Question

. On heating KCIO3, at a certain temperature, it
is observed that one mole of KCIO3 yields one
mole of 02. What is the mole fraction of KCIO4
in the final solid mixture containing only KCI
and KC104, the latter being formed by parallel
reaction?
(a) 0.50
(c) 0.33
(b) 0.25
(d) 0.67

Answer 1

Given,

KClO3 decomposes to give 1 mole O2

The reaction involved is 2KClO3

⟶2KCl+3O2

It is clear from above equation that 2 moles KClO3 produce 3 moles O2

but it is given that 1 mole O2 is produced.

⇒ Moles of KClO3

used= 31×2

=0.66 moles⟶(1)

∴ The left over moles of KClO3

are 1−0.66=0.34 moles ⟶(2)

Now the parallel reaction is

4KClO

3

⟶3KClO

4

+KCl

It is clear that 4 moles KClO

3

gives 3 moles KClO

4

& 1 mole KCl⟶(3)

Now, for determining mole fraction of KClO

4

first calculate moles of KClO

4

& KCl

From (2) & (3)

If 4 moles KClO

3

gives 3 moles KClO

4

⇒0.34 moles KClO

3

give

4

0.34×3

∴ No. of moles of KClO

4

produced=0.255 moles

Now, if 4 moles KClO

3

give 1 mole KCl

⇒0.34 moles KClO

3

give

4

0.34×1

moles

∴ No. of Moles of KCl produced=0.085 moles

∴ Total no. of Moles in final mix=0.255+0.085

=0.34 moles

∴ Mole fraction of KClO

4

=

Totalmoles

MolesofKClO

4

=

0.34

0.255

=0.75