D€coi]ipositio7i of Potassic Chlorate. 5 



& Phil Soc.—Vo\. xvii.). Suppose P molecules of KCIO3 

 to interact and give rise to Qi molecules of KCIO4, O2 

 molecules of KCl, and O3 molecules of O, then we shall 

 have the equation 



(2) PKCIO3 = Q1KCIO4 + QsKQ + QsOa 

 equating the coefficients of K on each side 



P = Qi + Q2, 



the coefficients of CI when equated give a similar equation. 

 From the coefficients of O we derive the equation 



3P = 4Qi + 2Q,. 

 Hence we have four unknown quantities and only two 

 equations. Substituting in (2) we get 



(3) 2PKCIO3 - 2Q1KCIO4 + 2(P - Qi)KCl + (3P - 4Q)02 

 an equation containing two variables with the condition that 

 P and Q are integers and 3P not less that 4Q. From the 

 equation it is manifest that from the interaction of a given 

 number of molecules of KCIO3 we may have several solu- 

 tions of the equation : the number of solutions may be found 



by finding the largest integer in ^— ; an additional solution 



may be found if we admit <? as a possible value of Qi. 

 P= II, the following equations are possible 

 22KC103 = 2KCIO4+ 20KCI + 29O2 

 = 4KCIO4 + 18KCI + 25O2 

 = 6KC104+i6KCl + 2i02 

 -8KCIO4+14KCI+17O2 

 = ioKC104+i2KCl+i302 

 -12KCIO4+10KCI + 9O2 

 = i4KC104 + 8KCl + 502 (7) 

 = i6KC104 + 6KCl+i02 

 Of these (7) is the equator given by Teed. 



If in (i) we make ;/ = 2, the equation becomes 

 4KC103 = 3KC104 + KC1; 

 if P = 2 in equation (3) we have 



4KC103 = 4KCl + 602, 

 and 



4KCIO3 = 2KCIO4 + 2KCI + 2O2 



