Final answer:
To produce 32.00 g of Fe₂O₃, 22.38 grams of iron are needed, based on the molar masses and the stoichiometric ratio from the balanced chemical equation.
Step-by-step explanation:
To determine the mass of Fe required to produce 32.00 g of Fe₂O₃, we start by using the molar mass of Fe₂O₃ to find the number of moles of Fe₂O₃ produced. One mole of Fe₂O₃ has a molar mass of 159.70 grams, so we can divide the mass of Fe₂O₃ by its molar mass to get the number of moles:
Number of moles of Fe₂O₃ = 32.00 g ÷ 159.70 g/mol = 0.2004 mol
The balanced chemical equation tells us that 4 moles of Fe produce 2 moles of Fe₂O₃. Using this stoichiometric ratio, we can find the number of moles of iron needed:
0.2004 mol Fe₂O₃ × (4 mol Fe / 2 mol Fe₂O₃) = 0.4008 mol Fe
Now, we multiply the number of moles of iron by the molar mass of iron to get the mass:
Mass of Fe = 0.4008 mol × 55.85 g/mol = 22.38 g
So, 22.38 grams of iron are required to produce 32.00 grams of Fe₂O₃, according to the given chemical reaction.