Question

The percentage loss in weight after heating a pure sample of potassium chlorate (mol. $wt.=122.5$) will be:

• A. 12.25
• B. 24.5
• C. 39.18
• D. 19.6

Answer 1

Answer: (C)

39.18

$2KCl{O}_3⟶2KCl+3O_2$

$2\times 122.5$ $g$ $3\times 32$ $g$ $O_2$

$⟹2\times 122.5$ $g$ $KCl{O}_3$ show loss of $3\times 32$ $g$ of $O_2$

$100$ $g$ $KCl{O}_3$ shows weight loss of $\frac{3\times 32\times 100}{2\times 122.5}=39.17$

$⟹\%$ Loss of weight $=39.17$