How many liters of oxygen are formed during the decomposition of 79 grams of potassium

How many liters of oxygen are formed during the decomposition of 79 grams of potassium permanganate, if the reaction yield is 60%?

To solve, we write down the equation for the decomposition of potassium permanganate:
1.2KMnO4 = K2MnO4 + MnO2 + O2 – OBP, oxygen obtained;
2. Calculations:
M (KMnO4) = 158 g / mol;
M (O2) = 32 g / mol.
3. Find the amount of the original substance by the formula:
Y (Cu) = m / M = 79/158 = 0.5 mol.
4. Proportion:
0.5 mol (KMnO4) – X mol (O2);
-2 mol -1 mol hence, X mol (O2) = 0.5 * 1/2 = 0.25 mol.
5. Find the theoretical volume for О2:
V (O2) = 0.25 * 22.4 = 5.6 liters.
Taking into account the product yield, we calculate the practical volume:
V (O2) = 0.60 * 5.6 = 3.36 liters.
Answer: the volume of oxygen is 3.36 liters.