Magnetite weighing 5.00 g was calcined with 1.00 g of aluminum. Find the mass of the resulting iron.

3Fe3O4 + 8Al⇒4Al2O3 + 9Fe
n (Fe3O4) = m / M = 5 g: (56 3 + 16 4) g / mol≈0.02 mol
n (Al) = m / M = 1 g: 27 g / mol≈0.04 mol
Since there is 3 in front of Fe3O4, it means that n (Fe3O4) must be divided by 3. It turns out ≈0.007 mol.
Since Al is preceded by 8, it means that n (Al) must be divided by 8. It turns out ≈0.005 mol.
We decide on a disadvantage.
n (Fe) = n (Al) 9 = 0.045 mol
m (Fe) = M n = 56 g / mol 0.045 mol = 2.52 g
Answer: 2.52 g