What mass and what volume of oxygen will be obtained under normal conditions of thermal decomposition

What mass and what volume of oxygen will be obtained under normal conditions of thermal decomposition of potassium permangomate with a mass of 3 g?

Thermal decomposition of potassium permanganate:
2KMnO4 = K2MnO4 + MnO2 + O2.
1) Find the molar mass of potassium permanganate according to the periodic table: 39 + 55 + 16 * 4 = 158.
2) Find the amount of permanganate substance. To do this, divide its mass by its molar mass: 3/158 = 0.02 mol.
4) According to the reaction equation, there is one mole of oxygen for two moles of potassium permanganate. This means that the amount of oxygen substance is: 2 * 0.02 = 0.04 mol.
5) Oxygen mass: 0.04 * 32 = 1.28 g – answer.
6) Oxygen volume: 0.04 * 22.4 = 0.096 l – the answer.