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.

Which reactant is in excess?

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

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.