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

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.

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

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

• 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 $\frac{3}{16}\times 0.054=0.010125$ 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 $\frac{8}{3}\times 0.0098=0.0261$ 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

Answer 2

The answer is 5.5g Fe