Formed 600 grams of aluminum oxide (III). How many liters of oxygen (O2) did you need?

Let’s implement the solution:

According to the condition, we will write the data:
4Al + 3O2 = 2Al2O3 – compounds, aluminum oxide was obtained.

Calculations:
M (Al2O3) = 101.8 g / mol.

Determine the amount of oxide:
Y (Al2O3) = m / M = 600 / 101.8 = 5.9 mol.

Proportion:
5.9 mol (Al2O3) – X mol (O2);

-2 mol – 3 mol from here, X mol (O2) = 5.9 * 3/2 = 8.85 mol.

We find the volume of the original substance:
V (O2) = 8.85 * 22.4 = 198.24 L

Answer: you need oxygen with a volume of 198.24 liters