The correct option is
C 0.75Given,
1 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= 1×23=0.66 moles⟶(1)
∴ The left over moles of KClO3 are 1−0.66=0.34 moles ⟶(2)
Now the parallel reaction is
4KClO3⟶3KClO4+KCl
It is clear that 4 moles KClO3 gives 3 moles KClO4 & 1 mole KCl⟶(3)
Now, for determining mole fraction of KClO4 first calculate moles of KClO4 & KCl
From (2) & (3)
If 4 moles KClO3 gives 3 moles KClO4
⇒0.34 moles KClO3 give 0.34×34
∴ No. of moles of KClO4 produced=0.255 moles
Now, if 4 moles KClO3 give 1 mole KCl
⇒0.34 moles KClO3 give 0.34×14 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 KClO4=MolesofKClO4Totalmoles
=0.2550.34
=0.75