Calculate the mass of aluminum required to react with volume oxygen. at 11.2 liters.

The reaction of aluminum with oxygen is described by the following chemical reaction equation.

4Al + 3O2 = Al2O3;

During the reaction, 4 moles of aluminum react with 3 moles of oxygen.

One mole of ideal gas under normal conditions takes a volume of 22.4 liters.

Let’s find the amount of the substance contained in 11.2 liters of oxygen.

N O2 = 11.2 / 22.4 = 0.5 mol;

The amount of aluminum substance will be:

N Al = N O2 x 4/3 = 0.5 x 4/3 = 0.667 mol;

Let’s find the mass of aluminum.

M Al = 27 grams / mol;

m Al = 27 x 0.667 = 18 grams;