CUET · CHEMISTRY · PYQ PAPER 2025
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\(KMnO _4\) is prepared by the fusion of \(MnO _2\) with an alkali metal hydroxide and an oxidizing agent like \(KNO _3\) to give a dark-green manganate ion which disproportionate to give permanganate as follows.
\(\begin{array}{l}2 MnO _2+4 KOH + O _2 \rightarrow 2 K_2 MnO _4+2 H _2 O \\ 3 K_2 MnO _4+4 H ^{+} \rightarrow 2 KMnO _4+ MnO _2+2 H _2 O \end{array}\)
On heating \(K _2 MnO _4\) decomposes at 513 K to give \(K _2 MnO _4\). Permanganate ion is tetrahedral and diamagnetic
Acidified \(KMnO _4\) acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
When one mole of \(KMnO _4\) is used for the oxidation of oxalic acid, how many moles of \(CO _2\) will be produced?
- A 5
- B 10
- C 5/2
- D 2
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
\(2 KMnO _4+5 H _2 C _2 O _4+3 H _2 SO _4 \rightarrow 2 MnSO _4+10 CO _2+K _2 SO _4+8 H _2 O\) From the balanced equation: \(2\) mol \(KMnO _4 \rightarrow 10\) mol \(CO _2\) \(1\) mol \(KMnO _4 \rightarrow \frac{10}{2} = 5\) mol \(CO _2\)
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