25.0 g Al2O3 * (1 mol Al2O3 / 101.96 g Al2O3) = 0.2454 mol Al2O3
According to the balanced chemical equation, 2 mol of AlCl3 are produced for every 2 mol of Al2O3. Therefore, the number of moles of AlCl3 produced can be determined:
0.2454 mol Al2O3 * (2 mol AlCl3 / 2 mol Al2O3) = 0.2454 mol AlCl3
Finally, we calculate the mass of AlCl3 produced by multiplying the number of moles by its molar mass:
0.2454 mol AlCl3 * (133.34 g AlCl3 / 1 mol AlCl3) = 32.68 g AlCl3
Therefore, 32.68 grams of AlCl3 is produced when 25.0 grams of Al2O3 reacts with HCl.
