Question

KCIO, may be prepared by means of following series of reactions
CI, + 2KOH → KCI + KCIO+H,O
3KCIO2KCI+KCIO,
4KCIO, 3KCIO, +KCI
How much Cl, is needed to prepare 400 g KCIO, by the above sequence?
(K = 39, Cl = 35.5, 0 = 16, H= 1)
was heated when its weight​

Answer 1

Answer:

The arrangement of conditions includes;

Cl

2

+2KOH→KCl+KClO+H

2

O......(1)

3KClO→2KCl+KClO

3

......(2)

4KClO

3

→3KClO

4

+KCl.......(3)

From the eqn(3)

The proportion

KCl

KClO

4

=

1

3

Implying that we have 3 moles of KClO

3

We have 1mole of KCl

Since the molar mass of KClO

4

is 138.55,

We have =

138.55

156

Moles delivered =1.126moles

On the off chance that we accept 100% yield these originated from (1.125×

3

4

) moles of KClO

3

=1.501 moles of KClO

3

required to deliver 156g of KClO

4

Moles of KClO expected to deliver 1.501 moles of KClO

3

=1.501×3=4.504 moles of KClO from eqn(2)

From eqn(1)

Moles of Cl2=moles of KClO =4.504 moles.

Along these lines, the mass of Cl

2

=4.504×71

Mass of Cl

2

=319g to deliver 156g of KClO

4

if 100% yield is accepted in the three compound responses.

Hence, this is the answer.