`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)

Options

A. The amount of `Cl_(2)` required to prepare `277 g` of `KClO_(4)` by above series of reaction is `568 g`.
B. The volume of `KOH` in litres used by `Cl_(2)`, if `KOH` is `1.5 M`, is `1.067 L`.
C. The amount of `Cl_(2)` required to prepare `200 g "of" KClO_(4)` by above series of reactionis `284 g`
D. The volume of `KOH` in litres used by `Cl_(2)`, if `KOH` is `1.5 M`, is `10.76 L`

Answer

A, D

Explanation

`1 "mol" Cl_(2) -= 2 "mol" KOH -= 1 "mol" KClO -= (1)/(3) "mol" KClO_(3)`
`-= (1)/(4) "mol" KClO_(4)`
Moles of `KClO_(4) = (277)/(138.5) = 2`
`[{:("Moles of" Cl_(2) = 2 xx 4 = 8),("mass of" Cl_(2) = 8 xx 71 = 568 g):}]`
Moles of `KOH = 2 xx 8 = 16`
`V_(KOH) = (16)/(1.5) = 10.67 L`