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KClO(4) can be prepared by the following...

`KClO_(4)` can be prepared by the following reaction:
`Cl_(2)+2KOH to KCl + KClO+H_(2)O`
`3KClO to 2KCl +KClO_(3), 4KClO_(3) to 3KClO_(4)+KCl`
To prepare 200g `KClO_(4)`, the required amount of `Cl_(2)` is equivalent to-

A

8.95 equivalent `H_(2)SO_(4)`

B

129.02L `O_(2)` at STP

C

11.52 mol oxygen

D

410.1g chloride

Text Solution

Verified by Experts

The correct Answer is:
D
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