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?

1 Answer

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.