Reareproduced if the reaction begins with 3.0 grams of Al and 3.0 grams of FeO 2Al+3FeOarrow 1Al_(2)O_(3)+3Fe 1035 0.698 27.53 g 2335 Typoints

To determine the amount of product formed in the reaction, we need to follow these steps:<br /><br />1. Calculate the molar masses of the reactants and products.<br />2. Convert the masses of the reactants to moles.<br />3. Determine the limiting reactant.<br />4. Calculate the amount of product formed.<br /><br />Given reaction: \(2Al + 3FeO \rightarrow Al_2O_3 + 3Fe\)<br /><br />### Step 1: Calculate Molar Masses<br /><br />- Molar mass of Al: \(26.98 \, \text{g/mol}\)<br />- Molar mass of FeO: \(55.85 \, \text{g/mol} + 16.00 \, \text{g/mol} = 71.85 \, \text{g/mol}\)<br />- Molar mass of Al₂O₃: \(2 \times 26.98 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} = 101.96 \, \text{g/mol}\)<br />- Molar mass of Fe: \(55.85 \, \text{g/mol}\)<br /><br />### Step 2: Convert Masses to Moles<br /><br />- Moles of Al: \(\frac{3.0 \, \text{g}}{26.98 \, \text{g/mol}} = 0.111 \, \text{mol}\)<br />- Moles of FeO: \(\frac{3.0 \, \text{g}}{71.85 \, \text{g/mol}} = 0.042 \, \text{mol}\)<br /><br />### Step 3: Determine the Limiting Reactant<br /><br />From the balanced equation, the molar ratio of Al to FeO is 2:3.<br /><br />- Required moles of FeO for 0.111 moles of Al: \(0.111 \, \text{mol} \times \frac{3}{2} = 0.1665 \, \text{mol}\)<br /><br />Since we only have 0.042 moles of FeO, FeO is the limiting reactant.<br /><br />### Step 4: Calculate the Amount of Product Formed<br /><br />Using the limiting reactant (FeO):<br /><br />- Moles of Al₂O₃ produced: \(0.042 \, \text{mol FeO} \times \frac{1 \, \text{mol Al}_2\text{O}_3}{3 \, \text{mol FeO}} = 0.014 \, \text{mol Al}_2\text{O}_3\)<br /><br />- Mass of Al₂O₃ produced: \(0.014 \, \text{mol} \times 101.96 \, \text{g/mol} = 1.427 \, \text{g}\)<br /><br />Therefore, the amount of Al₂O₃ produced is \(1.427 \, \text{g}\).