Question

POINTS!! 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?

Answer 1

Hello!

1) What is the limiting reactant?

The chemical equation for the reaction is the following:

Fe + S → FeS

To determine the limiting reactant, we will determine the moles of each reactant and divide it by the coefficient in the chemical equation (In this case is I for each one)


`moles Fe=15,24 g Fe * (1 mol Fe)/(55,845 g Fe)=0,2729 mol Fe \\ \\ moles S=17,34 g S* (1 mol S)/(32,065 g S)= 0,5408 mol S`

Now, the reactant with the less number of moles is Fe, so Fe is the limiting reactant.

2) The reactant in excess.

From the chemical reaction:

Fe + S → FeS

The reactant in excess is the one that is in the highest amount. To determine which is the reactant in excess we need to calculate the moles of each reactant and divide by the coefficient in the chemical reaction:


`moles Fe=15,24 g Fe * (1 mol Fe)/(55,845 g Fe)=0,2729 mol Fe \\ \\ moles S=17,34 g S* (1 mol S)/(32,065 g S)= 0,5408 mol S`

The reactant in excess is S, the amount that is left after the reaction is:


`molS_(ex)=molS-molFe=0,5408molS-0,2729molFe \\ \\ = 0,2679 mol S`

So there are 0,2679 moles of S in excess after the reaction.

3) What mass of FeS is created?

To determine the mass of FeS created we need to go from moles of Fe (Limiting reactant) to grams of FeS (Product) using the following conversion factor:


`0,2729 moles Fe* (1 mol FeS)/(1 mol Fe) * (87,91 g FeS)/(1 mol FeS)=23,99 g FeS`

So, 23,99 grams of FeS are produced from this chemical reaction.

Have a nice day!