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KClO4 can be prepared by Cl2 and KOH by a series of reactions as given below :Cl2 + KOH ----> KCl + KClO + H2O ​KClO ----> KCl + KClO3KClO3 ----> KClO4 + KCl​ calculate mass of Cl2 in gm required to produce 1385 gm KClO4calculate the total mass of KCl produced in 1st, 2nd and 3rd reaction

KClO4 can be prepared by Cl2 and KOH by a series of reactions as given below :Cl2 + KOH ----> KCl + KClO + H2O ​KClO ----> KCl + KClO3KClO3 ----> KClO4 + KCl​ calculate mass of Cl2 in gm required to produce 1385 gm KClO4calculate the total mass of KCl produced in 1st, 2nd and 3rd reaction

Grade:11

2 Answers

Vikas TU
14149 Points
6 years ago
The arrangement of conditions includes; 
Cl2 + 2KOH → KCl + KClO + H2O - - (1) 
3KClO → 2KCl + KClO3 - - (2) 
4KClO3 → 3KClO4 + KCl - - (3) 
From the eqn(3) 
The proportion KClO4 :KCl = 3:1 
Implying that we have 3 moles of KClO3 We have 1mole of KCl 
Since the molar mass of KClO4 is 138.55, we have 156/138.55 
Moles delivered = 1.126 moles 
On the off chance that we accept 100 % yield these originated from (1.125 *4/3) moles of KClO3 
=1.501 moles of KClO3 required to deliver 156g of KClO4 
Moles of KClO expected to deliver 1.501 moles of KClO3 
= 1.501 * 3 =4.504 moles of KClO from eqn(2) 
From eqn(1) 
Moles of Cl2 = moles of KClo =4.504moles. 
Along these lines, the mass of Cl2 = 4.504 * 71 
Mass of Cl2=319g to deliver 156g of KClO4 if 100% yield is accepted in the three compound responses.
Baba
13 Points
2 years ago
i) Let mole of Cl2
 required = x
Cl2+ 2KOH−> KCl + KClO + H2O
            (X)                      (x)      (x) 
 
3KClO −> 2KCl + KClO
x               2x/3      x/3
4KClO3−>3KClO4 + KCl
 x/3            x/4               x/12
Mole of KClO4 formed = 1385/138.5(molar weight) 
=10
 
x/4 =10 , x = 40
weight of Cl2 required for the reaction = 40 × 71 = 2840 gm
 
(ii) KCl produced from Ist, IInd and IIIrd reaction = x+(2x/3) +(x/12) moles
=21x/12 mole=(21/12) ×40=70 mole
weight of KCl produced =70 × 74.5 = 5215 gram
 

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