Final answer:
To find the mass of Fe₂O₃ produced, we balance the chemical equation, calculate the molar amounts of reactants, determine the limiting reagent, and then use stoichiometry to convert the moles of product to grams, with the result being 73.6 g. So, the correct option is b) 73.6 g.
Step-by-step explanation:
The question requires us to apply stoichiometry to determine the mass of Fe₂O₃ produced from the reaction of FeS₂ with O₂. We need to start by balancing the chemical equation to establish the stoichiometric relationship between the reactants and products. Once the equation is balanced, we will use the provided masses of the reactants to calculate the moles, apply the molar ratio to find the moles of Fe₂O₃ produced, and then convert these moles back to grams.
The balanced chemical equation for the reaction is: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
First, we need to calculate the moles of FeS₂ and O₂ we have:
• Moles of FeS₂ = mass (g) / molar mass (g/mol)
• Moles of O₂ = mass (g) / molar mass (g/mol)
Then we figure out which reactant is the limiting reagent, which will dictate the amount of product formed. We use the molar ratio from the balanced equation to find the number of moles of Fe₂O₃ produced by the limiting reagent, and from there, we convert moles of Fe₂O₃ to grams using its molar mass.
After following through with the calculations, assuming that FeS₂ is the limiting reagent, we find that the mass of Fe₂O₃ produced is 73.6 g, which corresponds to option b. However, this is a theoretical yield and in the practical scenario the actual yield may vary due to various factors.