Question

In one experiment, 7.62 g of Fe are allowed to react with 8.67 g of S. What is the limiting reagent, what is the mass of FeS formed, and how much excess reagent is left, respectively?

I am not sure how to solve this.
I started by doing (7.62 g Fe/55.85 g)=.137 g

(8.67g S/32.07 g)=.270

I divided both by .137 because it was the lowest number.
I got a ratio of 1Fe:2S
Now what should I do?

Answer 1

I obtained 0.136 (and that's MOLES Fe and not g iron).
Then 0.270 mols S.

0.136 mols Fe x (1 mol S/1 mol Fe) = 0.136 mol S. Do you have that much S? Yes, so Fe is the limiting reagent and S is the excess reagent.

Part B. So 0.136 mol FeS formed.
g FeS = mols FeS x molar mass FeS.