Cum 1molcompus are 2 molCl ,rezulta ,la analiza, 2molAgCl.
miu= (14x12gC+16x1gH+2x35,5gCl+2x16gO)/mol= 287g/mol compus
miu=(108Ag+35,5gCl)/mol= 143,5g/mol AgCl
din287g compus............rezulta 2x143,5gAg Cl
10g...........................................m= calculeaza !!!!