Question

$KClO _4$ can be prepared by following reactions:
i. $Cl _2+2 KOH \rightarrow KCl + KClO + H _2 O$
ii. $3 KClO \rightarrow 2 KCl + KClO _3$
iii. $4 KClO _3 \rightarrow 3 KClO _4+ KCl$
(Atomic weight of $K , Cl$, and $O$ are $39,35.5$ and $16$ )

• A. The amount of $Cl _2$ required to prepare $277\, g$ of $KClO _4$ by above series 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 reaction is $204\, g$
• D. The volume of $KOH$ in litres used by $Cl _2$, if $KOH$ is $1.5\, M$, is $10.67\, L$

Answer 1

Answer: (A), (D)

$1 \,mol \,Cl _2 \equiv 2\, mol\, KOH \equiv 1 \,mol \,KClO \equiv \frac{1}{3} mol$
$KClO _3 \equiv \frac{1}{4} mol\, KClO _4$
Moles of $KClO _4=\frac{277}{138.5}=2$
$[$ Moles of $Cl_2 = 2\times 4 = 8$, mass of $Cl_2 = 8\times 71 = 568\,g ]$
Moles of $KOH =2 \times 8=16$
$V_{ KOH }=\frac{16}{1.5}=10.67\, L$