What amount of Al2O3 is produced from the reaction of 3.0 mol Al with 2.0 mol Fe2O3?
2Al + Fe2O3 è Al2O3 + 2Fe

Question

BlackTom AI · Accepted Answer

To tackle this stoichiometry question, first restate the balanced equation: 2 Al + Fe2O3 → Al2O3 + 2 Fe.
Option by option analysis:
Option 1: 1.5 mol. This aligns with the stoichiometry: 2 mol Al produce 1 mol Al2O3. Since Al is the limiting reagent (we have 3.0 mol Al but would need 4.0 mol Al to fully react with 2.0 mol Fe2O3), the amount of Al2O3 produced is (3.0 mol Al) × (1 mol Al2O3 / 2 mol Al) = 1.5 mol Al2O3. This reasoning makes 1.5 mol the correct theoretical yield.
Option 2: 2.0 mol. This would imply enough Al2O3 is formed if 4.0 mol Al were used or if Fe2O3 were limiting, but we do not have that much Al; hence this overestimates what can be produced given the actual limiting reagent. It ignores the 2:1 Al to Al2O3 ratio.
Option 3: 3.5 mol. This is inconsistent with the reaction stoichiometry; it would require more Al2O3 per mole of Al than the balanced equation allows. It also suggests a yield greater than what the 3.0 mol of Al can support.
Option 4: 6.0 mol. This is far beyond the theoretical maximum from the available reactants and ignores the 2:1 ratio. It would require either twice as much Al2O3 per Al than the reaction provides or overcounting the product.
In summary, the limiting reagent is aluminum (Al), and the correct theoretical yield for Al2O3 is (3.0 mol Al) ÷ (2 mol Al per 1 mol Al2O3) = 1.5 mol Al2O3, which matches Option 1.

What amount of Al2O3 is produced from the reaction of 3.0 mol Al with 2.0 mol Fe2O3? 2Al + Fe2O3 è Al2O3 + 2Fe

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