What can you conclude about the mass of iron (III) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of yellow sulfur solid, S8(s)?

Answer : The mass of iron(III)sulfide is, 5.4288 g

Solution : Given,

Mass of iron, Fe = 3 g

Mass of sulfur, = 2.5 g

Molar mass of Fe = 56 g/mole

Molar mass of = 256 g/mole

Molar mass of iron(III)sulfide, = 208 g/mole

The balanced chemical reaction is,

First we have to calculate the moles of iron and sulfur.

From the balanced reaction, we conclude that

16 moles of Fe react with 3 moles of

0.054 moles of Fe react with moles of

Therefore, the excess reagent in this reaction is, Fe and limiting reagent is,

Now we have to calculate the moles of FeS.

As, 3 moles of gives 8 moles of

So, 0.0098 moles of gives moles of

The moles of = 0.0261 moles

Now we have to calculate the mass of .

Mass of = Moles of × Molar mass of

Mass of = 0.0261 g × 208 g/mole = 5.4288 g

Therefore, the mass of iron(III)sulfide is, 5.4288 g

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