Question 39When heated, potassium permanganate decomposes according to the following equation:2KMnO4 ⟶ K2MnO4 + MnO2 + O2(a) Some potassium permanganate was heated in the 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. 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)

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Accepted Answer

2KMnO4 ⟶ K2MnO4 + MnO2 + O2Loss in mass = 1.32 g = 1 lit of oxygenVapour density of gas =Wt. of certain volume of gas Wt. of same volume of H2=1.320.0825=16 g\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \[0.5em] = \dfrac{1.32}{0.0825} \[0.5em] = 16 \text{ g}Wt. of same volume of H2​Wt. of certain volume of gas ​=0.08251.32​=16 gMolecular weight = 2 x Vapour density= 2 x 16 = 32 gHence, relative molecular mass of oxygen is 32 g(b) Molar mass of 2KMnO4 = 2[39 + 55 + 4(16)] = 2[39 + 55 + 64] = 316 g316 g of KMnO4 gives oxygen = 24 litres∴ 15.8 g of KMnO4 will give= 24316\dfrac{24}{316}31624​ x 15.8 = 1.2 L

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 40

Question 39