Final answer:
To find out how many grams of oxygen are needed to react with 8.74 g of iron to produce 12.5 g of iron(III) oxide (rust), subtract the mass of iron from the mass of rust, resulting in 3.76 grams of oxygen required.
Step-by-step explanation:
To determine how many grams of oxygen are needed to produce 12.5 g of iron(III) oxide from 8.74 g of iron, we use the law of conservation of mass which states that the mass of the reactants in a chemical reaction must equal the mass of the products. The question is related to stoichiometry, so first, we identify the molecular weights of iron (Fe) and iron(III) oxide (Fe2O3).
Iron has an atomic mass of 55.85 g/mol, while one mole of iron(III) oxide has a mass of approximately 159.69 g/mol. Stoichiometry calculations are done by starting with the mass of iron, converting that to moles, and using the molar ratio from the balanced equation to determine the moles of oxygen involved. However, since the desired information here is mass, we can directly calculate the mass of oxygen by subtracting the mass of iron from the mass of rust produced.
Therefore, the mass of oxygen needed is calculated as follows:
Mass of rust - Mass of iron = Mass of oxygen
12.5 g - 8.74 g = 3.76 g
So, 3.76 grams of oxygen are needed for the reaction to produce 12.5 g of iron(III) oxide.