Question

❗❗❗e urgent❗❗❗
Dau coroana
Determinati citi moli de gaz renultă din reactia a 13 g de zinc cu HCI si cate grame de acid se consumă in reactie.
(Masa atomică: H-1; C1-35,5, 2n-65)​

Answer 1

Zn+2HCl->ZnCl2+H2

nZn=mZn/MZn=>nZn=13 g/65 g=0,2 moli=>nH2=0,2 moli

nHCl=0,4 moli=>nHCl=mHCl/MHCl=>mHCl=0,4×36,5=14,6 g

Answer 2

Răspuns:

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

Zn + 2HCl => ZnCl2+ H2

0,2 moli..........................x moli

MZn= 65g/mol

n=m/M=13/65=,02 moli

x=0,2×1/1=0,2 moli H2

65g......2×36,5g

Zn + 2HCl => ZnCl2+ H2

13g........yg

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

y=13×2×36,5/65= calculează