Calculate moles of KMnO 4 used : | TestHub

Passage / Comprehension

$H_{2} O_{2} + 2 KI 40% yield I_{2} + 2 KOH H_{2} O_{2} + 2 KMnO_{4} + 3 H_{2} SO_{4} 50% yield K_{2} SO_{4} + 2 MnSO_{4} + 3 O_{2} + 4 H_{2} O$

100 ml of $H_{2} O_{2}$ sample was divided into two parts. First part was treated with KI and KOH formed required 200 ml of $M /2 H_{2} SO_{4}$ for complete neutralisation. Other part was treated with just sufficient $KMnO_{4}$ yielding 6.72 L of $O_{2}$ at 1 atm & 273 K.

Calculate moles of $KMnO_{4}$ used :

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Answer:

Solution:

Find moles of O₂: Moles of O₂ = $\frac{6.72 L}{22.4 L / m o l} = 0.3 m o l$ .

Identify stoichiometry: According to the balanced equation, 2 moles of KMnO₄ yield 3 moles of O₂ at 100% efficiency.

Account for 50% yield: Actual moles of O₂ = Theoretical moles $\times 0.50 ⟹ 0.3 = (\frac{3}{2} \times n_{K M n O_{4}}) \times 0.50$ .

Calculate final answer: $0.3 = 0.75 \times n_{K M n O_{4}} ⟹ n_{K M n O_{4}} = \frac{0.3}{0.75} = 0.4mol$ (Option D).

Chemistry - Concentration Terms Question with Solution | TestHub

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