a)
\(2M+2nHCl\rightarrow2MCl_n+nH_2\)
\(n_{HCl}=0,15.2=0,3\left(mol\right)\)
\(\rightarrow n_{MCln}=\frac{0,3.2}{2n}=\frac{0,3}{n}\left(mol\right)\)
\(M_{MCln}=14,25:\frac{0,3}{n}=47,5n\)
Ta có
\(M_M+35,5=47,5n\rightarrow M_M=12n\)
n=2 thì MM=24
\(\rightarrow\) M là Magie(Mg)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(V_{H2}=0,15.22,4=3,36\left(l\right)\)
\(CM_{MgCl2}=\frac{0,15}{0,15}=1M\)