Final answer:
The mass of iron oxide (Fe2O3) that reacted with excess carbon monoxide to produce 8.65 g of carbon dioxide is approximately 10.46 g, as found by using stoichiometric calculations based on the balanced chemical equation.
Step-by-step explanation:
To determine the mass of iron oxide (Fe2O3) that reacted with excess carbon monoxide (CO), we must first understand the stoichiometry of the reaction Fe2O3 + 3CO → 2Fe + 3CO2. The molar ratio of Fe2O3 to CO2 is 1:3. Therefore, for every mole of Fe2O3 that reacts, 3 moles of CO2 are produced.
To find the mass of Fe2O3 that reacted, we must first convert the mass of CO2 produced to moles. The molar mass of CO2 is approximately 44.01 g/mol, so:
8.65 g CO2 × (1 mol CO2 / 44.01 g CO2) = 0.1966 moles CO2
Using the molar ratio from the balanced equation (1 mole of Fe2O3 per 3 moles of CO2), the moles of Fe2O3 that reacted can be calculated as follows:
0.1966 moles CO2 × (1 mole Fe2O3 / 3 moles CO2) = 0.06553 moles Fe2O3
Finally, to get the mass of Fe2O3 that reacted, we multiply the moles of Fe2O3 by its molar mass (approximately 159.69 g/mol):
0.06553 moles Fe2O3 × 159.69 g/mol = 10.46 g Fe2O3
Therefore, the mass of iron oxide that reacted with excess carbon monoxide to produce 8.65 g of carbon dioxide is approximately 10.46 g.