Final answer:
The mass of iron required to react with 5.0 moles of sulfur is 2228 grams, and the mass of iron(II) sulfide produced is 439.55 grams.
Step-by-step explanation:
To answer the student's questions, we need to understand and apply stoichiometry, which is the calculation of reactants and products in chemical reactions.
a) To find the mass of iron needed to react with 5.0 moles of sulfur (S), we use the mole ratio from the balanced equation 8 Fe + S8 → 8 FeS. This ratio is 8:1 for Fe:S. We can then calculate the mass of iron required:
5.0 moles of S × (8 moles Fe / 1 mole S) × (55.85 g Fe / 1 mole Fe) = 2228 grams of Fe
b) To find out how many grams of FeS are produced, we again use stoichiometry. From the balanced equation, 1 mole of S produces 1 mole of FeS.
Therefore, 5.0 moles of S will produce 5.0 moles of FeS.
The molar mass of FeS is about 87.91 g/mol. Thus:
5.0 moles of FeS × (87.91 g FeS / 1 mole FeS) = 439.55 grams of FeS