations is repeated from the limiting reactant calculation to emphasize that the second part of the problem is simply an extension of the first part. There is no need to repeat the entire calculation as only the final step(s) times the answer of the first part will give the final answer to this part. Remaining O2 = 6.00 g O2 - 1.867166 g O2 = 4.13283 = 4.13 g O2 4.96 ∆ 3 Mg(s) + N2(g) → Mg3N2(s)  1 mol Mg   1 mol Mg 3 N 2  a) Mole Mg3N2 (from Mg) = ( 2.22 g Mg )    = 0.030440 = 0.0304 mol Mg3N2  24.31 g Mg   3 mol Mg   1 mol N 2  1 mol Mg3 N 2  Mole Mg3N2 (from N2) = ( 3.75 g N 2 )    = 0.13383 = 0.134 mol Mg3N2  28.02 g N 2  1 mol N 2  Mg is the limiting reactant; N2 is present in excess. b) 0.0304 mol Mg3N2  1 mol Mg   1 mol Mg 3 N 2   100.95 g Mg3 N 2  c) Mass Mg3N2 = ( 2.22 g Mg )      24.31 g Mg   3 mol Mg   1 mol Mg 3 N 2  = 3.07293 = 3.07 g Mg3N2 There will be no Mg remaining.  1 mol Mg   1 mol N 2   28.02 g N 2  Mass N2 = 3.75 g N2 - ( 2.22 g Mg )      24.31 g Mg   3 mol Mg   1 mol N 2  = 2.897067 = 2.90 g N2 present in excess. 4-16 4.97 ∆ 2 KClO3(s) → 2 KCl(s) + 3 O2(g) Mass KClO3 =  1 mol O 2 ( ( 0.900 − 0.700 ) g O2 )  32.00 gO   2 mol KClO3   122.55 g KClO3       3 mol O 2   1 mol KClO3  2  = 0.510625 g KClO3 (unrounded)  0.510625 g KClO3  Mass % KClO3 =   x 100% = 56.7361 = 56.7% KClO3  0.900 g Sample  4.98 ∆ CaCO3(s) → CaO(s) + CO2(g) Mass CaCO3 =  1 mol CO 2 ( ( 0.693 − 0.508) g CO2 )  44.01 g CO  1 mol CaCO3   100.09 g CaCO3      1 mol CO 2   1 mol CaCO3  2  = 0.420737 g CaCO3 (unrounded)  0.420737 g CaCO3  Mass % CaCO3 =   x 100% = 60.7124 = 60.7% CaCO3  0.693 g Sample  4.99 To find the mass of Fe, write a balanced equation for the reaction, determine whether Al or Fe2O3 is the limiting reactant, and convert to mass. 2 Al(s) + Fe2O3(s) → 2 Fe(s) + Al2O3(s) When the masses of both reactants are given, you must determine which reactant is limiting.  103 g   1 mol Al  2 mol Fe  = 37.064 mol Fe (unrounded) Mole Fe (from Al) = (1.00 kg Al )   1 kg   26.98 g Al  2 mol Al       2 mol Fe  Mole Fe (from Fe2O3) = ( 2.00 mol Fe2 O3 )   = 4.00 mol Fe  1 mol Fe 2 O3  Fe2O3 is limiting, so 4.00 moles of Fe forms.  55.85 g Fe  Mass = ( 4.00 mol Fe )   = 223.4 = 223 g Fe  1 mol Fe  Though not required by the problem, this could be converted to 0.223 kg. 4.100 In ionic compounds, iron has two common oxidation states, +2 and +3. First, write the balanced equations for the formation and decomposition of compound A. Then, determine which reactant is limiting and from the amount of the limiting reactant calculate how much compound B will form. Compound A is iron chloride with iron in the higher, +3, oxidation state. Thus, the formula for compound A is FeCl3 and the correct name is iron(III) chloride. The balanced equation for formation of FeCl3 is: 2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s) The FeCl3 decomposes to FeCl2 with iron in the +2 oxidation state. The balanced equation for the decomposition of FeCl3 is: 2 FeCl3(s) → 2 FeCl2(s) + Cl2(g) To find out whether 50.6 g Fe or 83.8 g Cl2 limits the amount of product, calculate the number of moles of iron(III) chloride that could form based on each reactant.  1 mol Fe   2 mol FeCl3  Mole FeCl3 (from Fe) = ( 50.6 g Fe )    = 0.905998 FeCl3 (unrounded)  55.85 g Fe   2 mol Fe   1 mol Cl2  2 mol FeCl3  Mole FeCl3 (from Cl2) = ( 83.8 g Cl2 )    = 0.787964 FeCl3 (unrounded)  70.90 g Cl2  3 mol Cl2  Since fewer moles of FeCl3 are produced from the available amount of chlorine, the chlorine is limiting and we calculate the amount of compound B from the amount of chlorine gas. 4-17  1 mol Cl2   2 mol FeCl3   2 mol FeCl2   126.75 g FeCl2  Mass FeCl2 = ( 83.8 g Cl2 )       70.90 g Cl2   3 mol Cl2   2 mol FeCl3   1 mol FeCl2  = 99.874 = 99.9 g FeCl2 4.101 The equilibrium state is dynamic, because the forward and reverse processes continue even after apparent change has ceased. 4.102 The system must be closed so the gaseous product cannot escape. 4.103 Acetic acid molecules are constantly and randomly colliding with each other and with water molecules. Occasionally, an H+ is transferred from an acetic acid molecule to a water molecule because of one of these collisions. At first, this process occurs at a larger rate than the reverse transfer of an H+ from a H3O+ to a CH3COO-. As the concentration of H+ and CH3COO- build up, the rate of transfer of H+ between them equals the rate of transfer of H+ between CH3COOH and H2O and equilibrium is reached when about 2% of the CH3COOH molecules have ionized. 4.104 On a molecular scale, chemical reactions are dynamic. If NO and Br2 are placed in a container, molecules of NO and molecules of Br2 will react to form NOBr. Some of the NOBr molecules will decompose and the resulting NO and Br2 molecules will recombine with different NO and Br2 molecules. In this sense, the reaction is dynamic because the original NO and Br2 pairings do not remain permanently attached to each other. Eventually, the rate of the forward reaction (combination of NO and Br2) will equal the rate of the reverse reaction (decomposition of NOBr) at which point the reaction is said to have reached equilibrium. If you could take “snapshot” pictures of the molecules at equilibrium, you would see a constant number of reactant (NO, Br2) and product molecules (NOBr), but the pairings would not stay the same. 4.105 The reaction is N2(g) + 3 H2(g) → 2 NH3(g). a) First, the limiting reactant must be found.  1 mol H 2  2 mol NH3  Mole NH3 (from H2) = (10.0 g H 2 )    = 3.306878 mol NH3 (unrounded)  2.016 g H 2  3 mol H 2   1 mol N 2  2 mol NH3  Mole NH3 (from N2) = ( 20.0 g N 2 )    = 1.42755 mol NH3 (unrounded)  28.02 g N 2  1 mol N 2  N2 is the limiting reactant.  1 mol N 2   2 mol NH3   17.03 g NH3  Mass NH3 = ( 20.0 g N 2 )   = 24.311 g NH3 (unrounded)    28.02 g N 2   1 mol N 2   1 mol NH3  % yield = (15.0 g actual)/( 24.311 g theoretical) x 100% = 61.700 = 61.7% b) Initially, 4.96 moles of H2 and 0.714 moles of N2 were present. 15.0 g (= 0.881 moles) NH3 was formed, requiring that 0.440 moles of N2 and 1.32 moles of H2 be consumed. This leaves 0.274 moles N2 and 3.64 moles H2 at equilibrium. 4.106 4.107 Fe(s) + 2 H+(aq) → Fe2+(aq) + H2(g) 0 +1 +2 0  49 mg Fe   10−3 g   1 mol Fe   1 mol Fe2 + b) Fe2+ ions = ( 737 g Sauce )         125 g Sauce   1 mg   55.85 g Fe   1 mol Fe = 3.11509 x 1021 = 3.1 x 1021 Fe2+ ions per jar of sauce a) O.N.: The reactions are: ∆ CaCO3(s) → CaO(s) + CO2(g) ∆ CaO(s) + SO2(g) → CaSO3(s) ∆ S(s) + O2(g) → SO2(g) These reactions sum to: CaCO3(s) + S(s) + O2(g) → CaSO3(s) + CO2(g) 4-18  6.022 x 1023 Fe 2 + ions     1 mol Fe2 +   Mass CaCO3 =  103 g   0.33%   1 mol S  1 mol CaCO3   100.09 g CaCO3   100%  kg Coal           1 kg   100%   32.07 g S  1 mol S  1 mol CaCO3   70%  = 1.25062 x 106 = 1.3 x 106 g CaCO3 needed Note: The percent ratio yields the mass of sulfur in the compound. (8.5 x 10 4.108 4 )  1 mol C6 H12 O6  2 mol C2 H5 OH   46.07 g C2 H5 OH  Mass of C2H5OH = (10.0 g C6 H12 O6 )      180.16 g C6 H12 O6  1 mol C6 H12 O6   1 mol C2 H5 OH  = 5.1143 = 5.11 g C2H5OH  1 mol C6 H12 O6   2 mol CO 2  22.4 L CO 2  Volume CO2 = (10.0 g C6 H12 O6 )      180.16 g C6 H12 O6   1 mol C6 H12 O6   1 mol CO 2  = 2.4866785 = 2.49 L CO2 4.109 4.110  10−3 L   0.03190 mol MnO −4   5 mol Fe2 +   55.85 g Fe  Mass Fe = ( 39.32 mL )       2+  −    L  1 mL     1 mol MnO 4   1 mol Fe  = 0.3502655 g Fe (unrounded) Mass % Fe = [(0.3502655 g Fe) / (1.1081 g Sample)] x 100% = 31.60956 = 31.61% Fe a) A solution of Na2C2O4 (soluble salt) is added to a solution containing 1.9348 g of road salt (NaCl and CaCl2, both soluble salts) and an insoluble precipitate is formed (CaC2O4). The Na+ and Cl– ions are spectators in this reaction because all Na+ salts are soluble and most Cl– salts are soluble. Ca2+(aq) + C2O42–(aq) → CaC2O4(s) b) You may recognize this reaction as a redox titration, because the permanganate ion, MnO4–, is a common oxidizing agent. The MnO4– oxidizes the oxalate ion, C2O42– to CO2. Mn changes from +7 to +2 (reduction) and C changes from +3 to +4 (oxidation). The equation that describes this process is: H2C2O4(aq) + MnO4–(aq) → Mn2+(aq) + CO2(g) H2C2O4 is a weak acid, so it cannot be written in a fully dissociated form (2 H+(aq) + C2O42–(aq)). KMnO4 is a soluble salt, so it can be written in its dissociated form. K+(aq) is omitted because it is a spectator ion. We will balance the equation using the oxidation number method (you can verify by doing the half reaction method) and first assigning oxidation numbers to all elements in the reaction. H2C2O4(aq) + MnO4–(aq) → Mn2+(aq) + 2 CO2(g) Oxidation numbers: +1 +3 -2 +7 -2 +2 +4 -2 Identify the oxidized and reduced species and multiply one or both species by the appropriate factors to make the electrons lost equal the electrons gained. 5 H2C2O4(aq) + 2 MnO4–(aq) → 2 Mn2+(aq) + 10 CO2(g) + Adding water and H (aq) to finish balancing the equation is appropriate since the reaction takes place in acidic medium. Add 8 H2O(l) to right side of equation to balance the oxygen and then add 6 H+(aq) to the left to balance hydrogen. 5 H2C2O4(aq) + 2 MnO4–(aq) + 6 H+(aq) → 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l) c) Oxidizing agent = KMnO4 d) Reducing agent = H2C2O4 (remember that the agent refers to the whole compound, not a particular element within a compound) e) The balanced equations provide the accurate molar ratios between species. Working backwards, we know that (Moles of MnO4–) is related to moles of C2O42– at the endpoint (5:2) (Moles of C2O42–) is related to moles of CaC2O4 (1:1) (Moles of CaC2O4) is related to moles of CaCl2 (1:1)  10−3 L   0.1019 mol KMnO4   5 mol H 2 C2 O4   1 mol CaCl2   110.98 g CaCl2  Mass CaCl2 = ( 37.68 mL )       1 mL   1L   2 mol KMnO4   1 mol H 2 C2 O4   1 mol CaCl2    = 1.06529 g CaCl2 (unrounded) Mass % CaCl2 = [(1.06529 g CaCl2) / (1.9348 g Sample)] x 100% = 55.0594 = 55.06% CaCl2 4-19 4.111 Write the balanced chemical equations: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) Determine the NaOH concentration from the reaction of NaOH with H2SO4.  10−3 L   0.0782 mol H 2SO 4   2 mol NaOH    1 mL  1 M NaOH = ( 50.0 mL )    −3        1 mL   L   1 mol H 2SO 4   18.4 mL   10 L    = 0.425 M NaOH Use the NaOH concentration and the reaction of HCl with NaOH to determine HCl concentration.  10−3 L   0.425 mol NaOH   1 mol HCl     1 mL  1 M HCl = ( 27.5 mL )        1 mL   L   1 mol NaOH   100.mL   10−3 L    = 0.116875 = 0.117 M HCl 4.112 a) b) c) Molecular: H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) Total ionic: 2 H+(aq) + SO42–(aq) + 2 Na+(aq) + 2 OH–(aq) → 2 Na+(aq) + SO42–(aq) + 2 H2O(l) Net ionic: H+(aq) + OH–(aq) → H2O(l) 0.020 mol of H2SO4 and 0.040 mol of NaOH reacted. M H2SO4 = (0.020 mol H2SO4/25mL) (1 mL / 10–3 L) = 0.80 M H2SO4 M NaOH = (0.040 mol NaOH/25 mL) (1 mL / 10–3 L) = 1.6 M NaOH 4.113 The balanced equation for this reaction is: CaMg(CO3)2(s) + 4 HCl(aq) → Ca2+(aq) + Mg2+(aq) + 2 H2O(l) + 2 CO2(g) + 4 Cl–(aq) To find mass %, convert the solution to moles of HCl, use the molar ratio to find moles of dolomite, convert to mass and divide by the mass of soil.  10−3 L   0.2516 mol HCl  1 mol CaMg(CO3 )2   184.41 g CaMg(CO3 )2  Mass CaMg(CO3)2 = ( 33.56 mL )      1 mL   1L 4 mol HCl    1 mol CaMg(CO3 )2    = 0.389275 g CaMg(CO3)2 (unrounded) Mass % CaMg(CO3)2 = [(0.389275 g CaMg(CO3)2) / (12.86 s Soil)] x 100% = 3.02702 = 3.027% CaMg(CO3)2 4.114 The balanced chemical equations for HA or HB with sodium hydroxide are the same. For HA it is: HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) The concentration of first HA solution is:  10−3 L   0.0906 mol NaOH   1 mol HA     1 mL  1 M HA = ( 87.3 mL )        1 mL   L   1 mol NaOH   43.5 mL   10−3 L    = 0.1818248 = 0.182 M HA (To finish the problem, it must be assumed that both HA solutions have the same concentration. This is not necessarily true based on the information given in the problem.) The titration of the acid mixture involves the reaction of NaOH with both of the acids. The amount reacting with HA may be related to the first titration as: Volume NaOH reacting with second HA = (87.3 mL NaOH) (37.2 mL HA / 43.5 mL HA) = 74.65655 mL (unrounded) Thus, (96.4 -74.65655) mL = 21.74345 mL NaOH reacted with HB. Calculating the molarity of HB is now similar to calculating the molarity of HA:  10−3 L   0.0906 mol NaOH   1 mol HB     1 mL  1 M HB = ( 21.74345 mL )        1 mL   L   1 mol NaOH   (50.0 − 37.2) mL   10−3 L    = 0.153902857 = 0.154 M HB 4-20 4.115 a) Assigning oxidation states to each element indicates which are the oxidizing and reducing agents. Step 1: N oxidized from -3 to +2 by O2, and O is reduced from 0 to -2. Oxidizing agent = O2 Reducing agent = NH3 Step 2: N oxidized from +2 to +4 by O2, and O is reduced from 0 to -2. Oxidizing agent = O2 Reducing agent = NO Step 3: N oxidized from +4 to +5 by NO2, and N is reduced from +4 to +2. Oxidizing agent = NO2 Reducing agent = NO2 b) This problem could be done as a stepwise set of calculations, however, if the equations are combined into an overall reaction the answer may be found in one calculation: 12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(l) + 4 NO(g)  103 g   1 mol HNO3  12 mol NH3  17.03 g NH3   1 kg  Mass NH3 = 3.0 x 104 kg HNO3   1 kg   63.02 g HNO  8 mol HNO  1 mol NH   103 g  3  3  3     4 4 = 1.21604 x 10 = 1.2 x 10 kg NH3 This assumes that the NO produced in step 3 is not recycled. ( ) 4.116 a) MnS(s) + 2 HBr(aq) → MnBr2(aq) + H2S(g) MnS(s) + 2 H+(aq) → Mn2+(aq) + H2S(g) b) K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2 KNO3(aq) CO32–(aq) + Sr2+(aq) → SrCO3(s) c) KNO2(aq) + HCl(aq) → HNO2(aq) + KCl(aq) NO2–(aq) + H+(aq) → HNO2(aq) d) Ca(OH)2(aq) + HNO3(aq) → Ca(NO3)2(aq) + 2 H2O(l) OH–(aq) + H+(aq) → H2O(l) e) Ba(C2H3O2)2(aq) + FeSO4(aq) → BaSO4(s) + Fe(C2H3O2)2(aq) Ba2+(aq) + SO42–(aq) → BaSO4(s) f) ZnCO3(s) + H2SO4(aq) → ZnSO4(aq) + H2O(l) + CO2(g) ZnCO3(s) + 2 H+(aq) → Zn2+(aq) + H2O(l) + CO2(g) g) Cu(NO3)2(aq) + H2S(aq) → CuS(s) + 2 HNO3(aq) Cu2+(aq) + H2S(aq) → CuS(s) + 2 H+(aq) h) Mg(OH)2(s) + 2 HClO3(aq) → Mg(ClO3)2(aq) + 2 H2O(l) Mg(OH)2(s) + 2 H+(aq) → Mg2+(aq) + 2 H2O(l) i) KCl(aq) + (NH4)3PO4(aq) → No Reaction j) Ba(OH)2(aq) + 2 HCN(aq) → Ba(CN)2(aq) + 2 H2O(l) OH–(aq) + HCN(aq) → CN–(aq) + H2O(l) 4.117 a) 4 KOH(aq) + 3 H2O2(aq) + 2 Cr(OH)3(s) → 2 K2CrO4(aq) + 8 H2O(l) H2O2 is the oxidizing agent (O.N.(O) goes from -1 to -2). Cr(OH)3 is the reducing agent (O.N.(Cr) goes from +3 to +6). b) 4 MnO4–(aq) + 3 ClO2–(aq) + 2 H2O(l) → 4 MnO2(s) + 3 ClO4–(aq) + 4 OH–(aq) MnO4– is the oxidizing agent (O.N.(Mn) goes from +7 to +4). ClO2– is the reducing agent (O.N.(Cl) goes from +3 to +7). c) 2 KMnO4(aq) + 3 Na2SO3(aq) + H2O(l) → 2 MnO2(s) + 3 Na2SO4(aq) + 2 KOH(aq) KMnO4 is the oxidizing agent (O.N.(Mn) goes from +7 to +4). Na2SO3 is the reducing agent (O.N.(S) goes from +4 to +6). d) 2 CrO42–(aq) + 3 HSnO2–(aq) + H2O(l) → 2 CrO2–(aq) + 3 HSnO3–(aq) + 2 OH–(aq) CrO42– is the oxidizing agent (O.N.(Cr) goes from +6 to +3). HSnO2– is the reducing agent (O.N.(Sn) goes from +2 to +4). e) 2 KMnO4(aq) + 3 NaNO2(aq) + H2O(l) → 2 MnO2(s) + 3 NaNO3(aq) + 2 KOH(aq) KMnO4 is the oxidizing agent (O.N.(Mn) goes from +7 to +4). NaNO2 is the reducing agent (O.N.(N) goes from +3 to +5). 4-21 f) 4 I– (aq) + O2(g) + 2 H2O(l) → 2 I2(s) + 4 OH–(aq) O2 is the oxidizing agent (O.N.(O) goes from 0 to -2). I– is the reducing agent (O.N.(I) goes from -1 to 0). 4.118 The reactions are: 1) Cl(g) + O3(g) → ClO(g) + O2(g) 2) ClO(g) + ClO(g) → Cl2O2(g) 3) Cl2O2(g) →  2 Cl(g) + O2(g) 4) ClO(g) + O(g) → Cl(g) + O2(g) a) In reaction 1, Cl is oxidized (0 to +1), and O is reduced (0 to -1). In reaction 2, no oxidation or reduction occurs. In reaction 3, Cl is reduced (+1 to 0) and O is oxidized (-1 to 0). In reaction 4, Cl is reduced (+1 to 0) and O (from ClO) is oxidized (-1 to 0). The oxidation-reduction reactions are 1, 3, and 4. b) Combining the equations requires multiplying equation 1 by 2, and then adding the three equations to give: 2 O3(g) → 3 O2(g) light 4.119 4.120 4.121 4.122 The reaction is: 2 Na2O2(s) + 2 CO2(g) → 2 Na2CO3(s) + O2(g).  1 mol Na 2 O 2   2 mol CO 2  44.01 g CO 2    L Air Volume = ( 80.0 g Na 2 O 2 )       77.98 g Na 2 O 2   2 mol Na 2 O 2  1 mol CO 2   0.0720 g CO 2  = 627.08 = 627 L Air 3  103 m   1 cm 3  1 mL Mass Mg = 1.00 km3    −2   3   1 km   10 m   1 cm 9 9 = 1.3520 x 10 = 1.4 x 10 kg Mg ( )   1.04 g   0.13% Mg   1 kg       3    1 mL   100%   10 g  A 1.00 kg piece of glass of composition 75% SiO2, 15% Na2O, and 10% CaO would contain (1.00 kg) (75% / 100%) = 0.75 kg SiO2 (1.00 kg) (15% / 100%) = 0.15 kg Na2O (1.00 kg) (10% / 100%) = 0.10 kg CaO In this example, the SiO2 is added directly while the sodium oxide comes from decomposition of sodium carbonate and the calcium oxide from decomposition of calcium carbonate: Na2CO3(s) → Na2O(s) + CO2(g) CaCO3(s) → CaO(s) + CO2(g)  103 g   1 mol Na 2 O   1 mol Na 2 CO3   105.99 g Na 2 CO3   1 kg  Mass Na2CO3 = ( 0.15 kg Na 2 O )   1 kg   61.98 g Na O   1 mol Na O   1 mol Na CO   103 g  2  2 2 3      = 0.25651 = 0.26 kg Na2CO3  103 g   1 mol CaO  1 mol CaCO3   100.09 g CaCO3   1 kg  Mass CaCO3 = ( 0.10 kg CaO )   1 kg   56.08 g CaO  1 mol CaO   1 mol CaCO   103 g    3     = 0.178477 = 0.18 kg CaCO3 Combine 0.75 kg SiO2, 0.26 kg Na2CO3, and 0.18 kg CaCO3 and heat to make 1.00 kg glass.  103 g   1 mol C2 H 4  a) Moles C2H4 = ( 0.100 kg C2 H 4 )  = 3.56506 mol C2H4 (unrounded)  1 kg   28.05 g C H  2 4     103 g   1 mol HCl  = 2.74273 mol HCl (unrounded) Moles HCl = ( 0.100 kg HCl )   1 kg   36.46 g HCl     The HCl is limiting. 4-22 Moles HCl remaining = 0 mol Moles C2H4 remaining = 3.56506 mol C2H4 - (2.74273 mol HCl) (1mol C2H4 / 1 mol HCl) = 0.822328 = 0.82 mol C2H4 Moles C2H5Cl formed = (2.74273 mol HCl) (1 mol C2H5Cl / 1 mol HCl) = 2.74273 = 2.74 mol C2H5Cl Molecules of gas = Total mol x Avogadro’s Number = [(0 + 0.822328 + 2.74273) mol](6.022 x 1023 molecules / mol) = 2.1468779 x 1024 = 2.15 x 1024 molecules b) This will still be based on the HCl. Moles HCl = (2.74273 mol HCl) / 2 = 1.371365 mol HCl (unrounded) Moles C2H4 = 3.56506 mol C2H4 - [(1.371365 mol HCl) (1 mol C2H4 / 1 mol HCl)] = 2.193695 mol C2H4 (unrounded) Moles C2H5Cl = (1.371365 mol HCl) (1 mol C2H5Cl / 1 mol HCl) = 1.371365 mol C2H5Cl (unrounded) Total moles = (1.371365 + 2.193695 + 1.371365) mol = 4.936425 = 4.94 mol total 4.123 Let x = grams of NaCl, then 35.0 - x = g of MgSO4. Moles NaCl = x grams (1 mol NaCl / 58.44 g NaCl) = x / 58.44 Moles MgSO4 = (35 - x) grams (1 mol MgSO4 / 120.38 g MgSO4) = (35 - x) / 120.38 Given that moles of NaCl is twice as large as moles of MgSO4 x / 58.44 = 2 (35 - x) / 120.38 x / 58.44 = (70 - 2 x) / 120.38 120.38 x = 58.44 (70 - 2 x) = 4090.8 - 116.88 x (120.38 + 116.88) x = 237.26 x = 4090.8 x = 4090.8 / 237.26 = 17.2418 = 17.2 g NaCl 35.0 - x = 35.0 - 17.2418 = 17.7582 = 17.8 g MgSO4 4.124 a) There is not enough information to write complete chemical equations, but the following equations can be written: C5H11I4NO4(s) + Na2CO3(s) → 4 I–(aq) + other products I–(aq) + Br2(l) + HCl(aq) → IO3–(aq) + other products For every mole of thyroxine reacted, 4 moles IO3– are produced. b) IO3–(aq) + H+(aq) + I–(aq) → I2(aq) + H2O(l) Oxidation numbers: +5 -2 +1 -1 0 +1 -2 This is a difficult equation to balance because the iodine species are both reducing and oxidizing. Start balancing the equation by placing a coefficient of 5 in front of I–(aq), so the electrons lost equal the electrons gained. Do not place a 5 in front of I2(aq), because not all of the I2(aq) comes from oxidation of I–(aq). Some of the I2(aq) comes from the reduction of IO3–(aq). Place a coefficient of 3 in front of I2(aq) to correctly balance iodine. The reaction is now balanced from a redox standpoint, so finish balancing the reaction by balancing the oxygen and hydrogen. IO3–(aq) + 6 H+(aq) + 5 I–(aq) → 3 I2(aq) + 3 H2O(l) IO3– is the oxidizing agent, and I– is the reducing agent. If 3 moles of I2 are produced per mole of IO3–, and 4 moles of IO3– are produced per mole of thyroxine, then 12 moles of I2 are produced per mole of thyroxine. c) Using Thy to represent thyroxine The balanced equation for this reaction is I2(aq) + 2 S2O32–(aq) → 2 I–(aq) + S4O62–(aq).  10−3 L   0.1000 mol S2 O32 −  1 mol I2   1 mol Thy  776.8 g Thy  Mass Thy = (17.23 mL )       2−     L  1 mL   2 mol S2 O3   12 mol I2   1 mol Thy  = 0.055767766 g Thy (unrounded) Mass % Thy = [(0.055767766 g Thy) / (0.4332 g extract)] x 100% = 12.8734 = 12.87% thyroxine 4.125 The reactions are 2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l) Mg(OH)2(s) + CO2(g) → MgCO3(s) + H2O(l) 2 Al(OH)3(s) + 3 CO2(g) → Al2(CO3)3(s) + 3 H2O(l) 4-23  103 g   1 mol LiOH   1 mol CO 2   44.01 g CO 2  a) Mass CO2 = ( 3.50 kg LiOH )   1 kg   23.95 g LiOH   2 mol LiOH   1 mol CO    2    3 = 3215.762 = 3.22 x 10 g CO2 b) Mass CO2 absorbed by 1.00 g LiOH  1 mol LiOH  1 mol CO 2   44.01 g CO 2  Mass = (1.00 g LiOH )      23.95 g LiOH  2 mol LiOH   1 mol CO 2  = 0.918789 = 0.919 g CO2 Mass CO2 absorbed by 1.00 g Mg(OH)2  1 mol Mg(OH) 2   1 mol CO 2  44.01 g CO 2  Mass = (1.00 g Mg(OH) 2 )      58.33 g Mg(OH) 2   1 mol Mg(OH) 2  1 mol CO 2  = 0.75450 = 0.754 g CO2 Mass CO2 absorbed by 1.00 g Al(OH)3  1 mol Al(OH)3   3 mol CO 2   44.01 g CO 2  Mass = (1.00 g Al(OH)3 )      78.00 g Al(OH)3   2 mol Al(OH)3   1 mol CO 2  = 0.846346 = 0.846 g CO2 4.126 Balance the equation to obtain the correct molar ratios. This is not a redox reaction as none of the O.N.’s change. Here is a suggested method for approaching balancing the equation. — Since PO42– remains as a unit on both sides of the equation, treat it as a unit when balancing. — On first inspection, one can see that Na needs to be balanced by adding a “2” in front of NaHCO3. This then affects the balance of C, so add a “2” in front of CO2. — Hydrogen is not balanced, so change the coefficient of water to “2,” as this will have the least impact on the other species. — Verify that the other species are balanced. ∆ Ca(H2PO4)2(s) + 2 NaHCO3(s) → 2 CO2(g) + 2 H2O(g) + CaHPO4(s) + Na2HPO4(s) a) Determine whether Ca(H2PO4)2 or NaHCO3 limits the production of CO2. In each case calculate the moles of CO2 that might form.  31%   1 mol NaHCO3   2 mol CO 2  Mole CO2 (NaHCO3) = (1.00 g )      100%   84.01 g NaHCO3   2 mol NaHCO3  = 3.690 x 10–3 mol CO2 (unrounded)  2 mol CO 2  35%   1 mol Ca(H 2 PO 4 ) 2  Mole CO2 (Ca(H2PO4)2) = (1.00 g )       100%   234.05 g Ca(H 2 PO 4 ) 2  1 mol Ca(H 2 PO 4 ) 2  = 2.9908 x 10–3 mol CO2 (unrounded) Since Ca(H2PO4)2 is limiting, 3.0 x 10–3 mol CO2 will be produced. b) Volume CO2 = (2.9908 x 10–3 mol CO2) (37.0 L/mol CO2) = 0.1106596 = 0.11 L CO2 4.127 The chemical equation is: HNO3(g) + NaOH(aq) → NaNO3(aq) + H2O(l) a) It takes a total of (20.00 + 3.22) mL = 23.22 mL NaOH to titrate a total of (50.00 + 30.00) mL = 80.00 mL of acid.  10−3 L   0.0502 mol NaOH   1 mol HNO3     1 mL  1 M HCl = ( 23.22 mL )        1 mL   L   1 mol NaOH   80.00 mL   10−3 L    = 0.01457055 = 0.0146 M HCl b) First calculate the moles of the acid and base present. The difference will give the excess NaOH.  10−3 L   0.0502 mol NaOH  Moles NaOH = ( 20.00 mL )  = 1.004 x 10–3 mol NaOH (unrounded)   1 mL   L    4-24  10−3 L   0.01457055 mol HCl  Moles HCl = ( 50.00 mL )    1 mL   L    –4 = 7.285275 x 10 mol HCl (unrounded) Moles excess NaOH = 1.004x 10–3 mol NaOH - (7.285275 x 10–4 mol HCl) (1 mol NaOH / 1 mol HCl) = 2.754725 x 10–4 = 2.8 x 10–4 mol NaOH 4.128 a) Determine the moles of each element present. The sample was burned in an unknown amount of O2, therefore, the moles of oxygen must be found by a different method.  1 mol CO 2   1 mol C  –3 Moles C = ( 0.1880 g CO 2 )    = 4.271756 x 10 mol C (unrounded) 44.01 g CO 1 mol CO 2  2    1 mol H 2 O   2 mol H  –3 Moles H = ( 0.02750 g H 2 O )    = 3.052164 x 10 mol H (unrounded)  18.02 g H 2 O   1 mol H 2 O   1 mol Bi 2 O3  2 mol Bi  –4 Moles Bi = ( 0.1422 g Bi 2 O3 )    = 6.103004 x 10 mol Bi (unrounded) 466.0 g Bi O 1 mol Bi O 2 3 2 3    Subtracting the mass of each element present from the mass of the sample will give the mass of oxygen originally present in the sample. This mass is used to find the moles of oxygen. 0.22105 g sample - [12.01 g/mol (4.271756 x 10–3 mol C)] - [1.008 g/mol (3.052164 x 10–3 mol H)] - [(209.0 g/mol (6.103004 x 10–4 mol Bi)] = 0.0391168 g O (unrounded) Moles O = (0.0391168 g O) (1 mol O / 16.00 g O) = 2.44480 x 10–3 mol O (unrounded) Divide each of the moles by the smallest value (moles Bi). C = (4.271756 x 10–3 mol) / (6.103004 x 10–4 mol) = 7 H = (3.052164 x 10–3 mol) / (6.103004 x 10–4 mol) = 5 O = (2.44480 x 10–3 mol) / (6.103004 x 10–4 mol) = 4 Bi = (6.103004 x 10–4 mol) / (6.103004 x 10–4 mol) = 1 Empirical formula = C7H5O4Bi b) The empirical formula mass is 362 g/mol. Therefore, there are 1086 / 362 = 3 empirical formula units per molecular formula making the molecular formula = 3 x C7H5O4Bi = C21H15O12Bi3. c) Bi(OH)3(s) + 3 HC7H5O3(aq) → Bi(C7H5O3)3(s) + 3 H2O(l)  10−3 g   1 mol Active   3 mol Bi   1 mol Bi(OH)3   260.0 g Bi(OH)3   1 mg   100%  d) ( 0.600 mg )      −3        1086 g   1 mol Active   1 mol Bi   1 mol Bi(OH)3   10 g   88.0%   1 mg   = 0.48970 = 0.490 mg Bi(OH)3 4.129 a) Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq) 2 Na+(aq) + CO32–(aq) + Ca2+(aq) + 2 Cl–(aq) → CaCO3(s) + 2 Na+(aq) + 2 Cl–(aq) CO32–(aq) + Ca2+(aq) → CaCO3(s) b) Mass = 2 (0.050 mol ion) (1 mol CaCO3 / 1 mol ion) (100.09 g CaCO3 / 1 mol CaCO3) = 10.009 = 10. g CaCO3 CO32– = 3 (0.050) mol c) Original moles: Na+ = 6 (0.050) mol 2+ Cl– = 4 (0.050) mol Ca = 2 (0.050) mol + – The moles of Na and Cl do not change. The moles of Ca2+ goes to zero, and removes 2 (0.050) mol of CO32–. M Na+ = ( ) 6 0.050 mol Na +  1 mL  +  −3  = 0.60 M Na 250. 250. mL + ( )  10 L  ( ) 4 0.050 mol Cl−  1 mL  M Cl = = 0.40 M Cl– ( 250. + 250.) mL  10−3 L  – M CO32– = ( 3 − 2 ) ( 0.050 mol CO32 − )  1 mL    ( 250. + 250.) mL  10−3 L  = 0.10 M CO32– 4-25 4.130 a) Complete combustion of hydrocarbons involves heating the hydrocarbon in the presence of oxygen to produce carbon dioxide and water. Ethanol: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) Gasoline: 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) b) The amounts of each fuel must be found: Gasoline = (1.00 L) (90.0% / 100%) (1 mL / 10–3 L) (0.742 g /mL) = 667.8 g gasoline (unrounded) Ethanol = (1.00 L) (10.0% / 100%) (1 mL / 10–3 L) (0.789 g /mL) = 78.9 g ethanol  1 mol C8 H18   25 mol O 2   32.00 g O 2  Mass O2 (gasoline) = ( 667.8 g C8 H18 )      114.22 g C8 H18   2 mol C8 H18   1 mol O 2  = 2338.64 g O2 (unrounded)  1 mol C 2 H5 OH   3 mol O 2   32.00 g O 2  Mass O2 (ethanol) = ( 78.9 g C2 H5 OH )      46.07 g C2 H5 OH   1 mol C 2 H5 OH   1 mol O 2  = 164.41 g O2 (unrounded) Total O2 = 2338.64 g O2 + 164.41 g O2 = 2503.05 = 2.50 x 103 g O2 c) (2503.05 g O2) (1 mol O2 / 32.00 g) (22.4 L / mol O2 = 1752.135 = 1.75 x 103 L d) (1752.135 L O2) (100% / 20.9%) = 8383.42 = 8.38 x 103 L air 4.131 1) 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) 2) 2 C8H18(l) + 17 O2(g) → 16 CO(g) + 18 H2O(g) a) Assume 100 grams of gasoline, thus 95.0 grams will react by equation 1, and 5.0 grams will react by equation 2.  1 mol C8 H18   16 mol CO 2   6.022 x 1023 CO 2  Molecules CO2 = ( 95.0 g C8 H18 )      1 mol CO 2  114.22 g C8 H18   2 mol C8 H18    = 4.00693 x 1024 molecules CO2 (unrounded)  1 mol C8 H18   16 mol CO   6.022 x 1023 CO  Molecules CO = ( 5.0 g C8 H18 )      1 mol CO  114.22 g C8 H18   2 mol C8 H18    = 2.10891 x 1023 molecules CO (unrounded) Ratio = (4.00693 x 1024 molecules CO2) / (2.10891 x 1023 molecules CO) = 18.99998 = 19 b) Again, we may assume 100 grams of gasoline.  1 mol C8 H18  16 mol CO 2   44.01 g CO 2  Mass CO2 = ( 95.0 g C8 H18 )      114.22 g C8 H18  2 mol C8 H18   1 mol CO 2  = 292.83 g CO2 (unrounded)  1 mol C8 H18   16 mol CO   28.01 g CO  Mass CO = ( 5.0 g C8 H18 )      114.22 g C8 H18   2 mol C8 H18   1 mol CO  = 9.8091 g CO (unrounded) Ratio = (292.83 g CO2) / (9.8091 g CO) = 29.852889 = 30 (x)(44.01) =1 c) For the mass ratio to be equal y(28.01) Where x and y are % so, x + y = 1 or y = 1 - x (x)(44.01) Substituting =1 (1 − x)(28.01) 44.01x = 28.01 - 28.01x 72.02x = 28.01 x = 0.39 and y = 1 - 0.39 = 0.61 Thus, 61% of the gasoline must form CO. 4-26 4.132 4.133 Combine the two equations to get an overall equation (this will require multiplying the second equation by 2). The overall equation is: C6H8O6 + Br2 + 2 NaOH → C6H6O6 + 2 NaBr + 2 H2O This leads to:  10−3 L   0.1350 mol NaOH   1 mol C6 H8 O6   176.12 g C6 H8 O6   1 mg  Mass vitamin C = ( 43.20 mL )    2 mol NaOH   1 mol C H O   −3   1 mL   L   6 8 6   10 g    = 513.5659 = 513.6 mg C6H8O6 Yes, the tablets have the quantity advertised.  1 mol NaCl   2 mol ions   1 mol MgCl2   95.21 g MgCl2  a) Mass MgCl2 = (12.4 g NaCl )       58.44 g NaCl   1 mol NaCl   3 mol ions   1 mol MgCl2  = 13.46799 = 13.5 g MgCl2  1 mol NaCl  2 mol ions   1 mol CaS   72.15 g CaS  b) Mass CaS = (12.4 g NaCl )       58.44 g NaCl  1 mol NaCl   2 mol ions   1 mol CaS  = 15.3090 = 15.3 g CaS c) The CaS solution dissolves the most protein. All three solutions have equal ion concentrations, but CaS will form two divalent ions. 4.134 a) The second reaction is a redox process because the O.N. of iron changes from 0 to +2 (it oxidizes) while the O.N. of hydrogen changes from +1 to 0 (it reduces). b) Determine the moles of HCl present and use the balanced chemical equation to determine the appropriate quantities.  3.00 mol HCl   1 mol Fe 2 O3   159.70 g Fe 2 O3  Mass Fe2O3 = 2.50 x 103 L    6 mol HCl   1 mol Fe O  L   2 3   ( ) = 199625 = 2.00 x 105 g Fe2O3  3.00 mol HCl   2 mol FeCl3   162.20 g FeCl3  Mass FeCl3 = 2.50 x 103 L    6 mol HCl   1 mol FeCl  L   3   = 405500 = 4.06 x 105 g FeCl3 c) Use reaction 2 like reaction 1 was used in part b.  3.00 mol HCl   1 mol Fe   55.85 g Fe  Mass Fe = 2.50 x103 L    2 mol HCl   1 mol Fe  L     ( ( ) ) = 209437.5 = 2.09 x 105 g Fe  3.00 mol HCl   1 mol FeCl2   126.75 g FeCl2  Mass FeCl2 = 2.50 x 103 L    2 mol HCl   1 mol FeCl  L   2   5 = 475312.5 = 4.75 x 10 g FeCl2 d) Use 1.00 g Fe2O3 to determine the mass of FeCl3 formed (reaction 1), and 0.280 g Fe to determine the mass of FeCl2 formed (reaction 2).  1 mol Fe 2 O3   2 mol FeCl3   162.20 g FeCl3  Mass FeCl3 = (1.00 g Fe 2 O3 )      159.70 g Fe 2 O3   1 mol Fe 2 O3   1 mol FeCl3  = 2.0313 g FeCl3 (unrounded)  1 mol Fe  1 mol FeCl2   126.75 g FeCl 2  Mass FeCl2 = ( 0.280 g Fe )      55.85 g Fe  1 mol Fe   1 mol FeCl2  = 0.635452 g FeCl2 (unrounded) Ratio = (0.635452 g FeCl2) / (2.0313 g FeCl3) = 0.312830 = 0.313 ( ) 4-27 a) 3 NH4ClO4(s) + 3 Al(s)  → Al2O3(s) + AlCl3(s) + 6 H2O(g) + 3 NO(g) catalyst Oxidizing agent = ammonium perchlorate reducing agent = aluminum  103 g   1 mol NH 4 ClO 4   9 mol Gas  b) Moles of gas = ( 50.0 kg NH 4 ClO 4 )   1 kg   117.49 g NH ClO   3 mol NH ClO  4 4  4 4    3 = 1276.70 = 1.28 x 10 mol gas c) Initial volumes:  103 g   1 cc   1 mL   10−3 L  = 25.6541 L (unrounded) V NH4ClO4 = ( 50.0 kg )   1 kg   1.95 g   1 cc   1 mL       V Al =  103 g   1 mol NH 4 ClO 4    26.98g Al   1 cc   1 mL   10−3 L  3 mol Al ( 50.0 kg NH 4 ClO4 )            1 kg   117.49 g NH 4 ClO4   3 mol NH 4 ClO 4   1 mol Al   2.70 g Al   1 cc   1 mL  = 4.2525 L (unrounded) Initial volume = 25.6541 L + 4.2525 L = 29.9066 L (unrounded) Final volume: Volume Al2O3 =  103 g   1 mol NH 4 ClO4   1 mol Al2 O3   101.96 g Al2 O3    1 mL   10−3 L  1 cc ( 50.0 kg NH 4 ClO4 )            1 kg   117.49 g NH 4 ClO 4   3 mol NH 4 ClO 4   1 mol Al2 O3  3.97 g Al2 O3   1 cc   1 mL  = 3.6432 L (unrounded) Volume AlCl3 =  103 g   1 mol NH 4 ClO4  1 mol AlCl3   133.33 g AlCl3    1 mL   10−3 L  1 cc ( 50.0 kg NH 4ClO4 )            1 kg   117.49 g NH 4 ClO4  3 mol NH 4 ClO4   1 mol AlCl3  2.44 g AlCl3   1 cc   1 mL  = 7.7515 L (unrounded) V gas = (1276.70 mol gas) (22.4 L / mol gas) = 28598.08 L (unrounded) Final volume = 3.6432 L + 7.7515 L + 28598.08 L = 28609.4747 L (unrounded) Volume change = (28609.4747 L) - (29.9066 L) = 28579.5681 = 2.86 x 104 L The volumes of all solids (before and after) are insignificant. 4.135 4.136 1) Cu(s) + Br2(aq) → CuBr2(aq) Cu(s) + Br2(aq) → Cu2+(aq) + 2 Br–(aq) 2) CuBr2(aq) + 2 NaOH(aq) → Cu(OH)2(s) + 2 NaBr(aq) Cu2+(aq) + 2 OH–(aq) → Cu(OH)2(s) 3) Cu(OH)2(s) → CuO(s) + H2O(l) Cu(OH)2(s) → CuO(s) + H2O(l) 4) CuO(s) + 2 HNO3(aq) → Cu(NO3)2(aq) + H2O(l) CuO(s) + 2 H+(aq) → Cu2+(aq) + H2O(l) 5) 3 Cu(NO3)2(aq) + 2 Na3PO4(aq) → Cu3(PO4)2(s) + 6 NaNO3(aq) 3 Cu2+(aq) + 2 PO43–(aq) → Cu3(PO4)2(s) 6) Cu3(PO4)2(s) + 3 H2SO4(aq) → 3 CuSO4(aq) + 2 H3PO4(aq) Cu3(PO4)2(s) + 6 H+(aq) → 3 Cu2+(aq) + 2 H3PO4(aq) 7) CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) 4-28