Calculate the mass and amount of oxygen that is formed during the decomposition of 120 g of KM4O4.

Let’s implement the solution according to the plan:
1. Let’s write down the reaction equation:
2KMnO4 = K2MnO4 + MnO2 + O2 – redox reaction, hydrogen gas is released:
2. Determine the molecular weights of substances:
M (KMnO4) = 39.1 + 54.9 + 16 * 4 = 158 g / mol;
M (O2) = 16 * 2 = 32 g / mol;
3. Let’s calculate the number of moles of KMnO4:
Y (KMnO4) = m / M = 120/158 = 0.759 mol;
4. Let’s make the proportion:
0.759 mol (KMnO4) – X mol (O2);
-2 mol -1 mol hence, X mol (O2) = 0.759 * 1/2 = 0.379 mol;
5. Find the mass and volume of oxygen:
m (O2) = Y * M = 0.379 * 32 = 12.15 g;
V (O2) = 0.379 * 22.4 = 8.48 liters.
Answer: during the decomposition reaction, oxygen with a mass of 12.15 g was released.