Question

The volume occupied by oxygen at STP obtained on heating the 490 g of KClO₃ is 130 L. Calculate the percentage purity of KClO₃ (KClO₃ on heating produces KCl and O₂).

Answer 1

Step-1: Calculate the amount of Oxygen gas obtained by heating 490 g of KClO₃

$$\begin{gathered} 2{\mathrm{KClO}}_3\left(\mathrm{s}\right)\stackrel{∆}{\to }2\mathrm{KCl}\left(\mathrm{s}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right) \\ (\mathrm{Potassium}\mathrm{chlorate})(\mathrm{Potassium}\mathrm{chloride})\left(\mathrm{Oxygen}\right) \end{gathered}$$

Gram molecular weight of Potassium chlorate = 39+ 35.5+ 3(16)

The gram molecular weight of Potassium chlorate = 122.5 g mol^-1

The gram molecular weight of Oxygen gas= 22.4 L at STP (Avogadro's law)

According to the balanced chemical equation,

2×122.5 g of potassium chlorate produces= 3x22.4 L of Oxygen gas

1 g of potassium chlorate produces= $\frac{3\times 22.4}{2\times 122.5}$L of Oxygen gas

490 g of potassium chlorate produces= $\frac{3\times 22.4}{2\times 122.5}\times 490=134.5$ L of Oxygen gas

Step-2: Calculate the Percentage purity

$Percentagepurity=\frac{Givenvolumeofoxygengas}{Calculatedvolumeoftheoxygen}\times 100$

$Percentagepurity=\frac{130}{134.5}x100=96.726\%$