Calculate the amount of iron that can be obtained with the interaction of 21.6 g al and 46.4 g fe3o4.

Let us write down the reaction equation
8Al + 3Fe3O4 = 3Fe3 + 4Al2O3
First, let’s determine how much aluminum is required to reduce iron from oxide.
Let’s use the balance of the amount of substance:
ν = m / M
ν (Al) = 8, ν (Fe3O4) = 3
Let’s equalize the expressions for the amount of substance:
m (Al) / (8 * M (Al)) = m (Fe3O4) / (3 * M (Fe3O4))
Let’s define the molar masses:
M (Al) = 26 g / mol
M (Fe3O4) = 56 * 3 + 16 * 4 = 232 g / mol
M (Fe3) = 56 * 3 = 168 g / mol
Let us express the mass of aluminum from the equation:
m (Al) = m (Fe3O4) * (8 * M (Al) / (3 * M (Fe3O4)) = 46.4 * 8 * 26 / (3 * 232) = 13.6 g
It turns out that we have taken aluminum in excess, only 13.6 g will be required, so further calculations will be carried out for iron oxide.
Let’s equalize the expressions for the amount of substance:
m (Fe3O4) / (3 * M (Fe3O4)) = m (Fe3) / (3 * M (Fe3))
Let us express the mass of iron from this expression:
m (Fe3) = m (Fe3O4) * 3 * M (Fe3) / (3 * M (Fe3O4) = 46.4 * 3 * 168 / (3 * 232) = 33.6 g
Let’s find its amount of substance:
ν = m / M = 33.6 / 168 = 0.2 mol
Answer: you get about, 2 mol of iron or 33.6 g.