C=n/V; 1=n/o,1--> n=o,1mol NaOH
CnH2n+1-COOH+NaOH--> CnH2n+1COONa+H2O
0,1MOL.................O,1MOL
0,1=7,4/M---M=74g/mol
74= 14n+1+12+32+1--> n=2 acidul acetic CH3COOH
OMOLOGUL INFERIOR ESTE ACIDUL FORMIC HCOOH
REACTIILE SCRIE-LE DUPA MODELUL REACTIEI din problema , dar
2CH3COOH+Fe----->(CH3COO)2Fe +H2
3CH3COOH+Al(OH)3----> (CH3COO)3Al+3H2O
