Question

Sulfuric acid reacts with aluminum hydroxide by double replacement. if 34 g of sulfuric acid react with 33 g of aluminum hydroxide, identify the limiting reactant.

Answer 1

To determine the limiting reactant between sulfuric acid and aluminum hydroxide, calculate the moles of each reactant using their molar masses. Comparisons based on the stoichiometric ratios from the balanced chemical equation reveal that sulfuric acid is the limiting reactant.

Step-by-step explanation:

To identify the limiting reactant when 34 g of sulfuric acid reacts with 33 g of aluminum hydroxide, we need to calculate the moles of each reactant based on their molar masses. The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and aluminum hydroxide (Al(OH)₃) is:

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

First, determine the molar mass of H₂SO₄, which is approximately 98 g/mol, and Al(OH)₃, which is approximately 78 g/mol. Then, calculate the moles of each reactant:

• Moles of H₂SO₄ = 34 g / 98 g/mol = 0.347 moles
• Moles of Al(OH)₃ = 33 g / 78 g/mol = 0.423 moles

The balanced equation shows that 3 moles of H₂SO₄ react with 2 moles of Al(OH)₃. We then set up ratios to see which reactant is limiting:

• 0.347 moles H₂SO₄ × (2 moles Al(OH)₃ / 3 moles H₂SO₄) = 0.231 moles Al(OH)₃ required
• 0.423 moles Al(OH)₃ × (3 moles H₂SO₄ / 2 moles Al(OH)₃) = 0.635 moles H₂SO₄ required

Since we only have 0.347 moles of H₂SO₄ but require 0.635 moles to react with all the available Al(OH)₃, sulfuric acid is the limiting reactant.

Answer 2

Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.