Răspuns:
Explicație:
3250 g Zn
VCl2= 560 mL= 0,56L
cantit. ZnCl2
- se afla nr. de moli de Cl2
n= 0,56 L : 22,4 l/moli= 0,025 moli Cl2
mCl2= 0,025moli . 71g/moli=1,775 gCl
x g 1,775g yg
Zn + Cl2 = ZnCl2
65g 71g 136g
x= 65 . 1,775 : 71 = 1,625 g Zn ------> deci Zn este in exces
3250 g - 1,625g = 3248, 375 g Zn in exces
y= 1,775 . 136 : 71=3,4 g Zn Cl2