If you start with 390 grams of Al2(SO3)3 , how many grams of Na2SO3 would be produced?

Molar mass of Al₂(SO₃)₃ = (26.98×2 + 32.06×3 + 16.00×9) g/mol = 294.14 g/mol
Molar mass Na₂SO₃ = (22.99×2 + 32.06 + 16.00×3) g/mol = 126.04 g/mol

Al₂(SO₃)₃ + 6 NaOH → 3 Na₂SO₃ + 2 Al(OH)₃
Mole ratio Al₂(SO₃)₃ : Na₂SO₃ = 1 : 3

No. of moles of Al₂(SO₃)₃ reacted = (390 g) / (294.14 g/mol) = 1.326 mol
No. of moles of Na₂SO₃ produced = (1.326 mol) × 3 = 3.978 mol
Mass of Na₂SO₃ produced = (3.978 mol) × (126.04 g/mol) = 501 g