To produce aluminum oxide Al2O3 from aluminum and oxygen gas, the balanced chemical equation is:4Al + 3O2 2Al2O3Now, let's calculate how many grams of Al2O3 can be produced from 25 grams of aluminum reacting with excess oxygen gas.First, we need to determine the molar mass of aluminum Al and aluminum oxide Al2O3 .Aluminum Al has a molar mass of 26.98 g/mol.Aluminum oxide Al2O3 has a molar mass of 2 26.98 + 3 16.00 = 53.96 + 48.00 = 101.96 g/mol.Next, we'll use stoichiometry to find out how many grams of Al2O3 can be produced from 25 grams of Al.1. Convert the mass of Al to moles:moles of Al = mass of Al / molar mass of Almoles of Al = 25 g / 26.98 g/mol = 0.926 moles of Al2. Use the balanced chemical equation to find the moles of Al2O3 produced:From the balanced equation, 4 moles of Al produce 2 moles of Al2O3.So, 0.926 moles of Al will produce 0.926 * 2 / 4 = 0.463 moles of Al2O3.3. Convert the moles of Al2O3 to grams:mass of Al2O3 = moles of Al2O3 molar mass of Al2O3mass of Al2O3 = 0.463 moles 101.96 g/mol = 47.22 gTherefore, 47.22 grams of Al2O3 can be produced from 25 grams of aluminum reacting with excess oxygen gas.