Final answer:
To produce 1.00 g of iron, 0.3507 g of aluminum is needed, based on the stoichiometry of a single replacement reaction between aluminum and iron (III) oxide.
Step-by-step explanation:
To determine the mass of aluminum metal that reacts to give 1.00 g of iron, you would use the stoichiometry of the reaction between aluminum (Al) and iron (III) oxide (Fe₂O₃), where aluminum displaces iron according to the following equation:
Fe₂O₃(s) + 2Al(s) → 2Fe(s) + Al₂O₃(s)
From the balanced equation, you can see that 2 moles of aluminum produce 2 moles of iron. Given that 0.013 moles of iron have a mass of 1.00 g, and the molar mass of Fe (iron) is 55.85 g/mol, you'd need 0.013 moles of aluminum, because the mole ratio between Fe and Al is 1:1. Since the molar mass of Al (aluminum) is 26.98 g/mol, the mass of aluminum required to produce 1.00 g of Fe would be calculated as:
Mass of Al = Moles of Al × Molar mass of Al = 0.013 moles × 26.98 g/mol = 0.3507 g
Therefore, to get 1.00 g of iron, 0.3507 g of aluminum is needed.