Question

The metathesis reaction between iron (111) nitrate and potassium hydroxide is shown below.
Fe(NO3)3(aq) +3KOH(aq) arrow 3KNO3(aq)+ Fe(OH)3(s)
How many grams of solid iron (111) hydroxide can be produced by the reaction of 400.0ml of 0.350M iron(111) nitrate solution with 225.0ml of 0.400M potassium hydroxide solution ?
determine the final concentration of each iron in solution at the end of the precipitation reaction

K,NO3,Fe,OH

Answer 1

Fe(NO3)3(aq) +3KOH(aq) -------> 3KNO3(aq)+ Fe(OH)3(s)

Moles of Fe(NO3)3 = Molarity*V in L = 400*0.35 / 1000 = 0.14

Moles of KOH = 225*0.4 / 1000 = 0.09

1 mole of Fe(NO3)3 reacts with 3 moles of KOH and forms 1 mole of Fe(OH)3

0.09 moles of KOH reats with 0.09/3 = 0.03 moles of Fe(NO3)3 and forms 0.03 moles of Fe(OH)3

Mass of Fe(OH)3 produced = moles *MOlar mass = 0.03*88.86 = 2.6658 gms

Moles of Fe(NO3) left = 0.14 - 0.03 = 0.11

Concentration of Fe(NO3)3 = moles / Total VOlume = 0.11 / (225+400) = 1.76*10^-4

[Fe+3] = 1.76*10^-4 M

[NO3-] = 3*1.76*10^-4 = 5.28*10^-4 M

as KOH is nothing left ---> K+ and OH- concentration is zero