$\rm n_{H_2}=\dfrac{79,16}{24,79}=3(mol)$

$\rm PTHH: M+2HCl \to MCl_2+H_2$

$\rm m_{MCl_2}=\dfrac{125.30,4}{100}=38(g)$

$\rm n_{MCl_2}=\dfrac{38}{3}=12,67(amu) \to sai$

`=>` sửa: $\rm V_{H_2}=7,916(l)$
$\rm n_{H_2}=0,3(mol)$

$\rm M_{MCl_2}=\dfrac{38}{0,3}=127(amu)$

$\rm \to M+35,5.2=127 \to M=56(amu)$

`=>` $\rm M:Fe(iron)$