Question

6,5 g de zinc se trateaza cu o solutie de HCl de concentratie 20%.

a. ce cantitate de solutie de HCl de concentratie 20% se consuma?
b. ce cantitate de sare rezultă?(g, moli)
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Answer 1

Răspuns:

a)

MZn=65g/mol

MHCl=1+35,5=36,5g/mol

65g....2x36,5g

Zn  +  2HCl  →  ZnCl2  +  H2↑

6,5g.....xg

x=6,5x2x36,5/65 = 7,3g HCl

C = mdx100/ms

ms = 100xmd/C = 100x7,3/20 = 36,5g.

b).

calculam in grame

MZnCl2=65+2x35,5=65+71=136g/mol

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

Zn  +  2HCl  →  ZnCl2  +  H2↑

6,5g.......................yg

y=6,5x136/65=13,6g ZnCl2

calculam in moli

n=m/M=6,5/65=0,1moli Zn

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

Zn  +  2HCl  →  ZnCl2  +  H2↑

0,1moli...............zmoli

z=0,1x1/1=0,1 moli ZnCl2

Explicație: