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Question

What volume of O2 measured at standard conditions will be formed by the action of 100 mL of 0.5 N KMnO4 on hydrogen peroxide in an acidic solution?

A
0.12 L
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B
0.28 L
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C
0.56 L
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D
1.12 L
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Solution

The correct option is B 0.56 L
The reaction is as follows:
2KMnO4+5H2O2+3H2SO42MnSO4+5O2+K2SO4+8H2O
So, number of moles of KMnO4=0.1 (given)
So, 100×0.11000=0.01 mol of KMnO4 will produce (52)×0.01=0.025 mol of O2=0.025×22.4=0.56 L.

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