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. 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

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular 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 limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.