Question

The $\%$ loss in weight after heating a pure sample of potassium chlorate (Molecular weight $=122.5$ g/mol) will be:

• A. $12.25$
• B. $24.5$
• C. $39.2$
• D. $19.6$

Answer 1

Answer: (C)

$39.2$

The balanced chemical equation for the thermal decomposition of potassium chlorate is,

$2KCl{O}_3\stackrel{\Delta }{\to }2KCl+3O_2$

2 moles of potassium chlorate corresponds to 3 moles of oxygen.

Thus, $2\times 122.5=245g$ of potassium chlorate corresponds to $3\times 32=96g$ of oxygen.

The $\%$ loss in weight after heating a pure sample of potassium chlorate (Mol. wt. $=122.5$ g/mol) will be $\frac{96}{245}\times 100=39.2$%.