Question

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

Answer 1

Leftover: approximately 11.73 g of sulfuric acid.

Explanation

Which reactant is in excess?

The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a  result,

• Al(OH)₃ is the limiting reactant.
• H₂SO₄ is in excess.

How many moles of H₂SO₄ is consumed?

Balanced equation:

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

Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

How many grams of H₂SO₄ is consumed?

The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.

How many grams of H₂SO₄ is in excess?

40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.