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  4. Oxygen is prepared by catalytic decompostion of potassium chlorate, KClO3. Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2). If 2.4 mole of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?

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Q. Oxygen is prepared by catalytic decompostion of potassium chlorate, $KClO_3$. Decomposition of potassium chlorate gives potassium chloride $(KCl)$ and oxygen $(O_2)$. If $2.4$ mole of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?

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Solution:

$2KClO_{3(s)} \to 2KCl_{(s)} + 3O_{2(g)}$
$\therefore 3 $ mol of oxygen requires $2$ mol of $KClO_3$.
For $2.4$ mol of oxygen, we need $= 1.6 \,mol$ of $KClO_3$
Therefore, mass of $KClO_3$ required =
$1.6\, mol \times 122.5 \,g \,mol^{-1} (M.W.$ of $KClO_3 = 122.5)$
$= 196.0 \,g$