Question

What is the volume of oxygen liberated at STP when 12.25 g of potassium chlorate is subjected to heating?

• A. 4.2 L
• B. 1.68 L
• C. 2.24 L
• D. 3.36 L

Answer 1

The correct option is D 3.36 L

The chemical equation for decomposition of $KCl{O}_3$ can be written as:

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

$245$ g of $KCl{O}_3$ gives $67.2$ L of $O_2$ at STP.

$12.25$ g of $KCl{O}_3$ gives $\frac{67.2}{245}\times 12.25$ $=3.36$ L.

Hence, option D is correct.