Answers to More Chapter 04 Study Questions

1. $$\mathrm{FeCl_3(aq) + 3 NaOH(aq) \rightarrow Fe(OH)_3(s) + 3 NaCl(aq)}$$.

2. $$\mathrm{30.0\: mL\: \times\dfrac{0.500\:moles\:NaOH}{1000\:mL\:solution}\times\dfrac{1\:mol\:FeCl_3}{3\:mol\:NaOH}\times\dfrac{1000\:mL\:solution}{0.200\:mol\:FeCl_3}=25.0\: mL}$$

3. $$\mathrm{30.0\: mL\: \times\dfrac{0.500\:moles\:NaOH}{1000\:mL\:solution}\times\dfrac{1\:mol\:Fe(OH)_3}{3\:mol\:NaOH}\times\dfrac{106.8\:g\:Fe(OH)_3}{1\:mol\:Fe(OH)_3}=0.534\:g}$$

1. $$\mathrm{3 AgNO_3(aq) + AlCl_3(aq) \rightarrow Al(NO_3)_3(aq) + 3 AgCl(s)}$$

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

$$\mathrm{25.0\: mL\:\times\dfrac{0.200\:mol\:AgNO_3}{1000\:mL\:solution}\times\dfrac{3\:mole\:AgCl}{3\:mol\:AgNO_3}\times\dfrac{143.3\:g\:AgCl}{1\:mole\:AgCl}=0.716\: g\: AgCl}$$

$$\mathrm{10.0\: mL\:\times\dfrac{0.150\:mol\:AgCl_3}{1000\:mL\:solution}\times\dfrac{3\:mole\:AgCl}{1\:mol\:AlCl_3}\times\dfrac{143.3\:g\:AgCl}{1\:mole\:AgCl}=0.645\: g\: AgCl}$$

Therefore, 0.645 g $$\ce{AgCl}$$ is formed.

1. $$\mathrm{Ba(OH)_2(aq) + 2 HNO_3(aq) \rightarrow 2 H_2O(l) + Ba(NO_3)_2(aq)}$$

$$\mathrm{25.0\: mL\:\times\dfrac{0.300\:mol\:HNO_3}{1000\:mL\:solution}\times\dfrac{1\:mole\:Ba(OH)_2}{2\:mol\:HNO_3}\times\dfrac{1000\:mL\:solution}{0.0500\:mole\:Ba(OH)_2}=75.0\: mL}$$

1. $$\mathrm{50.0\: mL\:\times\dfrac{2.00\:mol\:KCl}{1000\:mL\:solution}\times\dfrac{74.5\:g\:KCl}{1\:mol\:KCl}=7.45\: g\: KCl}$$

2. $$\mathrm{V_1 \times M_1 = V_2 \times M_2}$$        $$\mathrm{V_1 \times 6.00\: M\: HCl = 30.0\: mL \times 0.500\: M\: HCl}$$

$$\mathrm{V_1 = \dfrac{(30.0\: mL \times 0.500\: M)}{6.00\: M} = 2.50\: mL}$$

1. $$\mathrm{Ba(NO_3)_2(aq) + K_2SO_3(aq) \rightarrow BaSO_3(aq) + 2 KNO_3(aq)}$$

$$\mathrm{40.0\: mL\: \times\dfrac{0.250\:mol\:Ba(NO_3)_2}{1000\:mL\:solution}\times\dfrac{1\:mol\:BaSO_3}{1\:mol\:Ba(NO_3)_2}\times\dfrac{217.4\:g\:BaSO_3}{1\:mol\:BaSO_3}=2.17\:g}$$