Question

A way of generating dry O2 in the lab is to heat potassium chlorate. If you determined that 25 grams of O, are needed, how much potassium chlorate do you need to start with? 2KCI0, ->(heat)2KCI + 30,

Answer 1

About 64 grams of KClO₃.

Step-by-step explanation:

Potassium chlorate dissociates according to the equation:

`\displaystyle 2\text{KClO$_3$} \xrightarrow{\text{heat}} 2\text{KCl} + 3\text{O$_2$}`

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:

`\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$_3$}`

In conclusion, about 64 (to two significant digits) grams of KClO₃ is needed.