What volume of O₂(g) measured at 1 atm and 273 K will be formed by action of 100 mL of 0.5N KMnO₄ on hydrogen peroxide in an acid solution ? The skeleton equation for the reaction is
${\mathrm{KMnO}}_4+{\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2{\mathrm{O}}_2⟶{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{MnSO}}_4+{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$

2KMnO₄​+3H_2​SO_4​+5H₂​O_2​⟶K_2​SO₄​+2MnSO_4​+5O₂​+8H_2​O⟶(1)

Normality of KMnO_4​ solution=0.5

⇒ Molarity of KMnO_4​ solution=50/.5​=0.1M

[∵ Normality=Molarity × n factor

Here, for KMnO_4​ n factor is 5 [as e− change is 5]

Now, Molarity =0.1M, which means

0.1 moles of KMnO_4​ dissolved in 1000 ml solution

∴ 1n 100ml solution, moles of KMnO₄​ is 100×0.1/ 1000  ​=0.01 moles

From (1) , it is clear that

2 moles KMnO_4​ produce 5 moles O_2​

⇒0.01 moles KMnO₄​ will give 0.025 moles of O_2​

∴ Molar volume of 0.025 moles of O₂​ gas is as 1 mole=22.4L of O₂​

⇒0.025 moles=0.025×22.4L

=0.56 L