Question

2Al(OH)₃(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6H₂O(l)

There are 0.500 moles of Al(OH)₃ and 0.500 moles of H₂SO₄.
Find the limiting reagent in this reaction using the steps given in class.
How many moles of Al₂(SO₄)₃ can form under these conditions?

Answer 1

H₂SO₄ is the limiting reagent, allowing for the formation of 0.167 moles of Al₂(SO₄)₃.

Step-by-step explanation:

To find the limiting reagent in the reaction 2Al(OH)₃(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6H₂O(l), we compare the mole ratios of the reactants provided to those required by the balanced chemical equation.

According to the equation, 2 moles of Al(OH)₃ react with 3 moles of H₂SO₄. As we have 0.500 moles of each reactant, we can calculate the amount of product formed from each reactant separately using the stoichiometry of the balanced equation.

For Al(OH)₃: 0.500 moles Al(OH)₃ × (1 mole Al₂(SO₄)₃ / 2 moles Al(OH)₃) = 0.250 moles Al₂(SO₄)₃

For H₂SO₄: 0.500 moles H₂SO₄ × (1 mole Al₂(SO₄)₃ / 3 moles H₂SO₄) = 0.167 moles Al₂(SO₄)₃

H₂SO₄ produces fewer moles of Al₂(SO₄)₃, so it is the limiting reagent. Therefore, only 0.167 moles of Al₂(SO₄)₃ can form under these conditions.