Solution Stoichiometry

 

66) [Fe(NO3)2] = 0.500 M          mNaOH = ?

      V = 25.0 mL

      

      Fe(NO3)2(aq) + 2NaOH(aq) 2NaNO3(aq) + Fe(OH)2(s)

      Fe2+(aq) + 2OH-(aq) Fe(OH)2(s)

 

      [Fe(NO3)2] = n/V

      nFe2+ =  [Fe(NO3)2] x V = 0.500 mol Fe(NO3)2/L x 25.0 mL x 1 L/103 mL x 1 mol Fe2+/1 mol Fe(NO3)2 = 0.0125 mol Fe2+

      nOH- = 0.0125 mol Fe2+ x 2 mol OH-/1 mol Fe2+ = 0.0250 mol OH-

      mOH- = 0.0250 mol OH- x 1 mol NaOH/1 mol OH- x 40.00 g NaOH/1 mol NaOH = 1.00 g NaOH