Question

How many moles of \ce{Fe2O3}FeX 2 ​ OX 3 ​ will be produced from 27.0 \text{ g}27.0 g27, point, 0, start text, space, g, end text of \ce{Fe}FeF, e, assuming \ce{O2}OX 2 ​ is available in excess?

Answer 1

Answer : The number of moles of $Fe_2O_3$  produced will be, 0.241 moles.

Solution : Given,

Mass of Fe = 27.0 g

Molar mass of $Fe$ = 56 g/mole

First we have to calculate the moles of $Fe$.

$\text{ Moles of }Fe=\frac{\text{ Mass of }Fe}{\text{ Molar mass of }Fe}=\frac{27.0g}{56g/mole}=0.482moles$

Now we have to calculate the moles of $Fe_2O_3$

The balanced chemical reaction is,

$4Fe+3O_2\rightarrow 2Fe_2O_3$

From the reaction, we conclude that

As, 4 mole of $Fe$ react to give 2 mole of $Fe_2O_3$

So, 0.482 moles of $Fe$ react to give $\frac{0.482}{4}\times 2=0.241$ moles of $Fe_2O_3$

Thus, the number of moles of $Fe_2O_3$  produced will be, 0.241 moles.