Question

Sulfur combines with iron to form iron(II) sulfide. In an experiment, 15.24 g of Fe are allowed to react with 17.34 g of sulfur What is the limiting reactant? The reactant in excess? What mass of FeS is created?

Im really struggling in chemistry right now and I have no idea how to solve these problems! Please help!

Answer 1

`24.0 g`

Explanation:

`Fe + S + FeS`
`n(Fe) = 15.24 / 55.8 = 0.273 mol`
`n(S) = 17.34 / 32.0 = 0.542 mol`

A simple way to check which is the limiting reagent is to divide moles of substances by its mole coefficient in the reaction. In this case we need not divide as its 1 for both `Fe` and `S`. After doing so, we see which is the bigger value and in this case its sulfur. Thus sulfur is the excess and iron is the limiting reagent. Since the limiting reagent is used up complete we use this moles to calculate the products of any reaction.

Ratio of `Fe` to `FeS` is `1:1`.

Hence `n(FeS) = 0.273 mol`
`m(FeS) = 0.273 * (55.8+32.0) = 24.0 g`