a. 180g--> 3 mol acid acetic
b 3mol 1mol 1mol
3CH3COOH +Al------> (CH3COO)3Al +3/2H2
3x30/100 1x30/100 1x30/100
n = 0,3molAl-->m=0,3x27g= 8,1gAl pur->m=8,1x100/80gAl impur=>10,125gAl impur
n= 0,3mol sare--->m=0,3X204g= 61,2g sare -teoretic--> m=61,2x75/100=
=45,9g -practic