Find the mass of AlCl₃ that is produced when 10.0 grams of Al₂O₃ react with 10.0 g of HCl according to the following equation. Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(aq)

Question

Quizplus · Accepted Answer

To solve this problem, we need to use stoichiometry. We are given the mass of Al₂O₃ which is a reactant and we need to find the mass of AlCl₃ which is a product. We are also given the balanced chemical equation. According to the balanced equation: 1 mole Al₂O₃ reacts with 6 moles of HCl to produce 2 moles of AlCl₃. We need to convert the masses of the reactant (Al₂O₃) and the given reagent (HCl) into moles and use the stoichiometric coefficients to determine the mass of AlCl₃ produced. First, we need to find the limiting reagent. The amount of AlCl₃ produced will be limited by the reagent that produces the least amount of moles of AlCl₃. To do that, we need to find the number of moles of each reactant: moles of Al₂O₃ = mass / molar mass = 10.0 g / 426.41 g/mol = 0.0235 mol moles of HCl = mass / molar mass = 10.0 g / 36.46 g/mol = 0.274 mol Based on the balanced equation, 6 moles of HCl react with 1 mole of Al₂O₃ to produce 2 moles of AlCl₃. So, if all the Al₂O₃ reacts, then the max number of moles of AlCl₃ that can be produced is: moles of AlCl₃ = (0.0235 mol Al₂O₃) x (2 moles AlCl₃ / 1 mole Al₂O₃) = 0.0470 mol On the other hand, if all the HCl reacts, then the max number of moles of AlCl₃ that can be produced is: moles of AlCl₃ = (0.274 mol HCl) x (2 moles AlCl₃ / 6 moles HCl) = 0.0913 mol Therefore, HCl is the limiting reagent since it produces the least number of moles of AlCl₃. Now we can use the amount of HCl to calculate the amount of AlCl₃ formed: moles of AlCl₃ produced = (0.274 mol HCl) x (2 moles AlCl₃ / 6 moles HCl) = 0.0913 mol mass of AlCl₃ produced = moles x molar mass = 0.0913 mol x 133.33 g/mol = 12.2 g Therefore, the mass of AlCl₃ produced is 12.2 g which corresponds to answer choice (C).

Find the Mass of AlCl₃ That Is Produced When 10