# How many moles of oxygen are produced by the decompositon of 6.0 moles of potassium chlorate, KCLO3?

Apr 18, 2018

$9.0 \textcolor{w h i t e}{l} \text{mol}$

#### Explanation:

Potassium chlorate ${\text{KClO}}_{3}$ decomposes to produce potassium chloride $\text{KCl}$ and oxygen ${\text{O}}_{2}$. Balancing the equation

${\text{KClO"_3 to "KCl" + "O}}_{2}$ (not balanced)

based on the fact that the number of moles of oxygen atoms should be the same on both side of the equation will give

${\text{KClO"_3 to "KCl" + 3/2 "O}}_{2}$
$\textcolor{b l u e}{2} {\text{KClO"_3 to 2"KCl" + color(green)(3) "O}}_{2}$

Hence the ratio of number of moles of particles

$\left(n \left({\text{O"_2))/(n("KClO}}_{3}\right)\right) = \frac{\textcolor{b l u e}{3}}{\textcolor{g r e e n}{2}}$

$n \left({\text{O"_2)=n("KClO"_3)*(n("O"_2))/(n("KClO"_3))=3/2*n("KClO}}_{3}\right)$

Therefore the decomposition of $6.0 \textcolor{w h i t e}{l} \text{mol}$ of ${\text{KClO}}_{3}$ would yield

6.0*(n("O"_2))/(n("KClO"_3))=6.0*3/2=9.0color(white)(l)"mol"

Note that the species $\text{KCl}$ is highly stable. Both the potassium cation ${\text{K}}^{+}$ and the chloride anion ${\text{Cl}}^{-}$ have attained a noble gas octet configuration of valence shell electrons. Chemical reactions tend to favor the most chemically stable combination. Therefore this decomposition reaction would eventually produce $\text{KCl}$ rather than other salts that contain oxygen.