What is the volume of oxygen released during the decomposition of 10.1 g. potassium permanganate

What is the volume of oxygen released during the decomposition of 10.1 g. potassium permanganate, if the reaction proceeds according to the scheme: KMnO4 = K2MnO4 + MnO2 + O2

Decision:
1) We compose the reaction equation:
8KMnO4 = 4K2MnO4 + 4MnO2 + 4O2 – redox reaction, oxygen gas is released.
2) Determine the amount of KMnO4 if its mass is known: m (KMnO4) = 10.1 g
Y (KMnO4) = 10.1 / 158 = 0.06 mol
2) Based on the data on the reaction equation, we make up the proportion:
0.06 mol KMnO4 – X mol О2
Hence, X mol (O2) = 0.06 * 4/8 = 0.03 mol;
3) Calculate the volume of oxygen according to Avogadro’s law:
V (O2) = 0.03 * 22.4 = 0.67 liters.
Answer: V (O2) = 0.67 liters is released during the decomposition of potassium permanganate.