Question

How many moles of Fe(OH), are produced when 85.0 L of iron(III)

sulfate at a concentration of 0.600 mol/L reacts with excess
NaOH?

Answer 1

102 mol Fe(OH)₃

Step-by-step explanation:

The reaction that takes place is:

• Fe₂(SO₄)₃ + 6NaOH → 2Fe(OH)₃ + 3Na₂SO₄

First we calculate how many Fe₂(SO₄)₃ moles reacted, using the given concentration and volume:

• 0.600 mol/L * 85.0 L = 51 mol Fe₂(SO₄)₃

Then we convert Fe₂(SO₄)₃ moles into Fe(OH)₃ moles, using the stoichiometric coefficients of the reaction:

• 51 mol Fe₂(SO₄)₃ *
  `(2molFe(OH)_3)/(1molFe_2(SO_4)_3)` = 102 mol Fe(OH)₃