What is the maximum mass of iron metal if 50.0 grams of ironIII oxide is mixed with 50.0 grams of aluminum?

Answer 1

Fe2O3 + 2Al --> 2Fe + Al2O3

Before:

50.0g + 50.0g > 0.0g + (not important)

159.69(g/mol) + 26.98(g/mol)

In mol (before reaction):

+0.3131 mol + 1.853 mol (excess)

Reaction (used reactant > formed Fe):

-0.3131 mol - 0.6262 mol > + 0.6262 mol Fe

Remaining (= before - used):

0.0 mol Fe2O3 + 1.227 mol Al > 0.6262 mol Fe, this should be multiplied by the molar mass of Fe to get mass in grams: 0.6262 (molFe) * 55.85 (g/molFe) = 34.97 = 35.0 g Fe

Answer 2

The balanced chemical equation shows a 2:2 mole ratio between ironIII oxide and aluminum. Since the molecular weight of ironIII oxide is twice that of aluminum, the maximum mass of iron metal that can be produced is 50.0 grams. This is because all the ironIII oxide will be consumed in the reaction and converted into iron metal.