Answer:
A. 1.88 mol H₂
B. 182 g Al₂(SO₄)₃
C. 54.8%
Explanation:
2 Al + 3 H₂SO₄ ⇒ Al₂(SO₄)₃ + 3 H₂
A. Convert grams of Al to moles. The molar mass is 26.98 g/mol.
(33.8 g)/(26.98 g/mol) = 1.253 mol Al
Use stoichiometry to convert moles of Al to moles of H₂. Looking at the equation, you can see that for every 2 mol of Al consumed, 3 moles of H₂ is produced. Use this relationship.
(1.253 mol Al) × (3 mol H₂)/(2 mol Al) = 1.879 mol H₂
You will produce 1.88 mol of H₂ gas.
B. Again, use stoichiometry. For every 3 moles of H₂SO₄ consumed, 1 mole of Al₂(SO₄)₃ is produced.
(1.60 mol H₂SO₄) × (1 mol Al₂(SO₄)₃/3 mol H₂SO₄) = 0.533 mol Al₂(SO₄)₃
Convert moles of Al₂(SO₄)₃ to grams. The molar mass is 342.15 g/mol.
(0.533 mol) × (342.15 g/mol) = 182.48 g Al₂(SO₄)₃
You will produce 182 g of Al₂(SO₄)₃.
C. Calculate percent yield by dividing the actual yield by the theoretical yield. Multiply by 100%.
(100.0/182.48) × 100% = 54.8%
The percent yield is 54.8%.
