Question

A 1 g sample of $KCl{O}_3$ was heated under such conditions that a part of it decomposed according to the equation.
$\left(i\right)2KCl{O}_3\to 2KCl+3O_2$
And the remaining underwent change according to the equation.
$\left(ii\right)4KCl{O}_3\to 3KCl{O}_4+KCl$

If the amount of oxygen evolved was 146.8 mL at NTP, calculate the percentage by weight of $KCl{O}_4$ in the residue.

Answer 1

$KCl{O}_3\to KCl+O_2$

Applying POAC for O atoms in the above equation,

$\text{Moles of O in}KCl{O}_3=2\times \text{moles of}{O}_2$

$3\times \text{moles of}KCl{O}_3=2\times \text{moles of}{O}_2$

$3\times \frac{\text{wt. of}KCl{O}_3}{\text{molecular wt. of}KCl{O}_3}=2\times \frac{\text{Volume at NTP (ml)}}{22400}$

$\text{wt. of}KCl{O}_3=\frac{2\times 146.8\times 122.5}{3\times 22400}=0.5358g$

In the second reaction:

$KCl{O}_3\text{amount left}=1–0.5358=0.4642g$ .

We have,
$KCl{O}_3\to KCl{O}_4+KCl$

Applying POAC for O atoms,

$\text{Moles of O in}KCl{O}_3=\text{Moles of}KCl{O}_4$

$3\times \text{moles of}KCl{O}_3=4\times \text{moles of}KCl{O}_4$

$3\times \frac{wt.ofKCl{O}_3}{molewt.ofKCl{O}_3}=4\times \frac{wt.ofKCl{O}_4}{molewt.ofKCl{O}_4}$

$Wt.ofKCl{O}_4=\frac{3\times 0.4642\times 138.5}{122.5\times 4}=0.3937g$

Weight of residue (w) = 1 – weight of Oxygen

$w=1-\frac{146.8}{24400}\times 32gm=0.7902g.$

$\therefore$ $\%KCl{O}_4\text{in the residue}=\frac{0.3937}{0.7902}\times 100=49.8\%$