Al + 3HCl---> AlCl3 + 3/2H2 ori a moli
Mg+2HCl---> MgCl2 + H2 ori b moli
fac bilantul de masa metale: 27a +24b= 8,7........ec.1
fac bilantul moli H2: 1,5a+4= 0,4............................ec/2
fac bilantul moli HCl consumat: 3a+2b....................ec.3
a)din ec.1 si 2. rezulta: a= 0,1mol b= 0,25mol-->-a/b= 1/2,5
b inlocuiesc a si b in ec 3 si aflu masa acid consumat si masa de acid ramasa
m,HCl=niuxM= (0,3+0,5)X36,5g= 29,2g consumat
m,HCl ramas=md-29,2, unde md se asfla din
c= mdx100/ms---> md-=36,5x200/100= 73g
m,HCl ramas= 43,8g
aceasta masa intra in reactia ,din care aflu masa de NaOH cu care reactioneaza, apoi masa de solutie
36,5g(1mol) 40g(1mol)
HCl + NaOH----> NaOH + HCl
43,8.......................................X= 48g
c= mdx100/ms---> ms= mdx100/c= 48x100/40g=120g
c) niu.acid consumat= 0,8mol
1mol............6,023x10²³molec
0,8mol...........................N=4,8x10²³ molecHCl
