Question

$0.05g$ of a commercial sample of $KCl{O}_3$ on decomposition liberated just sufficient oxygen for complete oxidation of $20mL$ $CO$ at ${27}^{o}C$ and $750mm$ pressure. Calculate $\%$ of $KCl{O}_3$ in sample.

• A. 60.6 %
• B. 65.4 %
• C. 75.2 %
• D. 70.4 %

Answer 1

Answer: (B)

65.4 %

$2CO+O_2\to 2C{O}_2$

$2$ $:$ $1$ molar ratio

Number of moles of $C{O}_2$ can be found

$\frac{PV}{RT}=n=\frac{\frac{750}{760}atm\times 20\times {10}^{-3}L}{0.0821\times 300K}$

$=0.8013\times {10}^{-3}moles$

$\therefore$ Moles of $O_2$ will be half the moles of $CO$

$n_{O_2}=\frac{0.8013\times {10}^{-3}}{2}moles=4\times {10}^{-4}moles$ of $O_2$ was liberated.

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

$2$ $:$ $3$ molar ratio

$x$ $:$ $4\times {10}^{-4}$

$x=\frac{4\times {10}^{-4}\times 2}{3}=2.66\times {10}^{-4}$ moles was used.

Molar weight of $KCl{O}_3$ is $122.5g/mole$

Weight used is $=122.5\times 2.66\times {10}^{-4}=0.0327g$

Weight of sample $=0.05g$

% $KCl{O}_3$ in sample $=\frac{0.0327}{0.05}\times 100=65.33$%