96 
Hence we have the system of equation 
Pi “ Q25 P2 = 2Q 4 , P2 — 2Q 2 + 2Q 3 , 3P 2 = Qi + 6Q 2 + 2Q 3 . 
Hence we get P 2 = fP l5 Q x = fP 4 , Q 2 = P 4 , Q 3 = f P 1? substituting 
in the equation and multiplying each side by ~ we get the 
usual equation 
3Cu + 8HNO s = 4H 2 0 + 2Cu(N0 3 ) 2 + 2N 2 0 3 
II. Potassium ferrocyanide, water, and sulphuric acid 
yield ferrous sulphate, ammonium sulphate, potassium sul- 
phate, and carbonic oxide. 
P^NeKJFe) + P 2 (H 2 0) + P 3 (H 2 S0 4 ) 
= Qi(CO) + Q 2 (FeS0 4 ) + Q 3 (NH 4 ) a S0 4 + Q 4 (K 2 S0 4 ) 
Here the group S0 4 may be treated as an element. From 
the above equation we get the subsidiary equations 
6PF=Q b 6Py=2Q 3 , 4P 1 =2Q 4 , P 4 ^Q 2 , 2P 2 +2P 3 =8Q 3 , P 3 =Q 2 +Q 3 -f Q* 
Heuce we get 
Qi = 6P 4 , Q 2 = Pi, Q 3 jj= 3P 4 , Q 4 = 2P 4 , P a = 6P 4 , P 3 = 6P 4 
Replacing these values in the equation and dividing both 
sides by P 4 we get 
C 6 !s T 6 K 4 Fe+ 6H 2 0 + 6H 2 S0 4 - 6C0+FeS0 4 -f 3(NH 4 ) a S0 4 + 2(K 2 S0 4 ) 
which is the equation given by Fownes. 
