Respuesta :
Explanation:
The given reaction will be as follows.
[tex]KClO_{3}(s) \rightarrow KCl(s) + \frac{3}{2}O_{2}(g)[/tex]
Molar mass of [tex]KClO_{3}[/tex] = 122.5 g/mol
Molar mass of KCl = 74.5 g/mole
Molar mass of [tex]O_{2}[/tex] = 32 g/mole
We assume that the mass of [tex]KClO_{3}[/tex] in the sample be x g
Therefore, the mass of KCl in the sample = 0.95 - x
Hence, moles of [tex]KClO_{3}[/tex] = [tex]\frac{x}{122.5}[/tex]
So, moles of KCl produced = moles of [tex]KClO_{3}[/tex] = [tex]\frac{x}{122.5}[/tex]
or, mass of KCl produced = [tex](\frac{x}{122.5} ) \times 74.5[/tex]
Therefore, total mass of KCl = mass of residue = (0.95 - x) + [tex]{(\frac{x}{122.5}) \times 74.5}[/tex] = 0.820 g
-0.39x = 0.820 - 0.95
x = 0.33
Thus, % of [tex]KClO_{3}[/tex] in the original sample will be calculated as follows.
[tex]\frac{x}{0.95} \times 100[/tex]
= 34.73%
Thus, we can conclude that mass % of [tex]KClO_{3}[/tex] is 34.73%.
