Question

Determine the limiting reactant when 4.2 mol Fe(OH)3 reacts with 6.5 mol H2SO4 in the following reaction.

2 Fe(OH)3 + 3 H2SO4 --> Fe2(SO4)3 + 6 H2O

Answer 1

2 Fe(OH)₃ + 3 H₂SO₄ $\rightarrow$ Fe₂(SO₄)₃ + 6 H₂O

As you can see from the balanced equation, 2 molecules of Fe(OH)₃ react with 3 molecules of H₂SO₄. According to this stoichiometry, the number of the moles that are consumed relate as:

$\frac{n(Fe(OH)_3)}{2} = \frac{n(H_2SO_4)}{3}$

Let's evaluate the left and the right side of this equation separately, using the quantity of the substances given:

$n(Fe(OH)_3)/2 = \frac{4.2}{2} = 2.1$

$n(H_2SO_4)/3 = \frac{6.5}{3} = 2.17$

2.17>2.1, so Fe(OH)₃ is the limiting reactant. In other words, there is an excess of H₂SO₄.

Answer: For the conditions given, the limiting reactant is Fe(OH)₃.