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