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

Question

KClO 4 can be prepared by following reactions: i. Cl 2+2 KOH arrow KCl + KClO + H 2 O ii. 3 KClO arrow 2 KCl + KClO 3 iii. 4 KClO 3 arrow 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

Tardigrade Admin · Accepted Answer

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 Cl2 = 2× 4 = 8, mass of Cl2 = 8× 71 = 568 g ] Moles of KOH =2 × 8=16 V KOH =(16/1.5)=10.67 L

Q. $K Cl O_{4}$ can be prepared by following reactions:
i. $C l_{2} + 2 K O H \to K Cl + K ClO + H_{2} O$
ii. $3 K ClO \to 2 K Cl + K Cl O_{3}$
iii. $4 K Cl O_{3} \to 3 K Cl O_{4} + K Cl$
(Atomic weight of $K, Cl$ , and $O$ are $39, 35.5$ and $16$ )

The amount of $C l_{2}$ required to prepare $277 g$ of $K Cl O_{4}$ by above series of $K Cl O_{4}$ by above series of reaction is $568 g$

The volume of $K O H$ in litres used by $C l_{2}$ , if $K O H$ is $1.5 M$ , is $1.067 L$

The amount of $C l_{2}$ required to prepare $200 g$ of $K Cl O_{4}$ by above series of reaction is $204 g$

The volume of $K O H$ in litres used by $C l_{2}$ , if $K O H$ is $1.5 M$ , is $10.67 L$

Q. KClO4​ can be prepared by following reactions: i. Cl2​+2KOH→KCl+KClO+H2​O ii. 3KClO→2KCl+KClO3​ iii. 4KClO3​→3KClO4​+KCl (Atomic weight of K,Cl, and O are 39,35.5 and 16 )

The amount of Cl2​ required to prepare 277g of KClO4​ by above series of KClO4​ by above series of reaction is 568g

The volume of KOH in litres used by Cl2​, if KOH is 1.5M, is 1.067L

The amount of Cl2​ required to prepare 200g of KClO4​ by above series of reaction is 204g

The volume of KOH in litres used by Cl2​, if KOH is 1.5M, is 10.67L

1molCl2​≡2molKOH≡1molKClO≡31​mol KClO3​≡41​molKClO4​ Moles of KClO4​=138.5277​=2 [ Moles of Cl2​=2×4=8, mass of Cl2​=8×71=568g] Moles of KOH=2×8=16 VKOH​=1.516​=10.67L

Q. $K Cl O_{4}$ can be prepared by following reactions:
i. $C l_{2} + 2 K O H \to K Cl + K ClO + H_{2} O$
ii. $3 K ClO \to 2 K Cl + K Cl O_{3}$
iii. $4 K Cl O_{3} \to 3 K Cl O_{4} + K Cl$
(Atomic weight of $K, Cl$ , and $O$ are $39, 35.5$ and $16$ )

The amount of $C l_{2}$ required to prepare $277 g$ of $K Cl O_{4}$ by above series of $K Cl O_{4}$ by above series of reaction is $568 g$

The volume of $K O H$ in litres used by $C l_{2}$ , if $K O H$ is $1.5 M$ , is $1.067 L$

The amount of $C l_{2}$ required to prepare $200 g$ of $K Cl O_{4}$ by above series of reaction is $204 g$

The volume of $K O H$ in litres used by $C l_{2}$ , if $K O H$ is $1.5 M$ , is $10.67 L$