Final answer:
To produce 32.00 g of Fe₂O₃, 22.34 g of Fe is required. This is determined by converting the given mass of Fe₂O₃ to moles, using the stoichiometric mole ratio from the balanced chemical equation, and then converting moles of Fe to grams.
Step-by-step explanation:
To find the mass of Fe required to produce 32.00 g of Fe2O3, we begin by converting the mass of Fe2O3 to moles. The molar mass of Fe2O3 is 159.70 g/mol, so we use the following ratio to find the number of moles:
32.00 g Fe2O3 × (1 mol Fe2O3 / 159.70 g Fe2O3) = 0.200 mol Fe2O3
According to the balanced equation for the formation of iron(III) oxide:
4 Fe + 3 O2 → 2 Fe2O3
we see that 4 moles of Fe produce 2 moles of Fe2O3. This gives us a mole ratio of 4 moles Fe : 2 moles Fe2O3. Using this ratio, we can find the number of moles of Fe needed to produce 0.200 mol of Fe2O3:
0.200 mol Fe2O3 × (4 mol Fe / 2 mol Fe2O3) = 0.400 mol Fe
Finally, we convert moles of Fe to grams using the molar mass of Fe, which is 55.85 g/mol:
0.400 mol Fe × (55.85 g Fe / 1 mol Fe) = 22.34 g Fe
Therefore, 22.34 g of Fe is required to produce 32.00 g of Fe2O3.