Answer to Question #330211 in Chemistry for krist

Question

50.0 ml of 2.50M Sulfuric acid reacted with 43.5ml 2.75 Aluminum hydroxide. The reaction yielded Aluminum sulfate and water. (Al=27g/mol, S=32g/mol, H=1g/mol, O=16g/mol)

1. Write the balance chemical equation.

2. What is the limiting reactant?

3. How much reactant is in excess?

4. How many grams of Aluminum sulfate will be produced?

Answer 1

Solution:

Molar mass of sulfuric acid, H₂SO₄ = 2×A_r(H) + A_r(S) + 4×A_r(O) = 2×1 + 32 + 4×16 = 98 (g/mol)

The molar mass of H₂SO₄ is 98 g/mol

Molar mass of aluminum hydroxide, Al(OH)₃ = A_r(Al) + 3×A_r(O) + 3×A_r(H) = 27 + 3×16 + 3×1 = 78 (g/mol)

The molar mass of Al(OH)₃ is 78 g/mol

Molar mass of aluminum sulfate, Al₂(SO₄)₃ = 2×A_r(Al) + 3×A_r(S) + 12×A_r(O) = 2×27 + 3×32 + 12×16 = 342 (g/mol)

The molar mass of Al₂(SO₄)₃ is 342 g/mol

Calculate moles of each reactant:

Moles of H₂SO₄ = (2.50 M) × (50.0 mL) × (1 L / 1000 mL) = 0.125 mol H₂SO₄

Moles of Al(OH)₃ = (2.75 M) × (43.5 mL) × (1 L / 1000 mL) = 0.1196 mol Al(OH)₃

Balanced chemical equation:

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

According to stoichiometry:

3 mol of H₂SO₄ react with 2 mol of Al(OH)₃

Thus, 0.125 mol of H₂SO₄ react with:

(0.125 mol H₂SO₄) × (2 mol Al(OH)₃ / 3 mol H₂SO₄) = 0.0833 mol Al(OH)₃

However, initially there is 0.1196 mol of Al(OH)₃ (according to the task)

Therefore, H₂SO₄ acts as limiting reactant and Al(OH)₃ is excess reactant

Moles of Al(OH)₃ in excess = 0.1196 mol − 0.0833 mol = 0.0363 mol

Moles of Al(OH)₃ in excess = 0.0363 mol

Mass of Al(OH)₃ in excess = (0.0363 mol) × (78 g/mol) = 2.83 g

Mass of Al(OH)₃ in excess = 2.83 g

According to stoichiometry:

3 mol of H₂SO₄ produce 1 mol of Al₂(SO₄)₃

Thus, 0.125 mol of H₂SO₄ produce:

(0.125 mol H₂SO₄) × (1 mol Al₂(SO₄)₃ / 3 mol H₂SO₄) = 0.0417 mol Al₂(SO₄)₃

Moles of Al₂(SO₄)₃ = 0.0417 mol

The molar mass of Al₂(SO₄)₃ is 342 g/mol

Therefore,

Mass of Al₂(SO₄)₃ = (0.0417 mol) × (342 g/mol) = 14.26 g

Mass of Al₂(SO₄)₃ = 14.26 g

Answers:

1) Balanced chemical equation: 3H₂SO₄(aq) + 2Al(OH)₃(s) → Al₂(SO₄)₃(aq) + 6H₂O(l)

2) The limiting reactant is H₂SO₄

3) Moles of Al(OH)₃ in excess = 0.0363 mol; Mass of Al(OH)₃ in excess = 2.83 g

4) 14.26 grams of aluminum sulfate will be produced