Question

what is the perentage purity of 10g KMnO4 if it reacts with 100 ml of 10 volume H2O2 solution.

Answer 1

Question

What is the percentage purity of 10 g KMnO₄ if it reacts with 100 ml of 10% volume H₂O₂ solution?

Solution

$\mathrm{V(H_2O_2) = V(solution)\cdot 10 / 100 = 100\cdot 10 / 100 = 10ml}$

Density of pure $\mathrm{H}_2\mathrm{O}_2 - \rho (\mathrm{H}_2\mathrm{O}_2) = 1.45\mathrm{g / ml}$

$\mathrm{m(H_2O_2) = V(H_2O_2)\cdot\rho(H_2O_2) = 10\cdot 1.45 = 14.5g}$

Molar weight of hydrogen peroxide – $\mathrm{M(H_2O_2) = 34g / mol}$

$\mathrm{n(H_2O_2) = m(H_2O_2) / M(H_2O_2) = 14.5 / 34 = 0.43mol}$

The reaction equation is as follows

$$2 \mathrm{KMnO_4} + 3 \mathrm{H_2O_2} \rightarrow 2 \mathrm{MnO_2} + 3 \mathrm{O_2} + 2 \mathrm{KOH} + 4 \mathrm{H_2O}$$

2 mol of KMnO₄ react with 3 mol of H₂O₂

$\mathrm{n(KMnO_4) = n(H_2O_2)\cdot 2 / 3 = 0.43\cdot 2 / 3 = 0.28mol}$

Molar weight of potassium permanganate – $\mathrm{M(KMnO_4) = 158g / mol}$

$\mathrm{m(KMnO_4) = n(KMnO_4)\cdot M(KMnO_4) = 0.28\cdot 158 = 43.3g}$

So, the task condition is incorrect!

100 ml of 10% volume H₂O₂ solution reacts with much greater amount of potassium permanganate. If 10 g of KMnO₄ reacts with 100 ml of 10% volume H₂O₂ solution there is unreacted hydrogen peroxide.

(Even in acidic medium: $2\mathrm{KMnO_4} + 5\mathrm{H_2O_2} + 3\mathrm{H_2SO_4} \rightarrow 2\mathrm{MnSO_4} + \mathrm{K_2SO_4} + 5\mathrm{O_2} + 8\mathrm{H_2O}$

$\mathrm{n(KMnO_4) = n(H_2O_2)\cdot 2 / 5 = 0.43\cdot 2 / 5 = 0.14mol}$; $\mathrm{m(KMnO_4) = n(KMnO_4)\cdot M(KMnO_4) = 0.17\cdot 158 =}$

27.2 g but not 10 g !)

If the task condition would be correct the last step would be as follows

Purity, % = $\mathrm{m(KMnO_4) / m(KMnO_4)_{given}\cdot 100 = 43.3 / ... \cdot 100 = ...\%}$