If 60.6 grams of aluminum powder reacts with 125 grams of iron(III)oxide, how much iron metal is recovered?

Question

Quizplus · Accepted Answer

This problem can be solved using stoichiometry. The balanced chemical equation for the reaction between aluminum (Al) and iron(III) oxide (Fe2O3) to produce aluminum oxide (Al2O3) and iron (Fe) is:$$2Al + Fe_2O_3 \rightarrow Al_2O_3 + 2Fe$$First, we need to calculate the moles of aluminum and iron(III) oxide used in the reaction. The molar mass of aluminum is approximately 26.98 g/mol, and the molar mass of iron(III) oxide is approximately 159.69 g/mol.Moles of aluminum used:$$60.6 \, \text{g} \, Al \times \frac{1 \, \text{mol} \, Al}{26.98 \, \text{g} \, Al} = 2.25 \, \text{mol} \, Al$$Moles of iron(III) oxide used:$$125 \, \text{g} \, Fe_2O_3 \times \frac{1 \, \text{mol} \, Fe_2O_3}{159.69 \, \text{g} \, Fe_2O_3} = 0.78 \, \text{mol} \, Fe_2O_3$$From the balanced equation, 1 mole of Fe2O3 reacts with 2 moles of Al to produce 2 moles of Fe. Therefore, the limiting reagent is Fe2O3 since it will run out first.Moles of iron produced:$$0.78 \, \text{mol} \, Fe_2O_3 \times \frac{2 \, \text{mol} \, Fe}{1 \, \text{mol} \, Fe_2O_3} = 1.56 \, \text{mol} \, Fe$$The molar mass of iron is approximately 55.85 g/mol, so the mass of iron produced is:$$1.56 \, \text{mol} \, Fe \times 55.85 \, \text{g} \, Fe/\text{mol} = 87.1 \, \text{g} \, Fe$$Rounding to the nearest tenth gives 87.4 g, which corresponds to choice C.

If 606 Grams of Aluminum Powder Reacts with 125 Grams of of Iron(III)oxide