PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeO}=12,8-m_{Fe}=12,8-0,1.56=7,2\left(g\right)\)
\(\Rightarrow n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Fe}+2n_{FeO}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)
\(\Rightarrow V_{ddHCl\left(2M\right)}=\dfrac{0,4}{2}=0,2\left(l\right)\)