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At high temperatures, sulfur combines with iron to form the brown-black iron (II) sulfide: Fe (s) + S (I) ---> FeS (s) In one experiment, 7.62 g of Fe are allowed to react with 8.67 g of S. a. What is the limiting reagent, and what is the reactant in excess? b. Calculate the mass of FeS formed.

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Answer:

-  S, is the reactant in excess

-  Fe  is the limiting reactant

-  Mass of FeS formed: 12 g

Explanation:

This is the reaction:

Fe (s)  +  S (l)  →  FeS

Ratio in this reaction is 1:1

We have 7.62 g of Fe, and 8.67 g of S.

Let's convert to moles:

Mass / Molar mass = moles

Molar mass Fe =  55.8 g/m

Molar mass S = 32 g/m

Moles Fe : 7.62g / 55.8 g/m = 0.136 moles

Moles S: 8.67 g /32 g/m = 0.271 moles

If we have 0.136 moles of Fe, we need 0.136 moles of S and we have 0.271 . S, is the reactant in excess.

For 0.271 moles of S, we need the same amount of Fe and we only have 0.136, so Fe is the limiting reactant.

Ratio between Fe and FeS is also 1:1 so 0.136 moles of Fe generates 0.136 moles of FeS

Molar mass FeS =

Moles - molar mass = mass

87.9 g/m .  0.136m = 12 g

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