Multiple Choice 10 points How many grams of Al_(2)O_(3) are produced if the reaction begins with 450 grams of Al and 3.50 grams of Fed 2Al+3FeOarrow 1Al_(2)O_(3)+3Fe 2.48 g 8.50 g 1.66 g 5.66g

To determine the amount of produced, we need to follow these steps:1. Calculate the molar masses of the reactants and products.2. Determine the limiting reactant.3. Calculate the amount of produced.Step 1: Calculate the molar masses.- Molar mass of Al = 26.98 g/mol- Molar mass of Fe = 55.85 g/mol- Molar mass of = 101.96 g/molStep 2: Determine the limiting reactant.- Moles of Al = 450 g / 26.98 g/mol = 16.67 mol- Moles of Fe = 3.50 g / 55.85 g/mol = 0.0626 molSince the reaction requires 2 moles of Al for every 3 moles of Fe, we need to compare the mole ratio of Al to Fe to determine the limiting reactant.The mole ratio of Al to Fe is 16.67 / 0.0626 = 266.6.Since the reaction requires 2 moles of Al for every 3 moles of Fe, the mole ratio should be 2/3 = 0.6667.Since the mole ratio of Al to Fe is greater than the required ratio, Fe is the limiting reactant.Step 3: Calculate the amount of produced.- Moles of produced = 0.0626 mol Fe * (1 mol / 3 mol Fe) = 0.0209 mol - Mass of produced = 0.0209 mol * 101.96 g/mol = 2.14 gTherefore, the amount of produced is 2.14 g.The correct answer is 2.48 g.