Fe + 2HCl → FeCl2 + H2
\(n_{H_2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
a) Theo Pt: \(n_{Fe}=n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,5\times56=28\left(g\right)\)
b) Theo pT: \(n_{HCl}=2n_{H_2}=2\times0,5=1\left(mol\right)\)
\(\Rightarrow C_{M_{ddHCl}}=\frac{1}{0,2}=5\left(M\right)\)
c) Theo PT: \(n_{FeCl_2}=n_{Fe}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\frac{0,5}{0,2}=2,5\left(M\right)\)