Question

In the reaction,
$2KCl{O}_3⟶2KCl+3O_2$
when 36.75 g of $KCl{O}_3$ is heated, the volume of oxygen evolved at NTP will be:

• A. $9.74d{m}^3$
• B. $8.92d{m}^3$
• C. $10.08d{m}^3$
• D. $22.4d{m}^3$

Answer 1

Answer: (C)

$10.08d{m}^3$

$2KCl{O}_3⟶KCl+3O_2$

$245g$ of $KCl{O}_3$ gives $22.4\times 3d{m}^3$ of $O_2$

$\therefore 36.75g$ of $KCl{O}_3$ gives $\frac{22.4\times 3}{245}\times 36.75=10.08d{m}^3$ of $O_2$