Answer:
About 64 grams of KClO₃.
Explanation:
Potassium chlorate dissociates according to the equation:
[tex]\displaystyle 2\text{KClO$_3$} \xrightarrow{\text{heat}} 2\text{KCl} + 3\text{O$_2$}[/tex]
To determine the amount of KClO₃ necessary to produce 25 grams of O₂, we can convert grams of O₂ to moles; moles of O₂ to moles of KClO₃; and finally to grams of KClO₃.
The molecular weights of O₂ and KClO₃ are 16.00 g/mol and 122.55 g/mol.
Hence:
[tex]\displaystyle 25\text{ g O$_2$} \cdot \frac{1\text{ mol O$_2$}}{32.00\text{ g O$_2$}} \cdot \frac{2\text{ mol KClO$_3$}}{3\text{ mol O$_2$}} \cdot \frac{122.55\text{ g KClO$_3$}}{1\text{ mol KClO$_3$}} = 64\text{ g KClO$_3$}[/tex]
In conclusion, about 64 (to two significant digits) grams of KClO₃ is needed.
