What volume will take up all the oxygen released during the thermal decomposition of 1 mol of each

What volume will take up all the oxygen released during the thermal decomposition of 1 mol of each of the substances, the formulas of which are: KClO3, KMnO4, HgO?

Let’s compose the reaction equations and make the calculations:
1. 2KClO3 = 2KCl + 3O2 – decomposition reaction, oxygen is released;
Let’s make the proportion:
1 mol (KClO3) – X mol (О2);
-2 mol -3 mol from here, Hmol (O2) = 1 * 3/2 = 1.5 mol;
Let’s determine the volume of О2 according to Avogadro’s law:
V (O2) = 1.5 * 22.4 = 33.6 L;
Answer: V (O2) = 33.6 liters.
2. 2KMnO4 = K2MnO4 + MnO2 + O2 – gaseous oxygen evolved;
We make the proportion according to the reaction equation:
1 mol (KMnO4) – X mol (О2);
-2 mol -1 mol hence, X mol (O2) = 1 * 1/2 = 0.5 mol;
We find the volume of oxygen:
V (O2) = 0.5 * 22.4 = 11.2 liters.
Answer: V (O2) = 11.2 liters.

3. 2HgO = 2Hg + O2 -decomposition of mercury oxide, oxygen is obtained;
We make the proportion:
1 mol (Hg) – X mol (О2)
-2 mol -1 mol hence, X mol (O2) = 1 * 1/2 = 0.5 mol;
Let’s calculate the volume of oxygen:
V (O2) = 0.5 * 22.4 = 11.2 liters.
Answer: V (O2) = 11.2 liters.