Answers and show work please

Question

450mL of a 0.85M solution of H2SO4 react with aluminum. What mass of aluminum sulfate can be produced?

Answer 1

3H₂SO₄(aq) + 2Al(s) ==> Al₂(SO₄)₃ + 3H₂(g) ... balanced equation

moles H₂SO₄ present = 450 ml x 1 L / 1000 ml x 0.85 mol/L = 0.3825 mols H₂SO₄

mass Al₂(SO₄)₃ produced = 0.3825 mol H₂SO₄ x 1 mol Al₂(SO₄)₃ / 3 mol H₂SO₄ x 342 g/mol = 43.6 g Al₂(SO₄)₃ = 44 g Al₂(SO₄)₃ (to 2 sig. figs.)