#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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

Vikas TU
14149 Points
4 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
one month 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