Question

When $1.225\mathrm{kg}$ of potassium chlorate was subjected to decomposition, the oxygen evolved in the reaction occupied a volume of $224\mathrm{L}$. What would be the percentage purity of potassium chlorate?

• A. 82%
• B. 30%
• C. 66%
• D. 25%

Answer 1

The correct option is C

66%

Explanation for the correct option:

• The given compound in the question is Potassium perchlorate (${\mathrm{KClO}}_3$) and the amount of Potassium perchlorate is $1.225\mathrm{kg}=1225\mathrm{gm}$
• The chemical reaction for the decomposition of Potassium perchlorate is given below:
• $\underset{\mathrm{Potassium}\mathrm{perchlorate}}{2{\mathrm{KClO}}_3\left(\mathrm{g}\right)}\to \underset{\mathrm{Potassium}\mathrm{chloride}}{2\mathrm{KCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}$

In the chemical reaction the amount of Potassium perchlorate and Oxygen is:

$\underset{2\times 1225.5=245g}{2{\mathrm{KClO}}_3\left(\mathrm{g}\right)}\to \underset{\mathrm{Potassium}\mathrm{chloride}}{2\mathrm{KCl}\left(\mathrm{s}\right)}+\underset{3\times 22.4L=67.2L}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}$

From the equation, we can see that $245\mathrm{gm}$Potassium perchlorate (${\mathrm{KClO}}_3$) will produce $67.2\mathrm{L}$ Oxygen.

$224\mathrm{L}$ of Oxygen will be produced from = $\frac{245}{67.5}\times 224=813\mathrm{gm}$ of Potassium perchlorate (${\mathrm{KClO}}_3$)

The purity of the sample will be given as = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{pure}{\mathrm{KClO}}_3}{\mathrm{Mass}\mathrm{of}\mathrm{given}\mathrm{sample}}\times 100$

$=\frac{813}{1225}\times 100=66.1\%$

Hence, the purity of Potassium perchlorate is $66.1\%~66\%$

Therefore, the correct answer is option (C).