Fe + 2HCl-----> FeCl2 +H2
a) Ta có
n\(_{Fe}=\frac{1,2}{56}=0,02\left(mol\right)\)
n\(_{HCl}=0,2.1,5=0,3\left(mol\right)\)
=> HCl dư
Theo pthh
n\(_{H2}=n_{Fe}=0,02\left(mol\right)\)
V\(_{H2}=0,02.22,4=0,448\left(l\right)\)
b) Theo pthh
n\(_{HCl}=2n_{Fe}=0,04\left(mol\right)\)
n\(_{HCl}dư=0,3-0,04=0,26\left(mol\right)\)
C\(_{M\left(HCl\right)}=\frac{0,26}{0,2}=1,3\left(M\right)\)
Theo pthh
n\(_{FeCl2}=n_{Fe}=0,02\left(mol\right)\)
C\(_{M\left(FeCl2\right)}=\frac{0,02}{0,2}=0,1\left(M\right)\)
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