If 36 grams of aluminum react with an excess of oxygen, as shown in the balanced chemical equation below, how many grams of aluminum oxide can be formed? (1 point)
molar mass of Al = 26.98 molar mass of Al2O3 = 101.96 36 / 26.98 = 1.3343 moles Al set up a proportion : 1.3343 / 4 = x / 2 (let x = moles of Al2O3) x = 0.66716 moles of Al2O3 0.66716 * 101.96 = 69.024 g of Al2O3
