What amount of aluminum oxide was obtained by burning aluminum in 6.72 liters of oxygen.

The reaction of aluminum with oxygen proceeds according to the following chemical formula:
4Al + 3O2 = 2Al2O3;
To obtain two moles of aluminum oxide, three moles of oxygen are needed.
Let’s find the amount of the substance contained in 6.72 liters of oxygen:
1 mole of ideal gas under normal conditions takes a volume of 22.4 liters
N O2 = 6.72 / 22.4 = 0.3 mol;
The amount of aluminum oxide substance will be:
N Al2O3 = N O2 / 3 x 2 = 0.3 / 3 x 2 = 0.2 mol;
Let’s find the mass of this amount of aluminum oxide:
M Al2O3 = 27 x 2 + 16 x 3 = 102 grams / mol;
m Al2O3 = 0.2 x 102 = 20.4 grams;