Răspuns:
a,b) m NaOH = 10% * 16 = 1.6g
n NaOH = m / M = 0.04 moli NaOH
1 2 1
ZnSO4 + 2NaOH = Zn(OH)2 + Na2SO4
x 0.04 x
2 moli NaOH ................ 1 mol ZnSO4
0.04 moli NaOH ................. x
Obtinem x = 0.02 moli ZnSO4
m ZnSO4 = 3.22g
Formula cristalohidratului este ZnSO4*nH2O
5.74g cristalohidrat ................. 3.22g ZnSO4
(161 + 18n) ................................... 161g ZnSO4
Obtinem 161 + 18n = 287
18n = 126
n = 7 molecule H2O intr.o molecula de cristalohodrat
ZnSO4*7H2O
c) Notam cu w = masa H2O adaugata
m sol = w + 5.74
m ZnSO4 = 3.22g
100g sol .................. 16.1g ZnSO4
(w + 5.74) ................... 3.22g ZnSO4
Obtinem : w + 5.74 = 20
w = 14.26g H2O adaugata
d) n Zn(OH)2 = x = 0.02 moli
m Zn(OH)2 = 1.98g.
99g Zn(OH)2 ............... 100g precipitat
1.98g Zn(OH)2 .............. y
Obtinem : y = 2g Zn(OH)2
