Calculate the amount of oxygen required to react with 54 g of aluminum.

Let’s write the reaction equation:
4Al + 3O2 = 2AlO3
It can be seen from the reaction equation that:
ν (O2) / ν (Al) = 3/4
Let us determine the amount of substance (Al), having previously determined its molar mass:
M (Al) = 26 g / mol
ν (Al) = m (Al) / M (Al) = 54/26 = 2.07 mol.
Determine the amount of substance (O2):
ν (O2) = 3 * ν (Al) / 4 = 3 * 3.07 / 4 = 1.55 mol
Answer: 1.55 mol of oxygen.