Potassium chlorate "KClO"_3 decomposes to produce potassium chloride "KCl" and oxygen "O"_2. Balancing the equation
"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
"KClO"_3 to "KCl" + 3/2 "O"_2
color(blue)(2)"KClO"_3 to 2"KCl" + color(green)(3) "O"_2
Hence the ratio of number of moles of particles
(n("O"_2))/(n("KClO"_3))=color(blue)(3)/color(green)(2)
n("O"_2)=n("KClO"_3)*(n("O"_2))/(n("KClO"_3))=3/2*n("KClO"_3)
Therefore the decomposition of 6.0color(white)(l)"mol" of "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 "KCl" is highly stable. Both the potassium cation "K"^+ and the chloride anion "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 "KCl" rather than other salts that contain oxygen.
