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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?

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Answer:

102 mol Fe(OH)₃

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 * [tex]\frac{2molFe(OH)_3}{1molFe_2(SO_4)_3}[/tex] = 102 mol Fe(OH)₃

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