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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)
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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
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