Calculate how much oxygen can be obtained by decomposing potassium permanganate weighing 395 g.

Calculate how much oxygen can be obtained by decomposing potassium permanganate weighing 395 g. And problem 2 Calculate the mass of potassium permanganate required to obtain a portion of oxygen, the amount of material is 2 mol

Let’s implement the solution:
1. We compose the equation according to the condition of the problem:
m = 395 g. X l. -?
2KMnO4 = O2 + K2MnO4 + MNO2 – oxygen is released;
2. Calculations:
M (KMnO4) = 158 g / mol;
M (O2) = 32 g / mol;
Y (KMnO4) = m / M = 395/158 = 2.5 mol.
3. Proportion:
2.5 mol (KMnO4) – X mol (O2);
-2 mol -1 mol hence, X mol (O2) = 2.5 * 1/2 = 1.25 mol.
4. Find the volume of O2:
V (O2) = 1.25 * 22.4 = 28 liters.
Answer: oxygen was obtained with a volume of 28 liters.
2. X g-? Y = 2 mol;
2KMnO4 = O2 + K2MnO4 + MnO2 – OBP;
M = 158 g / mol;
M (O2) = 32 g / mol.
Proportion:
X mol (KMnO4) – 2 mol (О2);
-2 mol – 1 mol from here, X mol (KMnO4) = 2 * 2/1 = 4 mol.
We find the mass of the original substance:
m (KMnO4) = Y * M = 4 * 158 = 632 g.
Answer: to carry out the process, you will need potassium permanganate weighing 632 g.