a, PT: \(M+2HCl\rightarrow MCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_M=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow M_M=\dfrac{11,2}{0,2}=56\left(g/mol\right)\)
→ M là Fe.
c, Theo PT: \(n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)