din aliajul dat, doar Zn reactioneaza cu HCl; din ecuatia reactiei deduc m,Zn
M,Zn= 65g/mol\
Vm=22,4L/mol
1mol...........................................1mol sau
Zn + 2HCl----> ZnCl2 + H2
65g.....................................22,4 l
x..................................................1,37 l
m,Zn=3,975g
m,Cu= 10-3,975=>6,025g
10g aliaj........3,975gZn .....6,025gCu
100g..........x=39,75%...........y= 60,25%
cu clorul reactioneaza ambele metale
Zn + Cl2--> ZnCl2
Cu + Cl2===> CuCl2
verifica , cu masele obtinute, ca se folosesc 3,478 l clor !!