9.0 mol
Potassium chlorate KClO 3 decomposes to produce potassium chloride KCl and oxygen O2. Balancing the equation KClO3→KCl+O2 (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 KClO3→KCl+32O22KClO3→2KCl+3O2 Hence the ratio of number of moles of particles n(O2)n(KClO3)=32n(O2)=n(KClO3)⋅n(O2)n(KClO3)=32⋅
n(KClO3)Therefore the decomposition of 6.0l mol
of KClO 3 would yield 6.0⋅n
(O2)n(KClO 3)=6.0⋅ 32=9.0l molNote 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.