Question

In the equation, Fe2O3 + 2Al → Al2O3 + 2 Fe, if 52.6 g of aluminum are used with excess iron oxide, how many grams of Fe2O3 will be converted to aluminum oxide?

Answer 1

`155.85` g

Explanation:

Let's consider the equation first.

`Fe_2O_3 + 2Al = Al_2O_3 + 2Fe`

It says that 2 moles of aluminium will react with 1 mole of Iron (III) Oxide to form 1 mole of Aluminium Oxide and 2 moles of Iron.

So, Unitary method.

`54` g Al will react with `rarr` `160` g of `Fe_2O_3`

`1` g Al will react with `rarr` `160/54` g of `Fe_2O_3`

`52.6` g Al will react with `rarr` `(160 * 52.6)/54` g of `Fe_2O_3` `approx` `155.85` g of `Fe_2O_3`

Hence Explained.