Question

calculati m si niu ZnCl2 care sa format la interactiunea 8g de Zn cu Hcl​

Answer 1

Răspuns:

65g....................136g

Zn +2 HCl => ZnCl2 + H2

8g.......................xg

MZn=65g/mol

NZnCl2=65+ 2×35,5=136g/mol

x=8×136/65=16,73g ZnCl2

n=m/M=8/65=0,12mol

1mol.....................1mol

Zn +2 HCl => ZnCl2 + H2

0,12moli..............ymoli

y=0,12×1/1=0,12 moli ZnCl2

Answer 2

Răspuns:

16,73g ZnCl₂ (0,12 moli ZnCl₂)

Explicație:

M(ZnCl₂) = A(Zn) + 2 × A(Cl) = 65 + 2 × 35,5 = 136 g/mol

8g‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍x

Zn + 2HCl → ZnCl₂ + H₂↑

65g‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍ ‍ ‍ ‍‍‍‍‍136g

x = 8 × 136 : 65 = 16,73g ZnCl₂

1 mol ZnCl₂..........136g ZnCl₂

n moli ZnCl₂..........16,73g ZnCl₂

n = 16,73 : 136 = 0,12 moli ZnCl₂

-Mihai