Calculate the mass of potassium chlorate required to liberate 6.72...

The balanced chemical equation for the decomposition of potassium chlorate is:

$2KClO_3 \rightarrow 2KCl + 3O_2$

From the stoichiometry of the reaction, $2$ moles of $KClO_3$ produce $3$ moles of $O_2$.

At STP, the volume of $1$ mole of an ideal gas is $22.4$ dm^3$.<br><br>Volume of$3$moles of$O_2 = 3 \times 22.4 = 67.2$dm^3$.

Mass of $2$ moles of $KClO_3 = 2 \times 122.5 = 245$ g.

Thus, $67.2$ dm^3$of$O_2$is liberated by$245$g of$KClO_3$.<br><br>Mass of$KClO_3$required to liberate$6.72$dm^3$ of $O_2 = \dfrac{245}{67.2} \times 6.72 = 24.5$ g.

Answer: $24.5$ g