cu HCl reactioneaza doar Al---. > deduc a
n,mol = V/V0---> n,H2= 10,08l/22,4l/mol= 0,45mol
2mol 6mol 3mol
2Al + 6HCl---> 2AlCl3 + 3H2
n=................n'..........................o,45mol
n=0,3molAl-----> m= 0,3molx27g/mol=8,1mol
m,Cu= 16,2-8,1=8,1g
aliajul contine 50% Al si 50% Cu !
n'= 0,9mol HCl-------> m'=0,9molx 36,5g/mol = calculeaza !!!!!