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Answers to Chapter 03 Study Questions

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      1. \(\mathrm{3(12.0) + 8(1.01) + 3(16.0)=92.1\: g/mole}\)
      2. 92.1 g
      3. 6.02 × 1023 molecules
      4. \(\mathrm{0.217\: moles\times\dfrac{92.1\:g}{1\:mole}=20.0\:g}\)

      1. 4 atoms (one \(\ce{N}\) + \(\ce{3 H}\))
      2. \(\mathrm{1\: mole\: NH_3\:\times\dfrac{6.02\times10^{23}\:molecules}{1\:mole\:NH_3}\times\dfrac{4\:atoms}{1\:molecule}=2.41\times10^{24}\: atoms}\)
      3. \(\mathrm{3.40\: g\: NH_3\times\dfrac{1\:mole\:NH_3}{17.0\:g\:NH_3}\times\dfrac{6.02\times10^{23}\:molecules}{1\:mole\:NH_3}\times\dfrac{4\:atoms}{1\:molecule}=4.82 \times 10^{23}\: atoms}\)

    1. \(\mathrm{Molar\: mass\: of\: NaNO_2 = 23.0 + 14.0 + 2(16.0)=69.0\: g/mole}\)

    \(\mathrm{\%\: Na = \dfrac{23.0}{69.0} = 33.3\%\: Na}\); \(\mathrm{\%\: N = \dfrac{14.0}{69.0} = 20.3\%\: N}\); \(\mathrm{\%O = \dfrac{2(16.0)}{69.0} = 46.4\%\: O}\)

    33.3% \(\ce{Na}\), 20.3% \(\ce{N}\) and 46.4% \(\ce{O}\).

      1. In 100 g of this compound, there are 40.7 g \(\ce{C}\), 5.1 g \(\ce{H}\), and 54.2 g \(\ce{O}\)
    \(\mathrm{40.7\: g\: C\times\dfrac{1\:mole\:C}{12.0\:g\:C} = 3.39\: moles\: C}\) \(\mathrm{\dfrac{3.39}{3.39}= 1\times 2 = 2}\)
    \(\mathrm{5.1\: g\: H\times\dfrac{1\:mole\:H}{1.0\:g\:H} =5.1\: moles\: H}\) \(\mathrm{\dfrac{5.1}{3.39}= 1.5 \times 2=3}\)
    \(\mathrm{54.2\: g\: O\times\dfrac{1\:mole\:O}{16.0\:g\:O}=3.39\: moles\: O}\) \(\mathrm{\dfrac{3.39}{3.39}= 1\times 2 = 2}\)

    empirical formula = \(\ce{C2H3O2}\)

    1. \(\mathrm{Molar\: mass\: of\: C_2H_3O_2 = 2(12.0) + 3(1.0) + 2(16.0)=59.0\: g/mole}\)

    \(\mathrm{\dfrac{118}{59.0} = 2 \rightarrow molecular\: formula = C_4H_6O_4}\)

      1. In 25.0 g of compound, there are 7.20 g \(\ce{Mg}\), 3.55 g \(\ce{C}\) and \(\mathrm{25.0-(7.20+3.55) = 14.25 g O}\).
    \(\mathrm{7.20\: g\: Mg\times\dfrac{1\:mole\:Mg}{24.3\:g\:Mg}=0.296\: moles\: Mg}\) \(\mathrm{\dfrac{0.296}{0.296}= 1}\)
    \(\mathrm{3.55\: g\: C\times\dfrac{1\:mole\:C}{12.0\:g\:C}=0.296\: moles\: C}\) \(\mathrm{\dfrac{0.296}{0.296}= 1}\)
    \(\mathrm{14.25\: g\: O\:\times\dfrac{1\:mole\:O}{16.0\:g\:O}=0.891\: moles\: O}\) \(\mathrm{\dfrac{0.891}{0.296}= 3}\)

    \(\mathrm{formula = MgCO_3}\)

    1. \(\mathrm{\%\: Mg = \dfrac{7.20}{25.0} = 28.8\%\: Mg}\); \(\mathrm{\% C = \dfrac{3.55}{25.0} = 14.2 \% \:C}\); \(\mathrm{\% O = \dfrac{14.25}{25.0} = 57.0\%\: O}\). (You should get the same result using molar mass and atomic masses.)
    2. \(\mathrm{13.9\: g\: compound\times\dfrac{28.8\:g\:Mg}{100\:g\:compound}=4.00\: g\: Mg}\)
    3. \(\mathrm{0.290\: mol\: C\times\dfrac{12.0\:g\:C}{1\:mole\:C}\times\dfrac{100\:g\:compound}{14.2\:g\:C}=24.5\: g\: compound}\)


    1. \(\mathrm{mass\: of\: Zn=33.64\: g - 32.00\: g=1.64\: g\: Zn}\)

    \(\mathrm{mass\: of\: O = 34.04\: g - 33.64\: g =0.40\: g\: O}\).

    \(\mathrm{\#\: moles\: \textrm{Zn:}\:\: 1.64\: g\: Zn\times\dfrac{1\:mole\:Zn}{65.4\:g\:Zn}=0.0251\: moles\: Zn}\) \(\mathrm{\dfrac{0.251}{0.25}= 1}\)
    \(\mathrm{\#\: moles\: \textrm{O:}\:\: 0.40\: g\: O\times\dfrac{1\:mole\:O}{16.0\:g\:O}=0.025\: moles\: O}\) \(\mathrm{\dfrac{0.25}{0.25}= 1}\)

    \(\mathrm{formula = ZnO}\)

      1. \(\mathrm{B_2H_6(l) + 3O_2(g) \rightarrow B_2O_3(s)+ 3H_2O(l)}\)
      2. \(\mathrm{4PH_3(g)+8O_2(g) \rightarrow 6H_2O(l)+ P_4O_{10}(s)}\)

      1. \(\mathrm{6 Li(s) + N_2(g) \rightarrow 2 Li_3N(s)}\)
      2. \(\mathrm{Co(NO_­­3)_3(aq)+3 NaOH(aq)\rightarrow 3 NaNO­­_3(aq) + Co(OH)_3 (s)}\)
      3. \(\mathrm{Zn(s) + 2HCl(aq)\rightarrow ZnCl_2(aq) + H_2(g)}\)
      4. (a) is a combination (synthesis) reaction. (b) is a double replacement reaction.

      1. \(\mathrm{8.20\: moles\: SO_2\times\dfrac{2\:mol\:H_2S}{2\:mol\:SO_2}=8.20\: moles\: H_2S}\)
      2. \(\mathrm{1.00\: mole\: H_2S\times\dfrac{3\:mol\:O_2}{2\:mol\:H_2S}\times\dfrac{32.0\:g\:O_2}{1\:mol\:O_2}=48.0\: g\: O_2}\)
      3. \(\mathrm{6.82\: g\: H_2S\times\dfrac{1\:mol\:H_2S}{34.1\:g\:H_2S}\times\dfrac{2\:mol\:H_2O}{2\:mol\:H_2S}\times\dfrac{18.0\:g\:H_2O}{1\:mol\:H_2O}=3.60\: g\: H_2O}\)
      4. \(\mathrm{7.98\: g\: H_2S\times\dfrac{1\:mol\:H_2S}{34.1\:g\:H_2S} \times\dfrac{2\:mol\:SO_2}{2\:mol\:H_2S}\times\dfrac{64.1\:g\:SO_2}{1\:mol\:SO_2}=15.0\: g\: SO_2}\)

        \(\mathrm{\% \:yield=\dfrac{actual\:yield}{theoretical\:yield}\times100\% =\dfrac{12.0\:g\:SO_2}{15.0\:g\:SO_2}\times100\%=80.0\%}\)
      5. \(\mathrm{2.66\: g\: H_2S\times\dfrac{1\:mol\:H_2S}{34.1\:g\:H_2S}\times\dfrac{2\:mol\:SO_2}{2\:mol\:H_2S}\times\dfrac{64.1\:g\:SO_2}{1\:mol\:SO_2}=5.00\: g\: SO_2}\)

        \(\mathrm{3.00\: g\: O_2\times\dfrac{1\:mol\:O_2}{32.0\:g\:O_2}\times \dfrac{2\:mol\:SO_2}{3\:mol\:O_2}\times\dfrac{64.1\:g\:SO_2}{1\:mol\:SO_2}=4.01\: g\: SO_2}\)

        \(\ce{O2}\) is limiting; 4.01 g \(\ce{SO2}\) is produced

      1. \(\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)}\)
      2. \(\ce{H2}\) is limiting (\(\ce{H2}\) and \(\ce{Cl2}\) react in a ratio 1:1 ratio and there is less \(\ce{H2}\).)
      3. \(\mathrm{7.50\: mol\: H_2\times\dfrac{2\:mol\:HCl}{1\:mol\:H_2}=15.0\: mol\: HCl}\)
      4. \(\mathrm{9.00-7.50=1.50\: mol\: Cl_2\: left}\)

    This page titled Answers to Chapter 03 Study Questions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Delmar Larsen.

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