Question

How many moles of Zn(FeS2)2 can be made from 2 grams of Zn, 3 grams of iron and 4 grams of sulphur

Answer 1

Answer : The moles of $Zn(FeS_2)_2$ is, 0.027 moles.

Solution : Given,

Mass of Zn = 2 g

Mass of Fe = 3 g

Mass of S = 4 g

Molar mass of Zn = 65 g/mole

Molar mass of Fe = 56 g/mole

Molar mass of S = 32 g/mole

First we have to calculate the moles of Zn, Fe and S.

$\text{ Moles of Zn}=\frac{\text{ Mass of Zn}}{\text{ Molar mass of Zn}}=\frac{2g}{65g/mole}=0.030moles$

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

$\text{ Moles of S}=\frac{\text{ Mass of S}}{\text{ Molar mass of S}}=\frac{4g}{32g/mole}=0.125moles$

In the given molecule $Zn(FeS_2)_2$, 1 mole of Zn , 2 moles of Fe and 4 moles of S are present.

From this we conclude that,

As, 2 moles of Fe combine with 1 mole of Zn

So, 0.054 moles of Fe combine with $\frac{1}{2}\times 0.054=0.027$ moles of Zn

And,

As, 2 moles of Fe combine with 4 moles of S

So, 0.054 moles of Fe combine with $\frac{4}{2}\times 0.054=0.108$ moles of S

Hence, we conclude that the Fe is the limiting reagent.

Now we have to calculate the moles of $Zn(FeS_2)_2$.

As, 2 moles of Fe present in 1 mole of $Zn(FeS_2)_2$

So, 0.054 moles of Fe present in $\frac{1}{2}\times 0.054=0.027$ moles of $Zn(FeS_2)_2$

Therefore, the moles of $Zn(FeS_2)_2$ is, 0.027 moles.