8 Answer Key Chapter 4
Solution Videos
| 4.1 Writing and Balancing Chemical Equations | ||
| 1. | An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter. | |
| 3. | (a) | PCl5 (s) + H2O (l) → POCl3 (l) + 2 HCl (aq) |
| (b) | 3 Cu (s) + 8 HNO3 (aq) → 3 Cu(NO3)2 (aq) + 4 H2O (l) + 2 NO (g) | |
| (c) | H2 (g)+I2 (s) → 2 HI (s) | |
| (d) | 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) | |
| (e) | 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) | |
| (f) | (NH4)2Cr2O7 (s) → Cr2O3 (s)+N2 (g) + 4 H2O (g) | |
| (g) | P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) | |
| (h) | PtCl4 (s) → Pt (s) + 2 Cl2 (g) | |
| 5. | (a) | CaCO3 (s) → CaO (s) + CO2 (g) |
| (b) | 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g) | |
| (c) | MgCl2 (aq) + 2 NaOH (aq) → Mg(OH)2 (s) + 2 NaCl (aq) | |
| (d) | 2 H2O (g) + 2 Na (s) → 2 NaOH (s) + H2 (g) | |
| 7. | (a) | Ba(NO3)2, KClO3 |
| (b) | 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) | |
| (c) | 2 Ba(NO3)2 (s) → 2 BaO (s) + 2 N2 (g) + 5 O2 (g) | |
| (d) | 2 Mg (s) + O2 (g) → 2 MgO (s) | |
| 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s) | ||
| 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) | ||
| 9. | (a) | 4 HF (aq) + SiO2 (s) → SiF4 (g) + 2 H2O (l) |
| (b) | Complete ionic equation: | |
| 2 Na+ (aq) + 2 F– (aq) + 2 Ca2+ (aq) + 2 Cl– (aq) → CaF2 (s) + 2 Na+ (aq) + 2 Cl– (aq) | ||
| Net ionic equation: | ||
| 2 F– (aq) + Ca2+ (aq) → CaF2 (s) | ||
| 11. | (a) | Complete: |
| 2 K+ (aq) + C2O42- (aq) + Ba2+ (aq) + 2 OH– (aq) → 2 K+ (aq) + 2 OH– (aq) + BaC2O4 (s) | ||
| Net: | ||
| Ba2+ (aq) + C2O42- (aq) → BaC2O4 (s) | ||
| (b) | Complete: | |
| Pb2+ (aq) + 2 NO3– (aq) + 2 H+ (aq) + SO42- (aq) → PbSO4 (s) + 2 H+ (aq) + 2 NO3– (aq) | ||
| Net: | ||
| Pb2+ (aq) + SO42- (aq) → PbSO4 (s) | ||
| (c) | Complete: | |
| CaCO3 (s) + 2 H+ (aq) + SO42- (aq) → CaSO4 (aq) + CO2 (g) + H2O (l) | ||
| Net: | ||
| CaCO3 (s) + 2 H+ (aq) + SO42- (aq) → CaSO4 (aq) + CO2 (g) + H2O (l) | ||
| 4.2 Classifying Chemical Reactions | ||
| 13. | (a) | oxidation-reduction (addition) |
| (b) | acid-base (neutralization) | |
| (c) | oxidation-reduction (combustion) | |
| 15. | It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction. | |
| 17. | (a) | H: +1, P: +5, O: −2 |
| (b) | Al: +3, H: +1, O: −2 | |
| (c) | Se: +4, O: −2 | |
| (d) | K: +1, N: +3, O: −2 | |
| (e) | In: +3, S: −2 | |
| (f) | P: +3, O: −2 | |
| 19. | (a) | acid-base |
| (b) | oxidation-reduction: Na is oxidized, H+ is reduced | |
| (c) | oxidation-reduction: Mg is oxidized, Cl2 is reduced | |
| (d) | acid-base | |
| (e) | oxidation-reduction: P3− is oxidized, O2 is reduced | |
| (f) | acid-base | |
| 21. | (a) | 2 HCl (g) + Ca(OH)2 (s) → CaCl2 (s) + 2 H2O (l) |
| (b) | Sr(OH)2 (aq) + 2 HNO3 (aq) → Sr(NO3)2 (aq) + 2 H2O (l) | |
| 23. | (a) | 2 Al (s) + 3 F2 (g) → 2 AlF3 (s) |
| (b) | 2 Al (s) + 3 CuBr2 (aq) → 3 Cu (s) + 2 AlBr3 (aq) | |
| (c) | P4 (s) + 5 O2 (g) → P4O10 (s) | |
| (d) | Ca (s) + 2 H2O (l) → Ca(OH)2 (aq) + H2 (g) | |
| 25. | (a) | Mg(OH)2 (s) + 2 HClO4 (aq) → Mg2+ (aq) + 2 ClO4– (aq) + 2 H2O (l) |
| (b) | SO3 (g) + 2 H2O (l) → H3O+ (aq) + HSO4– (aq) | |
| (c) | SrO (s) + H2SO4 (l) → SrSO4 (s) + H2O (l) | |
| 27. | H2 (g) + F2 (g) → 2 HF (g) | |
| 29. | 2 NaBr (aq) + Cl2 (g) → 2 NaCl (aq) + Br2 (l) | |
| 31. | 2 LiOH (aq) + CO2 (g) → Li2CO3 (aq) + H2O (l) | |
| 33. | (a) | Ca(OH)2 (s) + H2S (g) → CaS (s) + 2 H2O (l) |
| (b) | Na2CO3 (aq) + H2S (g) → Na2S (aq) + CO2 (g) + H2O (l) | |
| 35. | (a) | Step 1: N2 (g) + 3 H2 (g) → 2 NH3 (g) |
| Step 2: NH3 (g) + HNO3 (aq) → NH4NO3 (aq) → NH4NO3 (s) (after drying) | ||
| (b) | H2 (g) + Br2 (l) → 2 HBr (g) | |
| (c) | Zn (s) + S (s) → ZnS (s) and | |
| ZnS (s) + 2 HCl (aq) → ZnCl2 (aq) + H2S (g) | ||
| 37. | (a) | Sn4+ (aq) + 2 e– → Sn2+ (aq) |
| (b) | [Ag(NH3)]+ (aq) + e– → Ag (s) + 2 NH3 (aq) | |
| (c) | Hg2Cl2 (s) + 2 e– → 2 Hg (l) + 2 Cl– (aq) | |
| (d) | 2 H2O (l) → O2 (g) + 4 H+ (aq) + 4 e– | |
| (e) | 6 H2O (l) + 2IO3- (aq) + 10 e– → I2 (s) + 12 OH– (aq) | |
| (f) | H2O (l) + SO32- (aq) → SO42- (aq) + 2 H+ (aq) + 2 e– | |
| (g) | 8 H+ (aq) + MnO4– (aq) + 5 e– → Mn2+ (aq) + 4 H2O (l) | |
| (h) | Cl– (aq) + 6 OH– (aq) → ClO3– (aq) + 3 H2O (l) + 6 e– | |
| 39. | (a) | Sn2+ (aq) + 2 Cu2+ (aq) → Sn4+ (aq) + 2 Cu+ (aq) |
| (b) | H2S (g) + Hg22+ (aq) + 2 H2O (l) → 2 Hg (l) + S (s) + 2 H3O+ (l) | |
| (c) | 5 CN– (aq) + 2 ClO2 (aq) + 3 H2O (l) → 5 CNO– (aq) + 2 Cl– (aq) + 2 H3O+ (l) | |
| (d) | Fe2+ (aq) + Ce4+ (aq) → Fe3+ (aq) + Ce3+ (aq) | |
| (e) | 2 HBrO (aq) + 2 H2O (l) → 2 H3O+ (l) + 2 Br– (aq) + O2 (g) | |
| 41. | (a) | 2 MnO4– (aq) + 3 NO2– (aq) + H2O (l)→ 2 MnO2 (s) + 3 NO3– (aq) + 2 OH– (aq) |
| (b) | 3 MnO42- (aq) + 2 H2O (l) → 2 MnO4– (aq) + 4 OH– (aq) + MnO2 (s) | |
| (in base) | ||
| (c) | Br2 (l) + SO2 (g) + 2 H2O (l) → 4 H+ (aq) + 2 Br- (aq) + SO42- (aq) | |
| 4.3 Reaction Stoichiometry | ||
| 43. | (a) | 0.435 mol Na, 0.217 mol Cl2, 15.4 g Cl2 |
| (b) | 0.005780 mol HgO, 2.890 × 10−3 mol O2, 9.248 × 10−2 g O2 | |
| (c) | 8.00 mol NaNO3, 6.8 × 102 g NaNO3 | |
| (d) | 1665 mol CO2, 73.3 kg CO2 | |
| (e) | 18.86 mol CuO, 2.330 kg CuCO3 | |
| (f) | 0.4580 mol C2H4Br2, 86.05 g C2H4Br2 | |
| 45. | (a) | 0.0686 mol Mg, 1.67 g Mg |
| (b) | 2.701 × 10−3 mol O2, 0.08644 g O2 | |
| (c) | 6.43 mol MgCO3, 542 g MgCO3 | |
| (d) | 768 mol H2O, 13.8 kg H2O | |
| (e) | 16.31 mol BaO2, 2762 g BaO2 | |
| (f) | 0.207 mol C2H4, 5.81 g C2H4 | |
| 47. | (a) | volume HCl → solution → mol HCl → mol GaCl3 |
| (b) | 1.25 mol GaCl3, 2.2 × 102 g GaCl3 | |
| 49. | (a) | 5.337 × 1022 molecules |
| (b) | 10.41 g Zn(CN)2 | |
| 51. | SiO2 + 3 C → SiC + 2 CO | |
| 4.50 kg SiO2 | ||
| 53. | 5.00 × 103 kg | |
| 55. | 1.28 × 105 g CO2 | |
| 57. | 161.4 mL KI solution | |
| 59. | 176 g TiO2 | |
| 4.4 Reaction Yields | ||
| 61. | The limiting reactant is Cl2. | |
| 63. | Percent yield = 31% | |
| 65. | g CCl4 → mol CCl4 → mol CCl2F2 → g CCl2F2 | |
| Percent yield = 48.3% | ||
| 67. | Percent yield = 91.3% | |
| 69. | Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6% | |
| 71. | mol Cr → mol H3PO2 | |
| Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant. | ||
| 73. | Na2C2O4 is the limiting reactant. Percent yield = 86.56% | |
| 75. | Only four molecules can be made. | |
| 77. | This amount cannot be weighted by ordinary balances and is worthless. | |
| 4.5 Quantitative Chemical Analysis | ||
| 79. | 3.4 × 10−3 M H2SO4 | |
| 81. | 9.6 × 10−3 M Cl− | |
| 83. | 22.4% | |
| 85. | The empirical formula is BH3. The molecular formula is B2H6. | |
| 87. | 49.6 mL | |
| 89. | 13.64 mL | |
| 91. | 0.0122 M | |
| 93. | 34.99 mL KOH | |
| 95. | The empirical formula is WCl4. | |
