Find the mass of AlCl₃ that is produced when 25.0 grams of Al₂O₃ react with excess HCl according to the following equation.
Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)

Question

Find the mass of AlCl₃ that is produced when 25.0 grams of Al₂O₃ react with excess HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)

Quizplus · Accepted Answer

First, calculate the moles of Al₂O₃: Molar mass of Al₂O₃ = 101.96 g/mol. Moles of Al₂O₃ = 25.0 g / 101.96 g/mol = 0.245 moles. From the balanced equation, 1 mole of Al₂O₃ produces 2 moles of AlCl₃. Therefore, 0.245 moles of Al₂O₃ will produce 0.245 * 2 = 0.49 moles of AlCl₃. Molar mass of AlCl₃ = 133.33 g/mol. Mass of AlCl₃ = 0.49 moles * 133.33 g/mol = 65.4 g.

Find the Mass of AlCl₃ That Is Produced When 25