Find the mass of AlCl3 that is produced when 25.0g of Al2O3 react with HCl

Answer 1

Al₂O₃ + 6HCl ==> 2AlCl₃ + 3H₂O ... balanced equation

moles Al2O₃ present = 25.0 g x 1 mol/101.96 g = 0.245 moles

moles AlCl₃ produced (assuming HCl not limiting) = 0.245 mol Al2O3 x 2 mol AlCl₃/mol Al₂O₃ = 0.490 moles

mass AlCl₃ produced = 0.490 moles AlCl₃ x 133 g/mol = 65.2 g AlCl₃ produced