Calculate the amount of oxygen required to burn 27 grams of aluminum.

The equation for the reaction of aluminum with oxygen:

4Al + 3O2 = 2Al2O3

Amount of Al substance:

v (Al) = m (Al) / M (Al) = 27/27 = 1 (mol).

According to the reaction equation, 4 mol of Al reacts with 3 mol of O2, therefore:

v (O2) = v (Al) * 3/4 = 1 * 3/4 = 0.75 (mol).

Thus, the required volume of oxygen, measured under normal conditions (n.o.):

V (O2) = v (O2) * Vm = 0.75 * 22.4 = 16.8 (l).

Answer: 16.8 liters.