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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-

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Balance the following equation : KOH + Cl_2 to KCl + KClO + H_2O .

Balance the following equation : KClO_(3) to KCl + O_(2)

KClO_(4) can be prepared by following reactions: i. Cl_(2) + 2 KOH rarr KCl + KClO + H_(2) O ii. 3KClO rarr 2KCl + KClO_(3) iii. 4KClO_(3) rarr 3KlI_(4) + KCl (Atomic weight of K, Cl , and O are 369,35.5 and 16)

KClO_(4) can be prepared by following reactions: i. Cl_(2) + 2 KOH rarr KCl + KClO + H_(2) O ii. 3KClO rarr 2KCl + KClO_(3) iii. 4KClO_(3) rarr 3KlI_(4) + KCl (Atomic weight of K, Cl , and O are 369,35.5 and 16)

Complete the following reactions : KClO_(3) rarr ?

From the following reaction sequence Cl_2 + 2KOH to KCl + KClO + H_2O 3KClO to 2KCl + KClO_3 4KClO_3 to 3KClO_4 + KCl Calculate the mass of chlorine needed to produce 100 g of KClO_4

Which catalyst are used in following reactions : 2KClO_3 to 2KCl + 3O_2