Question

The reaction betwen aluminum and iron (III) oxide can

Answer 1

Answer : The mass of $Al_2O_3$ formed will be, 468.18 grams.

Solution : Given,

Mass of Al = 124 g

Mass of $Fe_2O_3$ = 601 g

Molar mass of Al = 27 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole

Molar mass of $Al_2O_3$ = 102 g/mole

First we have to calculate the moles of Al and $O_2$.

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{124g}{27g/mole}=4.59moles$

$\text{ Moles of }Fe_2O_3=\frac{\text{ Mass of }Fe_2O_3}{\text{ Molar mass of }Fe_2O_3}=\frac{601g}{160g/mole}=3.76moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe$

From the balanced reaction we conclude that

As, 2 mole of $Al$ react with 1 mole of $Fe_2O_3$

So, 4.59 moles of $O_2$ react with $\frac{4.59}{2}=2.295$ moles of $Fe_2O_3$

From this we conclude that, $Fe_2O_3$ is an excess reagent because the given moles are greater than the required moles and $Al$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$

From the reaction, we conclude that

As, 2 mole of $Al$ react to give 2 mole of $Al_2O_3$

So, 4.59 moles of $O_2$ react to give $\frac{2}{2}\times 4.59=4.59$ moles of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$

$\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3$

$\text{ Mass of }Al_2O_3=(4.59moles)\times (102g/mole)=468.18g$

Therefore, the mass of $Al_2O_3$ formed will be, 468.18 grams.