HCl
M=1 + 35,5=36,5------> 1mol= 36,5 g
raport de masa = r . m.= AH :ACl =1 : 35,5
Comp. procentuala prin masa moleculara
36,5 gHCl-----------1g H------------35,5 g Cl
100 g------------------x------------------------y
x=100: 36,5=2,73 la suta H
Y=100.35,5 : 36,5=97,26 la suta Cl
Al ( OH )3
M=27+3.16 + 3=78-----> 1mol=78 g
r. m.= AAl :3AO : 3AH=27 :3.16 : 3
=27: 48 :3 [se simplifica cu 3 ]
=9: 16 : 1
Comp. proc. prin masa moleculara
M=27 +3.16 +3=78----->1moli=78g
78g Al(OH)3----------27 g Al--------48 g O--------3g H
100 g--------------------x -----------------y------------------z
x=100.27 : 78=
y=100.48:78=
z=100.3 : 78=
Dupa aceste exemple lucreaza-le pe urmatoarele doua. Daca mai ai nevoie
scrie.
