# $0.05\;g$ of a commercial sample $KCl_3$ on decomposition liberated just sufficient oxygen for complete oxidation of $20\;ml\; CO$ at $27^{\large\circ}C$ and $750\;mm$ pressure.Calculate % of $KClO_3$ in sample.

$(a)\;70.2\%\qquad(b)\;61.3\%\qquad(c)\;65.4\%\qquad(d)\;62.5\%$

Step 1:
$2KClO_3\rightarrow 2KCl+3O_2$
$CO+\large\frac{1}{2}$$O_2\rightarrow CO_2 For, CO PV=\large\frac{w}{m}$$RT$
$\therefore$ Mole of CO = $\large\frac{PV}{RT}$
$= \large\frac{750\times 20}{760\times 0.081\times 300\times 1000}$
$= 8.01\times 10^{-4}$
Step 2:
Mole ratio for combination $CO : O_2$ $= 2 : 1$
$\therefore$ Mole of $O_2$ required for CO $=\large\frac{1}{2}$$\times 8.01\times 10^{-4} 3 mole of O_2 are given by 2 mole of KClO_3 \therefore\large\frac{8.01\times 10^{-4}}{2} of O_2 are given by \large\frac{2\times 8.01\times 10^{-4}}{2\times 3} Mole of KClO_3=2.67\times 10^{-4} Mole of KClO_3 \therefore weight of KClO_3=2.67\times 10^{-7}\times 122.5g = 3.27\times 10^{-2}g \% of KClO_3 in sample =\large\frac{3.27\times 10^{-2}}{0.05}$$\times 100 =65.4$%
Hence (c) is the correct answer.
edited Mar 18, 2014