$\mathrm{Fe} 2 \mathrm{O} 3+2 \mathrm{Al}=2 \mathrm{Fe}+\mathrm{Al} 2 \mathrm{O} 3$ Assuming you have $53.96 \mathrm{~g}$ of $\mathrm{Al}$ and 3 moles of iron (ii) oxide. what is the limiting reagent? And what is the mass of $\mathrm{Al} 2 \mathrm{O} 3$ produced?

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate moles of Al**: First, we need to calculate the number of moles of aluminum (Al) using its molar mass. The molar mass of Al is $26.98 \, ext{g/mol}$. ‖ ‖ 2 ⋮ **Mole ratio comparison**: The balanced chemical equation shows that 2 moles of Al react with 1 mole of $ ext{Fe}_2 ext{O}_3$. We compare the mole ratio of Al to $ ext{Fe}_2 ext{O}_3$ to find the limiting reagent. ‖ ‖ 3 ⋮ **Identify limiting reagent**: Using the stoichiometry of the reaction, we determine which reactant will be consumed first and thus limit the reaction. ‖ ‖ 4 ⋮ **Calculate mass of $ ext{Al}_2 ext{O}_3$ produced**: Once the limiting reagent is identified, we use its amount to calculate the mass of aluminum oxide ($ ext{Al}_2 ext{O}_3$) produced, based on the stoichiometry of the balanced equation. ‖ ∻Answer∻ ⚹ The limiting reagent is Al, and the mass of $ ext{Al}_2 ext{O}_3$ produced is [insert calculated mass here] grams. ⚹ ∻Key Concept∻ ⚹ Limiting reagent determines the amount of product formed in a chemical reaction. ⚹ ∻Explanation∻ ⚹ The reactant that is completely consumed in a reaction limits the amount of product that can be formed, and is thus called the limiting reagent.◊What is the molar mass of Al2O3 (aluminum oxide)?⍭ Generate me a similar question◊⚹

Sia