PT: \(M+2HCl\rightarrow MCl_2+H_2\)
Ta có: nHCl = 0,5.0,4 = 0,2 (mol)
Theo PT: \(n_M=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)
\(\Rightarrow M_M=\dfrac{2,4}{0,1}=24\left(g/mol\right)\)
Vậy: M là Mg.
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)