Solutions to Chapter 6 Exercises

6.1 Solutions

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) + 2HCl(aq);
(b) 3Cu(s) + 8HNO₃(aq) ⟶ 3Cu(NO₃)₂(aq) + 4H₂O(l) + 2NO(g);
(c) H₂(g) + I₂(s) ⟶ 2HI(s);
(d) 4Fe(s) + 3O₂(g) ⟶ 2Fe₂O₃(s);
(e) 2Na(s) + 2H₂O(l) ⟶ 2NaOH(aq) + H₂(g);
(f) (NH₄)₂Cr₅2O₇(s) ⟶ Cr₂O₃(s) + N₂(g) + 4H₂O(g);
(g) P₄(s) + 6Cl₂(g) ⟶ 4PCl₃(l);
(h) PtCl₄(s) ⟶ Pt(s) + 2Cl₂(g)

5. (a) CaCO₃(s) ⟶ CaO(s) + CO₂(g);
(b) 2C₄H₁₀(g) + 13O₂(g) ⟶ 8CO₂(g) + 10H₂O(g);
(c) MgC₁₂(aq) + 2NaOH(aq) ⟶ Mg(OH)₂(s) + 2NaCl(aq);
(d) 2H₂O(g) + 2Na(s) ⟶ 2NaOH(s) + H₂(g)

7. (a) Ba(NO₃)₂, KClO₃;
(b) 2KClO₃(s) ⟶ 2KCl(s) + 3O₂(g);
(c) 2Ba(NO₃)₂(s) ⟶ 2BaO(s) + 2N₂(g) + 5O₂(g);
(d) 2Mg(s) + O₂(g) ⟶ 2MgO(s); 4Al(s) + 3O₂(g) ⟶ 2Al₂O₃(g);
4Fe(s) + 3O₂(g) ⟶ 2Fe₂O₃(s)

9. (a) 4HF(aq) + SiO₂(s) ⟶ SiF₄(g) + 2H₂O(l);
(b) complete ionic equation:
2Na⁺(aq) + 2F⁻(aq) + Ca²⁺(aq) + 2Cl⁻(aq) ⟶ CaF₂(s) + 2Na⁺(aq) + 2Cl⁻(aq),
net ionic equation: 2F⁻(aq) + Ca²⁺ (aq) ⟶ CaF₂(s)

11. (a) 2K⁺(aq) + C₂O₄²⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) ⟶ 2K⁺(aq)+2OH⁻(aq) + BaC₂O₄(s)(complete)
Ba²⁺(aq) + C₂O₄²⁻(aq) ⟶ BaC₂O₄(s)(net)
(b) Pb²⁺(aq)+2NO₃⁻(aq) + 2H⁺(aq) + SO₄²⁻(aq) ⟶ PbSO₄(s) + 2H⁺(aq) + 2NO₃⁻(aq)(complete)
Pb²⁺(aq) + SO₄²⁻(aq)⟶PbSO₄(s)(net)
(c) CaCO₃(s) + 2H⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s) + CO₂(g) + H₂O(l) (complete)
CaCO₃(s) + 2H⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s) + CO₂(g) + H₂O(l) (net)

6.2 Solutions

1. (a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)

3. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.

5.(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

7. (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

9. (a) 2HCl(g) + Ca(OH)₂(s) ⟶ CaCl₂(s) + 2H₂O(l);
(b) Sr(OH)₂(aq) + 2HNO₃(aq) ⟶ Sr(NO₃)₂(aq) + 2H₂O(l)

11. (a) 2Al(s) + 3F₂(g) ⟶ 2AlF₃(s);
(b) 2Al(s) + 3CuBr₂(aq) ⟶ 3Cu(s) + 2AlBr₃(aq);
(c) P₄(s) + 5O₂(g) ⟶ P₄O₁₀(s);
(d) Ca(s) + 2H₂O(l) ⟶ Ca(OH)₂(aq) + H₂(g)

13. (a) Mg(OH)₂(s) + 2HClO₄(aq) ⟶ Mg²⁺(aq) + 2ClO₄⁻(aq) + 2H₂O(l);
(b) SO₃(g) + 2H₂O(l) ⟶ H₃O⁺(aq) + HSO₄⁻(aq), (a solution of H₂SO₄);
(c) SrO(s) + H₂SO₄(l) ⟶ SrSO₄(s)+H₂O

15. H₂(g) + F₂(g) ⟶ 2HF(g)

17. 2NaBr(aq) + Cl₂(g) ⟶ 2NaCl(aq) + Br₂(l)

19. 2LiOH(aq) + CO₂(g) ⟶ Li₂CO₃(aq) + H₂O(l)

21. (a) Ca(OH)₂(s) + H₂S(g) ⟶ CaS(s) + 2H₂O(l);
(b) Na₂CO₃(aq) + H₂S(g) ⟶ Na₂S(aq) + CO₂(g) + H₂O(l)

23. (a) step 1: N₂(g) + 3H₂(g) ⟶ 2NH₃(g),
step 2: NH₃(g) + HNO₃(aq) ⟶ NH₄NO₃(aq) ⟶ NH₄NO₃(s)(after drying);
(b) H₂(g)+Br₂(l) ⟶ 2HBr(g);
(c) Zn(s) + S(s) ⟶ ZnS(s)
and ZnS(s) + 2HCl(aq) ⟶ ZnCl₂(aq) + H₂S(g)

25. (a) Sn⁴⁺(aq) + 2e⁻ ⟶ Sn²⁺(aq),
(b) [Ag(NH₃)₂]⁺(aq) + e⁻ ⟶ Ag(s) + 2NH₃(aq);
(c) Hg₂Cl₂(s) + 2e⁻ ⟶ 2Hg(l) + 2Cl⁻(aq);
(d) 2H₂O(l) ⟶ O₂(g) + 4H⁺(aq) + 4e⁻;
(e) 6H₂O(l) + 2IO₃⁻(aq) + 10e⁻ ⟶ I₂(s) + 12OH⁻(aq);
(f) H₂O(l) + SO₃²⁻(aq) ⟶ SO₄²⁻(aq) + 2H⁺(aq) + 2e⁻;
(g) 8H⁺(aq) + MnO₄⁻(aq) + 5e⁻ ⟶ Mn²⁺(aq) + 4H₂O(l);
(h) Cl⁻(aq) + 6OH⁻(aq) ⟶ ClO₃⁻(aq) + 3H₂O(l) + 6e⁻

27. (a) Sn²⁺(aq) + 2Cu²⁺(aq) ⟶ Sn⁴⁺(aq) + 2Cu⁺(aq);
(b) H₂S(g) + Hg₂²⁺(aq) + 2H₂O(l) ⟶ 2Hg(l) + S(s) + 2H₃O⁺(aq);
(c) 5CN⁻(aq) + 2ClO₂(aq) + 3H₂O(l) ⟶ 5CNO⁻(aq) + 2Cl⁻(aq) + 2H₃O⁺(aq);
(d) Fe²⁺(aq) + Ce⁴⁺(aq) ⟶ Fe³⁺(aq) + Ce³⁺(aq);
(e) 2HBrO(aq) + 2H₂O(l) ⟶ 2H₃O⁺(aq) + 2Br⁻(aq)+O₂(g)

29. (a) 2MnO₄⁻(aq) + 3NO₂⁻(aq) + H₂O(l) ⟶ 2MnO₂(s) + 3NO₃⁻(aq)+2OH⁻(aq);
(b) 3MnO₄²⁻(aq) + 2H₂O(l) ⟶ 2MnO₄⁻(aq) + 4OH⁻(aq) + MnO₂(s)(in base);
(c) Br₂(l) + SO₂(g) + 2H₂O(l) ⟶ 4H⁺(aq) + 2Br⁻(aq) + SO₄²⁻(aq)

6.3 Solutions

**1. (a) volume HCl solution⟶mol HCl⟶mol GaCl3; (b) 1.25 mol GaCl3, 2.2 × 102 g GaCl3

**3. (a) 5.337 × 1022 molecules; (b) 10.41 g Zn(CN)2

**5. SiO2+3C⟶SiC+2CO, 4.50 kg SiO2

**7. 5.00 × 103 kg

**9. 1.28 × 105 g CO2

11. 161.40 mL KI solution

**13. 176 g TiO2

6.4 Solutions

1. The limiting reactant is Cl2.

3. Percent yield = 31%

5. g CCl₄ ⟶ mol CCl₄ ⟶ mol CCl₂F₂ ⟶ g CCl₂F₂, percent yield = 48.3%

7. percent yield = 91.3%

9. 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%

11. The conversion needed is mol Cr ⟶ mol H₃PO₄. Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.

13. Na₂C₂O₄ is the limiting reactant. percent yield = 86.6%

15. Only four molecules can be made.

17. This amount cannot be weighted by ordinary balances and is worthless.

6.5 Solutions

1. 3.4 × 10⁻³ M H₂SO₄

3. 9.6 × 10⁻³ M Cl⁻

5. 22.4%

7. The empirical formula is BH₃. The molecular formula is B₂H₆.

9. 49.6 mL

11. 13.64 mL

13. 1.22 M

15. 34.99 mL KOH

17. The empirical formula is WCl₄.