Combining 0.280 mol of Fe2O3 with excess carbon produced 19.4 g of Fe.
To determine how many grams of Fe can be produced from 0.280 mol of Fe2O3, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation for the reaction between Fe2O3 and carbon (C) to produce Fe is:
Fe2O3 + 3C → 2Fe + 3CO
According to the balanced equation, we see that for every 2 moles of Fe2O3, we can produce 2 moles of Fe.
Step 1: Calculate the number of moles of Fe that can be produced from 0.280 mol of Fe2O3:
0.280 mol Fe2O3 × (2 mol Fe / 2 mol Fe2O3) = 0.280 mol Fe
Step 2: To find the mass of Fe produced, we need to use the molar mass of Fe, which is 55.85 g/mol.
Mass of Fe = Number of moles of Fe × Molar mass of Fe
Mass of Fe = 0.280 mol Fe × 55.85 g/mol
Mass of Fe ≈ 15.598 g
Therefore, 0.280 mol of Fe2O3 combined with excess carbon would produce approximately 15.598 g of Fe.