Question

Determine the mass (in g) of Fe(OH)₃ that is produced when 464 mL of a 1.72×10^-2 M Fe(NO₃)₃ solution completely reacts with 490 mL of a 6.96×10^-2 M KOH solution according to the following balanced chemical equation.



Fe(NO₃)₃(aq) + 3KOH(aq) → Fe(OH)₃(s) + 3KNO₃(aq)

Answer 1

moles of Fe(NO3)3 = 464 x 1.72 x 10^-2 / 1000 = 7.98 x 10^-3

moles of KOH = 490 x 6.96 x 10^-2 / 1000 = 0.0341

Fe(NO₃)₃(aq) + 3KOH(aq) → Fe(OH)₃(s) + 3KNO₃(aq)

      1                     3                     1

7.98 x 10^-3        0.0341

here limiting reagent is Fe(NO3)3. because it has less number of moles.

1 mol Fe(NO3)3   --------------> 1 mol Fe(OH)3

7.98 x 10^-3 mol Fe(NO3)3   ------------->   ??

moles of Fe(OH)3 = 7.98 x 10^-3 x 1 / 1 = 7.98 x 10^-3 mol

mass of Fe(OH)3 = 7.98 x 10^-3 x 106.87

mass of Fe(OH)3 = 0.853 g