4.30.
Presupunem V = 1L = 1000mL sol.
ro = ms/V => ms = ro×V = 1,18×1000 = 1180g CuSO4
C = md/ms×100 => md = C×ms/100 = 16×1180/100 = 188,8g CuSO4
C = md/M×Vs = 188,8/160×1 = 1,18M
4.31. ro = ms/V => ms = 1,22×20 =24,4g sol. HCl
C = md/ms×100 => md = 30×24,4/100 = 7,32g HCl
n HCl = 7,32/36,5 = 0,2 moli
Zn + 2HCl => ZnCl2 + H2
65g Zn...........2 moli HCl
xg Zn...............0,2 moli HCl
X = 6,5g Zn
n Zn = 6,5/65 = 0,1 moli
R: Cantitatea de HCl este suficienta pentru a dizolva complet 6,54g Zn