Question

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

Answer 1

• The given chemical reaction is 2KClO₃ (s) $\stackrel{∆}{\to }$ 2 KCl (s)+ 3O₂(g).
• From the equation, 2 moles of Potassium chlorate(KClO₃) on heating produce 3 moles of Oxygen(O₂).

Molar mass of potassium chlorate(KClO₃) =39+35.5+(3×16)=122.5g

To produce 3×22.4L O_2, required number of moles of Potassium chlorate(KClO₃)= 2 moles.

Therefore to produce 130L of O_2, the required number of moles of potassium chlorate(KClO₃) =$\frac{2\times 130}{3\times 22.4}$=3.87moles

Mass of potassium chlorate(KClO₃) =3.87×122.5= 473.96g.

• Percentage purity of sample=$\frac{\mathrm{Mass}\mathrm{of}\mathrm{given}\mathrm{sample}}{\mathrm{Mass}\mathrm{of}\mathrm{pure}\mathrm{sample}}\times 100$=$\frac{473.96}{490}\times 100$= 96.72%.