2Ag2O----> 4Ag + O2
4Al+3O2--->2Al2O3
m INTRODUSA: 4xM,Ag+4xM,O+ 4XM,Al
m,REZULTATA: 4XM,Ag +4xM,O +4XM,Al -----. sunt egale!! SE APLICA LEGEA CONSERVARII MASEI !!!
M=(2x108gAg+16g)0/mol= 232g/mol Ag2O
M=27g/mol Al
2X232+4X27g.................2X.232gAg2O...........4X27gAl
100g amestec...............x.....................y
CALCULEAZA x% Ag2o si y% Al