\(2M+2nHCl\rightarrow2MCl_n+nH_2\)
Ta có
mdd = mM + mHCl - mH2
\(\Rightarrow n_{H2}=0,25\left(mol\right)\)
\(\Rightarrow n_{H2}=\frac{0,25}{2}=0,125\left(mol\right)\)
\(\Rightarrow n_M=\frac{0,125.2}{n}=\frac{0,25}{n}\left(mol\right)\)
\(M_M=7:\frac{0,25}{n}=28n\)
\(n=2\Rightarrow M_M=56\)
Vậy M là Fe( sắt)