Final answer:
To dissolve a 14.7 g block of aluminum, the minimum amount of sulfuric acid needed, calculated by stoichiometry, is 80.17 g.
Step-by-step explanation:
Sulfuric acid reacts with aluminum metal according to the balanced chemical equation: 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g). To determine the minimum amount of sulfuric acid needed to dissolve a 14.7 g aluminum block, we employ stoichiometry and calculate the molar masses of aluminum (Al) and sulfuric acid (H₂SO₄).
First, we calculate the molar mass of Al, which is approximately 26.98 g/mol. With a mass of 14.7 g, the moles of Al is approximately 14.7 g / 26.98 g/mol = 0.5449 mol. Using the stoichiometry of the reaction where 2 moles of Al react with 3 moles of H₂SO₄, we find that we need 0.5449 mol Al × (3 moles H2SO4 / 2 moles Al) = 0.8174 mol H2SO4. Next, we calculate the molar mass of H₂SO₄, which is 2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.08 g/mol. Therefore, the mass of H₂SO₄ needed is 0.8174 mol × 98.08 g/mol = 80.17 g. Thus, 80.17 grams of sulfuric acid is the minimum amount needed to completely dissolve a 14.7 g block of aluminum.