Question

0.2g of a sample of H₂O₂ reduced 20 ml of 0.1M KMnO₄ solution in acidic medium. What is the percentage purity of H₂O₂?

Answer 1

No of moles of KMnO₄ present in 20ml of 0.1M

KMnOH solution = \(\frac{20}{1000}\) x 0.1 

= 2 x 10^-3

The balanced equation for the redox reactions is:

2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂ …(1)

From the equation, 2 moles of KMnO₄ = 5 moles of H₂O₂

2 x 10^-3 moles of KMnO₄ will react with H₂O₂ = \(\frac{5}{2}\) x 2 x 10^-3 = 5 x 10^-3 moles Molecular wt. of H₂O₂ = 34

Amount of H₂O₂ actually present 

= 34 x 5 x 10^-3 

= 0.17g

Percentage purity of in 20 ml is 

= \(\frac{0.17}{0.20}\) x 100 = 85

Comments

Plz solve it in some other simple way

Answer 2

No of moles of KMnO₄ present in 20 mL of 0.1M

KMnO₄ solution = \( \frac{20}{1000}L\times 0.1\,M\)

= 2 x 10^-3 mol.

The balanced chemical equation for the redox reaction is : 

2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 8H₂O + 5O₂ .....(1)

According to above balance chemical equation -

∵ We have required 5 mole pure H₂O₂ to reduce 2 mol KMnO₄.

∴ Number of moles of pure H₂O₂ required to reduce 2 x 10^-3 mol KMnO₄ 

= \(\frac{5}{2}\) x 2 x 10^-3 mol

= 5 x 10^-3 mol of pure H₂O₂

Molecular weight of H₂O₂ = 34 g/mol

∴ Mass of pure H₂O₂ = 5 x 10^-3 mol x 34g/mol

= 0.17 g

But,

Amount of H₂O₂ actually precent = 0.2g

∴ Percentage purity of H₂O₂ = \(\frac{0.17}{0.20}\times 100\)

= 85

Hence,

The purity of H₂O₂ was 85%.