# 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?

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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?

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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

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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%.