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

Al (OH) 3 + H2SO4 → Al2 (SO4) 3 + H2O

The table shows the calculated amounts of reactants and products when the reaction was conducted in a laboratory.

Initial Mass and Yield

Sulfuric Acid Aluminum Hydroxide

Initial Amount of Reactant 40 g 15 g

Theoretical Yield of Water from Reactant 14.69 g 10.38 g

What is the approximate amount of the leftover reactant?

11.73 g of sulfuric acid

10.33 g of sulfuric acid

11.12 g of aluminum hydroxide

13.67 g of aluminum hydroxide

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.