Calculate how much oxygen can be obtained by decomposition of potassium permanganate weighing 395 g?
| February 4, 2021 | education
2KMnO4 -> K2MnO4 + MnO2 + O2.
The amount of permanganate reacted is:
395 g: 158 g / mol = 2.5 mol.
Oxygen, as can be seen from the reaction equation, should be released 2 times less, i.e. 1.25 mol.
Oxygen volume: 1.25 mol * 22.4 l / mol = 28 liters.
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