Question

$1.0$ g sample of $KCl{O}_3$ was heated under such conditions that a part of it decomposed according to the equation:
$2KCl{O}_3\to 2KCl+3O_2$,
and the remaining underwent change according to the equation:
$4KCl{O}_3\to +3KCl{O}_4+KCl$.
If the amount of $O_2$ evolved was $146.8$ mL at STP, calculate the percentage by weight of $KCl{O}_4$ in the residue.

• A. $49.87$
• B. $50.87$
• C. $49.00$
• D. None of these

Answer 1

The correct option is A $49.87$

$\underset{245g}{2KCl{O}_3}\to \underset{149g}{2KCl}+\underset{96g}{3O_2}$
$\underset{490g}{4KCl{O}_3}\to \underset{415.5g}{3KCl{O}_4}+\underset{74.5}{KCl}$
Given, total weight of the sample $=1$ g
Let $x$ g of $KCl{O}_3$ be converted to $KCl$ and $O_2$ and $(1-x)$ g be converted to $KCl{O}_4$ and $KCl$.
Given, $146.8$ mL of $O_2$ was evolved at STP.
Weight of $O_2=\frac{146.8\times 32}{22400}=0.2097$ g
Weight of $KCl{O}_3$ converted to $KCl$ and $O_2=$ $x=\frac{245\times 0.2097}{96}=0.5352$ g
Weight of $KCl{O}_4$ formed from remaining $KCl{O}_3$ $=\frac{415.5\times (1-x)}{490}=0.3941$ g
Weight of residue $=1-$ weight of $O_2=1-0.2097$ g
$\therefore \%$ of $KCl{O}_4$ in the residue $=\frac{0.3941\times 100}{(1-0.2097)}=49.87\%$