Ta có: \(n_{H_2}=\dfrac{0,61975}{22,4}=\dfrac{2479}{89600}\left(mol\right)\)
\(PTHH:M+2HCl--->MCl_2+H_2\uparrow\)
Theo PT: \(n_M=n_{H_2}=\dfrac{2479}{89600}\left(mol\right)\)
\(\Rightarrow M_M=\dfrac{1,4}{\dfrac{2479}{89600}}\approx52\left(g\right)\)
Vậy M là nguyên tố crom (Cr)
\(n_M=\dfrac{1,4}{M_M}\left(mol\right);n_{H_2}=\dfrac{0,61975}{22,4}\approx0,03\left(mol\right)\\ PTHH:M+2HCl\rightarrow MCl_2+H_2\\ \Rightarrow n_M=n_{H_2}\approx0,03\\ \Rightarrow\dfrac{1,4}{M_M}\approx0,03\\ \Rightarrow M_M\approx47\left(g/mol\right)\)
Vậy đề sai