Question

If 14.9 kg of Al reacts with an excess of Fe2O3, how many kg of Al2O3 will be produced?

Answer 1

Answer: 28.144 g of
`Al_2O_3[tex] will be produced.</strong></p><p><strong>Explanation:</strong> We are given that Al reacts with [tex]Fe_2O_3`, then the balanced chemical reaction is:

`2Al(s)+Fe_2O_3(aq.)\rightarrow Al_2O_3(aq.)+Fe(s)`

It is given that
`Fe_2O_3` is present in excess, so it is an excess reagent and Al is the limiting reagent.

Now, 2 moles of Al produces 1 mole of
`Al_2O_3`, which means that (2×27g) = 54g of Al produces 102g of
`Al_2O_3`.

Therefore, 14.9g of Al will produce =
`(102g* 14.9g)/(54g)\text{ of }Al_2O_3`

= 28.144g of
`Al_2O_3`