M(HCl)=1+35.5=36.5
1mol HCl are.............36.5 g HCl..........35.5g Cl
4 moli....................................................x
x=4*35.5=142g Cl
M(CuCl2)=64+2*35.5=135
1mol CuCl2................135g CuCl2............2*35.5=71g Cl
y............................142g
y=135*142/71=270g CuCl2 vor contine aceeasi masa de Cl ca 4 moli de HCl