Question

Predict 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 1

the answer is 5.5g Fe

Answer 2

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

Solution : Given,

Mass of iron, Fe = 3 g

Mass of sulfur,
`S_8` = 2.5 g

Molar mass of Fe = 56 g/mole

Molar mass of
`S_8` = 256 g/mole

Molar mass of iron(III)sulfide,
`Fe_2S_3` = 208 g/mole

• The balanced chemical reaction is,

`16Fe(s)+3S_8(s)\rightarrow 8Fe_2S_3(s)`

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

`\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=(3g)/(56g/mole)=0.054moles`

`\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=(2.5g)/(256g/mole)=0.0098moles`

• From the balanced reaction, we conclude that

16 moles of Fe react with 3 moles of
`S_8`

0.054 moles of Fe react with
`(3)/(16)* 0.054=0.010125` moles of
`S_8`

Therefore, the excess reagent in this reaction is, Fe and limiting reagent is,
`S_8`

Now we have to calculate the moles of FeS.

As, 3 moles of
`S_8` gives 8 moles of
`Fe_2S_3`

So, 0.0098 moles of
`S_8` gives
`(8)/(3)* 0.0098=0.0261` moles of
`F_2eS_3`

The moles of
`Fe_2S_3` = 0.0261 moles

Now we have to calculate the mass of
`Fe_2S_3`.

Mass of
`Fe_2S_3` = Moles of
`Fe_2S_3` × Molar mass of
`Fe_2S_3`

Mass of
`Fe_2S_3` = 0.0261 g × 208 g/mole = 5.4288 g

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