A55.0g sample of iron (iii) filings is reacted with 23.8g of powdered sulfur (s8). how much iron (iii) sulfide in moles would be produced in this reaction?

equation:

convert to moles
of iron:

convert to moles
of sulfur:

calculate the
limiting reagent:

solve the problem:

0.74 moles iron (III) sulfide

Explanation:

From the balanced equation of reaction:

1 mole of Fe reacts with 1 mole of S to give 1 mole of FeS.

moles =

mole of Fe = 55/55.8 = 0.99 moles

mole of S = 23.8/32.07 = 0.74 moles

Sulfur is limited in quantity and will therefore determine the rate of reaction.

1 mole of sulfur gives 1 mole of FeS

0.74 moles of sulfur will therefore give 0.74 moles of FeS.

0.74 moles iron (III) sulfide will be produced.