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 percentage 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 $O_2$ is given by $2$ moles of $KCl{O}_3$.

Therefore, $\frac{1}{2}\times 8.01\times {10}^{-4}$ moles of $O_2$ $=\frac{2\times 1\times 8.01\times {10}^{-4}}{2\times 3}$ moles of $KCl{O}_3$

$=2.67\times {10}^{-4}$ moles of $KCl{O}_3$

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

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

So, the answer is $65$.