Answers to Chapter 04 Study Questions

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1. \(\mathrm{\textrm{strong acid: } HNO_3(aq) \rightarrow H^+(aq) + NO_3^-(aq)}\)
2. \(\mathrm{\textrm{weak acid: } HClO(aq) \rightleftharpoons H^+(aq) + ClO^-(aq)}\)
3. \(\mathrm{\textrm{weak base: } NH_3(aq) + H_2O \rightleftharpoons NH_4^+(aq) + OH^-(aq)}\)
4. \(\mathrm{\textrm{neutral: } NaNO_3(s) \rightarrow Na^+(aq) + NO_3^-(aq)}\)
5. \(\mathrm{\textrm{strong base: } Ba(OH)_2(s)\rightarrow Ba^{2+}(aq) + 2OH^-(aq)}\)

1. \(\mathrm{\dfrac{29.2\:g\:NaCl}{0.250\:L}\times\dfrac{1\:mol\:NaCl}{58.4\:g}=2.00\: M\: NaCl}\)
2. \(\mathrm{Volume_{solution\: A}\times Molarity_{solution\: A} = Volume_{solution\: B}\times Molarity_{solution\: B}}\)
  \(\mathrm{125\: ml \times 0.350\: M\: NaCl = x\: ml \times 2.00\: M\: NaCl}\)
  \(\mathrm{x=\dfrac{125\:ml\times0.350\:M}{2.00\:M}= 21.9\: ml}\)

\(\mathrm{200\: ml\: solution\: \times\dfrac{2.50\:moles\:C_6H_{12}O_6}{1000\:ml\:solution}\times\dfrac{180.0\:g\:C_6H_{12}O_6}{1\:mole\:C_6H_{12}O_6}=90.0\: g\: C_6H_{12}O_6}\)

1. \(\mathrm{Mg(s) + 2HCl(aq)\rightarrow MgCl_2(aq) + H_2(g)}\),
  [or \(\mathrm{Mg(s) + 2H^+(aq)\rightarrow Mg^{2+}(aq) + H_2(g)}\)]
2. \(\mathrm{125\: ml \times\dfrac{2.00\:moles\:HCl}{1000\:ml}\times\dfrac{1\:mole\:H_2}{2\:moles\:HCl}= 0.125\: moles\: H_2}\)

1. Precipate forms. \(\mathrm{Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)}\)
2. No Reaction. (\(\ce{(NH4)2CO3}\) and \(\ce{LiCl}\) are both soluble)
3. Precipate forms. \(\mathrm{Ni^{2+}(aq) + S^{2-}(aq) \rightarrow NiS(s)}\)

Lots of possibilities: Pick a soluble \(\mathrm{Sr^{2+}}\) and a soluble \(\mathrm{SO_4^{2-}}\), such as: \(\mathrm{Sr(NO_3)_2}\) and \(\mathrm{Na_2SO_4}\).

any strong acid + strong base: \(\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O}\)

At neutralization, \(\mathrm{moles_{acid} = moles_{base}}\)

\(\mathrm{Volume_{acid} \times Molarity_{acid} = Volume_{base} \times Molarity_{base}}\)

\(\mathrm{Volume_{base}\times 2.00\: M = 12.5\: ml \times 0.0800\: M}\)

\(\mathrm{Volume_{base}=\dfrac{12.5\:ml\times0.0800\:M}{2.00\:M}=0.500\:ml}\)

1. \(\mathrm{3Pb(NO_3)_2(aq) + 2AlCl_3(aq)\rightarrow 2Al(NO_3)_3(aq) + 3PbCl_2(s)}\).
2. \(\mathrm{2.48\: mL \times\dfrac{0.300\:moles\:AlCl_3}{1000\:mL\:solution}\times\dfrac{3\:moles\:Pb(NO_3)_2}{2\:mol\:AlCl_3}\times\dfrac{1000\:mL}{0.200\:mol\:Pb^{2+}}=5.58\: mL}\)
3. \(\mathrm{2.48\: mL \times\dfrac{0.300\:moles\:AlCl_3}{1000\:mL\:solution}\times\dfrac{3\:moles\:PbCl_2}{2\:mol\:AlCl_3}\times\dfrac{278.1\:g\:PbCl_2}{1\:mol\:PbCl_2}=0.310\: g}\)

\(\mathrm{2Fe(NO_3)_3(aq) + 3Na_2CO_3(aq)\rightarrow 6NaNO_3(aq) + Fe_2(CO_3)_3(s)}\)

This is a limiting reactant problem, so first determine which reactant is limiting.

\(\mathrm{71.3\: mL\: Fe(NO_3)_3 \times\dfrac{0.500\:moles\:Fe(NO_3)_3}{1000\:mL\:solution}\times\dfrac{1\:mole\:Fe_2(CO_3)_3}{2\:mol\:Fe(NO_3)_3}\times\dfrac{291.7\:g\:Fe_2(CO_3)_3}{1\:mol\:Fe_2(CO_3)_3}}\)

\(\mathrm{=5.20\: g\: Fe_2(CO_3)_3}\)

\(\mathrm{112\: mL\: Na_2CO_3\times\dfrac{0.800\:moles\:Na_2CO_3}{1000\:mL\:solution}\times\dfrac{1\:mole\:Fe_2(CO_3)_3}{3\:mol\:Na_2CO_3}\times\dfrac{291.7\:g\:Fe_2(CO_3)_3}{1\:mol\:Fe_2(CO_3)_3}}\)

\(\mathrm{=8.71\: g\: Fe_2(CO_3)_3}\)

Therefore, 5.20 g \(\ce{Fe2(CO3)3}\) is formed.

\(\mathrm{3.94\: g\: Cu\:\times\dfrac{1\:mole\:Cu}{63.54\:g}\times\dfrac{8\:mol\:HNO_3}{3\:mol\:Cu}\times\dfrac{1000\:ml\:solution}{2.50\:mol\:HNO_3}=66.1\: ml\: solution}\)