What is the mass of aluminum oxide formed by the interaction of 54 g of aluminum with oxygen?

Given:
m (Al) = 54 g
Find: m (Al₂O₃)
Decision:
4Al + 3O₂ = 2Al₂O₃
From the equation we see that from two moles of Al, 1 mole of Al₂O₃ is formed.
Let’s find the amount of substance in 54 grams of Al – 54/27 = 2 mol;
Amount of substance Al₂O₃: Al / 2 = 2/2 = 1 mol;
Let us find the mass of this amount of Al₂O₃:
m (Al2O3) = 1 * 102 = 102 g