Question

Al (s) + Fe2O3 (s) --> Al2O3 (s) + Fe (s) (needs balancing)

How many grams of Fe can be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of Fe2O3?

Answer 1

Answer: 134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of
`Fe_2O_3`.

Solution:

Given :
`Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+Fe(s)`

After balancing,
`2Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+2Fe(s)`

As we are given
`Fe_2O_3` is in excess. Hence, Al will be considered as limiting reagent because it limits the formation of product.

`\text{Number of moles of}Al=\frac{\text{mass of the compound}}{\text{Molecular mass of the compound}}=(65.2 g)/(26.98 g/mol)=2.4166 moles`

According to reaction ,2 moles of
`Al` reacts with one mole of
`Fe_2O_3` to give 2 moles of Fe then 2.4166 moles of
`Al` will give :

`=(2)/(2)* 2.4166\text{moles of}Fe= 2.4166 moles`

Mass of
`Fe` produced =
`\text{Number of moles of compound}* \text{Molecular mass of the compound}=2.4166 moles* 55.85g/mol=134.967 g`

134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of
`Fe_2O_3`.