A 200 mL H₂O₂ sample is split in two. First part reacts by H 2 O 2 + 2KI → I 2 + 2KOH at 50% yield; formed KOH needs 500 mL 0.5 M HClO₄. Second part reacts with excess permanganate by 5 H 2 O

Question

A 200 mL H₂O₂ sample is split in two. First part reacts by $H_{2} O_{2} + 2KI \to I_{2} + 2KOH$ at 50% yield; formed KOH needs 500 mL 0.5 M HClO₄. Second part reacts with excess permanganate by $5 H_{2} O_{2} + 2 KMn O_{4} + 3 H_{2} S O_{4} \to K_{2} S O_{4} + 2 MnS O_{4} + 5 O_{2} + 8 H_{2} O$ at 66.6% yield and gives 4.48 L O₂ at 1 atm and 0°C. Find the sample's volume strength.

Round your answer to the nearest integer.

TestHub · Accepted Answer

The first aliquot yields 0.500×0.5=0.25 mol KOH, hence 0.125 mol reacted H₂O₂. At 50% yield, initial H₂O₂=0.25 mol. Second aliquot yields 4.48/22.4=0.20 mol O₂, equal to reacted H₂O₂; at 66.6%≈2/3 yield initial amount=0.30 mol. Total concentration=0.55/0.200=2.75 M. The printed range 31.20-31.23 follows volume strength=2.75×22.7/2=31.2125 using the 0°C,1-bar molar volume. Measured gas used 22.4 at 1 atm; thus the key combines different reference conventions. A uniform 1-atm volume-strength convention gives 30.8. Preserve the source range and disclose this dependence.

Per the requested integer conversion, round 31.20-31.23 to the nearest integer: 31.

Chemistry - Redox Reactions Question with Solution | TestHub

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