Question

Oxygen is prepared by catalytic; decomposition of potassium chlorate $\left(KCl{O}_3\right)$. Decomposition, of potassium chlorate gives potassium chloride $\left(KCl\right)$ and oxygen $\left(O_2\right)$. How many moles and how many grams of $KCl{O}_3$ are required to produce 2.4 mole $O_2$ ?

Answer 1

Decomposition of $KCl{O}_3$ takes place as:
$2KCl{O}_3\left(s\right)\to 2KCl\left(s\right)+3O_2\left(g\right)$
2 mole of $KCl{O}_3=$3 mole of $O_2$
$\because$ 3 mole $O_2$ formed by 2 mole $KCl{O}_3$
$\therefore 2.4$ mole of $O_2$ will be formed by:

$(\frac{2}{3}\times 2.4)$ mol $KCl{O}_3$
= 1.6 mole of $KCl{O}_3$.
Mass of $KCl{O}_3$ = Number of moles $\times$ Molar mass
$=1.6\times 122.5=196g$