Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{FeCl_2\left(LT\right)}=n_{H_2\left(LT\right)}=n_{Fe}=0,2\left(mol\right)\)
Mà: H% = 75%
\(\Rightarrow n_{FeCl_2\left(TT\right)}=n_{H_2\left(TT\right)}=0,2.75\%=0,15\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)