V=5ooml=0,5l
-calculez moli Al(OH)3 din solutie
c= n,mol/ V,litri' 2= n/0,5--> n=1molAl(OH)3
-ecuatia reactiei de neutralizare
2Al(OH)3+3H2SO4 ----> Al2(SO4)3 +6H2O
2mol.... ...3mol
1mol........n= 3/2 molH2S04
-masa de acid
m= n,molxM,g/mol; m= 1,5mol(2gH+32gS+4x16gO)/mol= 147g;aceasta masa se afla dizolvata in solutia 98%
-masa de solutie 98%
c/100= md/ms; 98/100=147/ms--------------> calculeaza ms !!!!!