Question

A 1 g sample of $KCl{O}_3$ was heated under such conditions that a part of it decomposed according to the equation:
(i) $2KCl{O}_3=2KCl+3O_2$
and the remaining underwent change according to the equation
(ii) $4KCl{O}_3=3KCl{O}_4+KCl$
If the amount of $O_2$ evolved was 146.8 mL at STP, calculate the mass of $KCl{O}_4$ produced.

Answer 1

Given,
$KCl{O}_3\to KCl+O_2$
Applying POAC for oxygen atom is equation (i)
$\text{moles of O in}KCl{O}_3=\text{moles of O in}{O}_2$
$\Rightarrow 3\times \text{moles of}KCl{O}_3=2\times \text{moles of}{O}_2$
$\Rightarrow 3\times \frac{\text{mass of}KCl{O}_3}{\text{molar mass of}KCl{O}_3}=2\times \frac{\text{volume of}{O}_{2}atSTP}{22400}$
$\Rightarrow \text{mass of}KCl{O}_3=\frac{2\times 134.4\times 122.5}{3\times 22400}=0.49g$

In the second reaction,
$\text{Amount of}KCl{O}_3\text{left}=1-0.49g=0.51g$
Given, reaction is
$KCl{O}_3\to KCl{O}_4+KCl$
Applying POAC for oxygen atom is equation (ii)
$\text{moles of O in}KCl{O}_3=\text{moles of O in}KCl{O}_4$
$\Rightarrow 3\times \text{moles of}KCl{O}_3=4\times \text{moles of}KCl{O}_4$
$\Rightarrow 3\times \frac{\text{mass of}KCl{O}_3}{\text{molar mass of}KCl{O}_3}=4\times \frac{\text{mass of}KCl{O}_4}{\text{molar mass of}KCl{O}_4}$
$\Rightarrow \text{mass of}KCl{O}_4=\frac{3\times 0.51\times 138.5}{122.5\times 4}=0.432g$