Fe+2HCl->FeCl2+H2
0,025-0,05---0,025
nHCl=0,05 mol
=>VH2=0,025.22,4=0,56l
=>mFe=0,025.56=1,4g
=>mFeCl2=0,025.127=3,175g
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a) Ta có: \(n_{HCl}=0,25\cdot0,2=0,05\left(mol\right)\)
\(\Rightarrow n_{H_2}=0,01mol\) \(\Rightarrow V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\)
b) Theo PTHH: \(n_{Fe}=n_{H_2}=0,01mol\)
\(\Rightarrow m_{Fe}=0,01\cdot56=0,56\left(g\right)\)
c) Coi như thể tích dd không đổi
Theo PTHH: \(n_{Fe}=n_{FeCl_2}=0,01mol\)
\(\Rightarrow C_M=\frac{0,01}{0,2}=0,05\left(M\right)\)