Chapter 4

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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.

(a) (b) (c) (d) (e) (f) (g) (h)

(a) (b) (c) (d)

(a) Ba(NO₃)₂, KClO₃; (b) (c) (d)

(a) (b) complete ionic equation: net ionic equation:

11.

(a)
(b)
(c)

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) (b)

23.

(a) (b) (c) (d)

25.

(a) (b) (a solution of H₂SO₄); (c)

27.

29.

31.

33.

(a) (b)

35.

(a) step 1: step 2: (b) (c) and

37.

(a) (b) (c) (d) (e) (f) (g) (h)

39.

(a) (b) (c) (d) (e)

41.

(a) (b) (c)

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) (b) 1.25 mol GaCl₃, 2.2 10² g GaCl₃

49.

(a) 5.337 10²² molecules; (b) 10.41 g Zn(CN)₂

51.

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

61.

The limiting reactant is Cl2.

63.

65.

67.

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.

The conversion needed is 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.

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.

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