Final answer:
To determine the theoretical yield of aluminum oxide, we need to identify the limiting reactant in the given equation. The aluminum is the limiting reactant, and it can produce approximately 0.046 moles or 4.7 g of Al2O3.
Step-by-step explanation:
To determine the theoretical yield of aluminum oxide, we need to identify the limiting reactant in the given equation. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. In this case, we compare the amounts of aluminum oxide that can be produced from both the aluminum and the oxygen.
The balanced equation tells us that 4 moles of Al can produce 2 moles of Al2O3, while 30₂ can produce 2 moles of Al2O3. We can calculate the number of moles of Al2O3 that can be produced from both reactants using their respective masses.
From the given information, we know that the aluminum could produce 4.7 g of Al2O3 and the oxygen could produce 17 g of Al2O3. To convert these masses to moles, we need to use the molar mass of Al2O3, which is 101.96 g/mol. By dividing the given masses by the molar mass, we find that the aluminum could produce approximately 0.046 moles of Al2O3 (4.7 g / 101.96 g/mol) and the oxygen could produce approximately 0.166 moles of Al2O3 (17 g / 101.96 g/mol).
Since we need a balanced equation with equal number of moles of reactants for a complete reaction, the reactant that produces the lesser amount of moles of Al2O3 will be the limiting reactant. In this case, the aluminum produces fewer moles of Al2O3 compared to the oxygen, which means aluminum is the limiting reactant. Therefore, the theoretical yield of aluminum oxide is the amount produced from the limiting reactant, which is approximately 0.046 moles or 4.7 g of Al2O3.