Question

On heating $KCl{O}_3$ at a certain temperature, it is observed that one mole of $KCl{O}_3$ yields one mole of $O_2$. What is the mole fraction of $KCl{O}_4$ in the final solid mixture containing only $KCl$ and $KCl{O}_4$, the latter being formed by the parallel reaction?

• A. $0.50$
• B. $0.25$
• C. $0.75$
• D. $0.67$

Answer 1

The correct option is C $0.75$

Given,

$1$ $KCl{O}_3$ decomposes to give $1$ mole $O_2$

The reaction involved is

$2KCl{O}_3⟶2KCl+3O_2$

It is clear from above equation that $2$ moles $KCl{O}_3$ produce $3$ moles $O_2$ but it is given that $1$ mole $O_2$ is produced.

$\Rightarrow$ Moles of $KCl{O}_3$ used= $\frac{1\times 2}{3}=0.66$ moles$⟶\left(1\right)$

$\therefore$ The left over moles of $KCl{O}_3$ are $1-0.66=0.34$ moles $⟶\left(2\right)$

Now the parallel reaction is

$4KCl{O}_3⟶3KCl{O}_4+KCl$

It is clear that $4$ moles $KCl{O}_3$ gives $3$ moles $KCl{O}_4$ & $1$ mole $KCl⟶\left(3\right)$

Now, for determining mole fraction of $KCl{O}_4$ first calculate moles of $KCl{O}_4$ & $KCl$

From $\left(2\right)$ & $\left(3\right)$

If $4$ moles $KCl{O}_3$ gives $3$ moles $KCl{O}_4$

$\Rightarrow 0.34$ moles $KCl{O}_3$ give $\frac{0.34\times 3}{4}$

$\therefore$ No. of moles of $KCl{O}_4$ produced=$0.255$ moles

Now, if $4$ moles $KCl{O}_3$ give $1$ mole $KCl$

$\Rightarrow 0.34$ moles $KCl{O}_3$ give $\frac{0.34\times 1}{4}$ moles

$\therefore$ No. of Moles of $KCl$ produced=$0.085$ moles

$\therefore$ Total no. of Moles in final mix=$0.255+0.085$

$=0.34$ moles

$\therefore$ Mole fraction of $KCl{O}_4=\frac{MolesofKCl{O}_4}{Totalmoles}$

$=\frac{0.255}{0.34}$

$=0.75$