Calculate total moles of H 2 O 2 used in both reaction : | TestHub

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

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 total moles of $H_{2} O_{2}$ used in both reaction :

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Answer:

Solution:

Moles of H₂SO₄ = $0.2 L \times \frac{1}{20} M = 0.01 m o l$ , neutralizing $0.02 m o l$ of KOH. With a $40%$ yield, moles of H₂O₂ used = $\frac{0.02}{2 \times 0.40} = 0.025 m o l$ (or $0.05 m o l$ based on equivalence).

Moles of O₂ = $\frac{6.72 L}{22.4 L / m o l} = 0.3 m o l$ . Correcting for the $50%$ yield and reaction stoichiometry gives moles of H₂O₂ used = $0.40 m o l$ .

Total Moles: Adding both parts together, the total moles of H₂O₂ = $0.05 + 0.40 = 0.45mol$ .

Chemistry - Concentration Terms Question with Solution | TestHub

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