Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen 4Al(s) + 30_2(g) rightarrow 2A1_20_3(s) A mixture of 200.7 g of aluminum (M = 26.98 g/m

First, we need to determine which reactant is limiting. We can do this by calculating the moles of each reactant: Number of moles of aluminum = 200.7 g / 26.98 g/mol = 7.43 mol Number of moles of oxygen = 190.0 g / 32.00 g/mol = 5.94 mol Since the ratio of aluminum to oxygen in the balanced chemical equation is 4:3, we can see that aluminum is the limiting reactant because there are not enough moles of oxygen to react with all of the aluminum. Next, we need to use the stoichiometry of the balanced chemical equation to determine the amount of aluminum oxide that can be formed. From the equation, we see that 4 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminum oxide. Therefore: Number of moles of aluminum oxide = (7.43 mol / 4 mol) * 2 mol = 3.73 mol Finally, we can convert the number of moles of aluminum oxide to grams using the molar mass of aluminum oxide: Mass of aluminum oxide = 3.73 mol * 101.96 g/mol = 380.8 g Therefore, the mass of aluminum oxide that can be formed is 380.8 g.