Question

When heated, potassium permanganate decomposes according to the following equation:
2KMnO₄ $\stackrel{}{\to }$ K₂MnO₄ + MnO₂ + O₂
(a) Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
(b) 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

(a) Let molecular mass of oxygen be M.
According to the balanced chemical equation, for two moles of potassium permanganate, one mole of oxygen is formed.
Loss in mass = 1.32 g
Mass of oxygen evolved = Loss in mass = 1.32 g
Number of moles of oxygen in 1.32 g = $\frac{1.32}{M}$
Mass of one litre of hydrogen = 0.0825 g (at same temperature and pressure conditions)
So, according to the Avogadro's Law, number of moles of hydrogen will be same as that of oxygen.
Number of moles of hydrogen = Number of moles of oxygen
$$\begin{gathered} \frac{\mathrm{Mass}\mathrm{of}\mathrm{hydrogen}\mathrm{gas}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{hydrogen}\mathrm{gas}}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{oxygen}\mathrm{gas}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{oxygen}\mathrm{gas}} \\ \frac{0.0825}{2}=\frac{1.32}{M} \\ M=\frac{1.32}{0.04125} \\ M=32 \end{gathered}$$
Hence, relative molecular mass of oxygen is 32.

(b) Number of moles of potassium permanganate = $\frac{15.8}{158}=0.1$
Then, according to the balanced chemical equation, two moles of potassium permanganate evolve one mole of oxygen gas.
Number of moles of oxygen formed by 0.1 moles of potassium permanganate = $\frac{0.1}{2}=0.05$
Volume of 0.05 moles of oxygen = Number of moles $\times$ Molar volume = 0.05 $\times$ 24 = 1.2 L