Determine the amount of oxygen released during the thermal decomposition of 51.2 potassium permanganate containing 5.0% impurities.
November 11, 2020
2KMnO4 → K2MnO4 + MnO2 + O2 ↑
m (pure KMnO4) = 51.2 x 0.95 = 48.64 g
n (KMnO4) = 48.64 / (39 + 55 + 64) = 0.31 mol
n (O2) = 0.155 x 22.4 = 3.472 l
Answer: the volume of oxygen released during the reaction is 3.472 liters.
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