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Question
When heated, potassium permanganate decomposes according to the following equation:
\[\ce{2KMnO4 -> \underset{solid residue}{K2MnO4 + MnO2} + O2}\]
Given that the molecular mass of potassium permanganate is 158 g, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres). [K = 39, Mn = 55, O = 16]
Solution
| 2KMnO4 | → | K2MnO4 | + | MnO2 | + | O2 |
| 2[39 + 55 + 64] | 24 lit | |||||
| 2 × 158 = 316 g |
Vol. of O2 produced by 316 g of KMnO4 = 24 lit.
∴ Vol. of O2 produced by 15.8 of KMnO4 = `(24 xx 15.8)/316`
= 1.2 lit.
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