Question

How many moles of potassium chlorate should be decomposed completely to obtain $67.2$ litres of oxygen at $STP$?

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

The reaction is $2KCl{O}_3\to 2KCl+3O_2$.

The decomposition of two moles of potassium chlorate gives three moles of oxygen at STP.

At STP, one mole of oxygen occupies $22.4$ L.

Hence, $3$ moles of oxygen will occupy $67.2$ L.

Thus, to obtain $67.2$ L of oxygen at STP, $2$ moles of potassium chlorate should be decomposed.

Therefore, option D is correct.