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Answers to More Chapter 04 Study Questions

1. 

  1. FeCl3(aq)  +  3 NaOH(aq)   →    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. \(\mathrm{30.0\: mL\: \times\dfrac{0.500\:moles\:NaOH}{1000\:mL\:solution}\times\dfrac{1\:mol\:Fe(OH)_3}{3\:mol\:NaOH}\times\dfrac{106.8\:g\:Fe(OH)_3}{1\:mol\:Fe(OH)_3}=0.534\:g}\)

 

2.    3 AgNO3(aq)  +  AlCl3(aq)   →    Al(NO3)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 AgCl is formed.

 

3.     Ba(OH)2(aq)   +   2 HNO3(aq)    →    2 H2O(l)   +   Ba(NO3)2(aq)

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

 

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

 

5.    V1 x M1  =  V2 x M2        V1 x  6.00 M HCl  =  30.0 mL x 0.500 M HCl

       V1 =  (30.0 mL x 0.500 M)/6.00 M  =  2.50 mL

 

6.     Ba(NO3)2(aq)   +   K2SO3(aq)    →     BaSO3(aq)   +   2 KNO3(aq)

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