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?

102 mol Fe(OH)₃

Explanation:

The reaction that takes place is:

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

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

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