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]

Answer 1

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.