Question

$0.05$ g of a commercial sample of $KCl{O}_3$ on decomposition liberated just sufficient oxygen for complete oxidation of $20$ mL of $CO$ at $27$ ${}^{o}C$ and $750$ mm pressure. The $\%$ of $KCl{O}_3$ in the sample is (as the nearest integer) :

Answer 1

Moles of $CO$ used $\frac{PV}{RT}=\frac{750\times 20}{760\times 0.0821\times 300\times 1000}=8.01\times {10}^{-4}$

$\therefore$ Moles of $O_2$ required for oxidation of CO $=\frac{1}{2}\times 8.01\times {10}^{-4}$ [$\because CO+\frac{1}{2}{O}_2\to C{O}_2$]

Now, $2KCl{O}_3\to 2KCl+3O_2$

$\because$ $3$ moles of $O_2$ are given by $2$ moles of $KCl{O}_3$

$\frac{1}{2}\times 8.01\times {10}^{-4}$ mole of $O_2$

$=\frac{2\times 1\times 8.01\times {10}^{-4}}{2\times 3}moleofKCl{O}_3$ $=2.67\times {10}^{-4}$ mole of $KCl{O}_3$

Mass of $KCl{O}_3=2.67\times {10}^{-4}\times 122.5$ $=3.27\times {10}^{-2}$ g

% of $KCl{O}_3=\frac{3.27\times {10}^{-2}}{0.05}\times 100=65.4$

So, answer is $65$.