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22.14.4: Chapter 4

  • Page ID
    452934
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    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) (b) (c) (d) (e) (f) (g) (h)

    5.

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

    7.

    (a) Ba(NO3)2, KClO3; (b) (c) (d)

    9.

    (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.


    22.14.4: Chapter 4 is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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