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.

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

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