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.
