Correct option is C)
Given,
1 KClO
3
decomposes to give 1 mole O
2
The reaction involved is
2KClO
3
⟶2KCl+3O
2
It is clear from above equation that 2 moles KClO
3
produce 3 moles O
2
but it is given that 1 mole O
2
is produced.
⇒ Moles of KClO
3
used=
3
1×2
=0.66 moles⟶(1)
∴ The left over moles of KClO
3
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
