Question

A mixture of KClO3 and KCl with a mass of 0.950 g was heated to produce O2. After heating, the mass of residue was 0.820 g. Assuming all the KClO3 decomposed to KCl and O2, calculate the mass percent of KClO3 in the original mixture.

Answer 1

Explanation:

The given reaction will be as follows.

         $KClO_{3}(s) \rightarrow KCl(s) + \frac{3}{2}O_{2}(g)$

Molar mass of $KClO_{3}$ = 122.5 g/mol

Molar mass of KCl = 74.5 g/mole

Molar mass of $O_{2}$ = 32 g/mole

We assume that the mass of $KClO_{3}$ in the sample be x g

Therefore, the mass of KCl in the sample = 0.95 - x

Hence, moles of $KClO_{3}$ = $\frac{x}{122.5}$

So, moles of KCl produced = moles of $KClO_{3}$  = $\frac{x}{122.5}$

or, mass of KCl produced = $(\frac{x}{122.5} ) \times 74.5$

Therefore, total mass of KCl = mass of residue = (0.95 - x) + ${(\frac{x}{122.5}) \times 74.5}$ = 0.820 g

                         -0.39x = 0.820 - 0.95

                             x = 0.33

Thus, % of $KClO_{3}$ in the original sample will be calculated as follows.

                   $\frac{x}{0.95} \times 100$

                      = 34.73%

Thus, we can conclude that mass % of $KClO_{3}$ is 34.73%.