Zn + 2HCl = ZnCl2+ H2
0,02....0,04....0,02....0,02
nZn=0,06 mol
nHCl =0,04 mol
=> Zn dư
mZnCl2 = 0,02.136=2,72 g
V H2 =0,02.22,4=0,448 l
mZn dư =65.(0,06-0,02)=2,6 g
\(Zn+2HCl-->ZnCl2+H2\)
a) Ta có
n\(_{Zn}=\frac{3,9}{65}=0,06\left(mol\right)\)
n\(_{HCl}=\frac{1,46}{36,5}=0,04\left(mol\right)\)
=> Zn dư
Theo pthh
n\(_{ZnCl2}=\frac{1}{2}n_{HCl}=0,02\left(mol\right)\)
m\(_{ZnCl2}=0,02.136=2,72\left(g\right)\)
b) Theo pthh
n\(_{H2}=\frac{1}{2}n_{HCl}=0,02\left(mol\right)\)
V\(_{H2}=0,02.22,4=0,448\left(l\right)\)
c) theo pthh
n\(_{Zn}=\frac{1}{2}n_{HCl}=0,02\left(mol\right)\)
n\(_{Zn}dư=0,04-0,02=0,02\left(mol\right)\)
m\(_{Zn}dư=0,02.65=1,3\left(g\right)\)
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