# The decomposition of potassium chlorate (KClO_3) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen?

Feb 25, 2017

$K C l {O}_{3} \left(s\right) + \Delta \rightarrow K C l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right)$

#### Explanation:

Normally, this rxn must be calalyzed with a little $M n \left(I V\right)$ salt, but we address the question with this stoichiometry.

We spit out a $19.7 \cdot m o l$ quantity of dioxygen gas. Given the stoichiometry there were $\frac{2}{3} \times 19.7 \cdot m o l$ $\text{potassium chlorate}$.

This represents a mass of $\frac{2}{3} \times 19.7 \cdot m o l \times 122.55 \cdot g \cdot m o {l}^{-} 1$ of $\text{potassium chlorate}$ $\approx$ $1.6 \cdot k g$.

This is not a reaction that you do at home.