Given the following balanced equation, determine how many moles of iron (III) sulfide, Fe2S3 (also known as “fool’s gold) form when 0.876 mol of Fe react. 2 Fe (s) + 3 S (s) --> Fe2S3 (s)

Answer 1

Hello, Mahnoor,

2 Fe (s) + 3 S (s) --> Fe₂S₃ (s)

The balanced equation tells us we should expect to produce 1 mole of iron sulfide for every 2 moles of iron. This is essentialy the conversion factor we need to answer the question.

(1 mole Fe₂O₃)/(2 moles Fe)

Therefore,

(0.876 moles Fe)*(1 mole Fe₂O₃)/(2 moles Fe) = 0.438 moles Fe₂O₃

Bob