What is the volume of 6.0M sulfuric acid needed to neutralize the excess hydroxide?

Molar mass of Al = 27.0 g/mol

Consider the reaction of Al with KOH :
2Al + 2KOH + 6H₂O → 2K[Al(OH)₄] + 3H₂
OR: Mole ratio Al : KOH = 2 : 2 = 1 : 1

Initial number of moles of Al = (1.00 g) / (27 g/mol) = 0.037 mol
Initial number of moles of KOH = (3.0 mol/L) × (25.00/1000 L) = 0.075 mol > 0.037 mol
Hence, KOH is in excess, and Al is the limiting reactant (limiting reagent).

Number of moles of KOH reacted = Number of moles of Al reacted = 0.037 mol
Number of moles of excess KOH = (0.075 - 0.037) mol = 0.038 mol

Consider the reaction of excess KOH and H₂SO₄ :
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
OR: Mole ratio KOH : H₂SO₄ = 2 : 1

Number of moles of KOH reacted = 0.038 mol
Number of moles of H₂SO₄ needed = (0.038 mol) × (1/2) = 0.019 mol
Volume of H₂SO₄ needed = (0.019 mol) / (6.0 mol/L) = 0.0032 L = 3.2 mL