Question

The reaction betwen aluminum and iron (III) oxide can

generatetemperatues approaching 3000 degrees Celsius and is used in
weldingmetals:
2Al + Fe2O3
----->Al2O3 + 2 Fe
In one process, 124 g of Al are reacted with 601 g
ofFe2O3. Calculate the mass in grams
ofAl2O3 formed.

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}=(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}=(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
`(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
`(2)/(2)* 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* \text{ Molar mass of }Al_2O_3`

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

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