\(n_{HCl}=\dfrac{P\cdot V}{T\cdot R}=\dfrac{1\cdot2,479}{\left(25+273\right)\cdot0,082}=0,1mol\)
\(R+2HCl\rightarrow RCl_2+H_2\)
\(\dfrac{5,6}{\overline{M_R}}\) 0,1
Theo pt: \(\Rightarrow\dfrac{5,6}{\overline{M_R}}=0,1\Rightarrow M_R=56\left(Fe\right)\)
Kim loại M là sắt (ll) (Fe)