Gọi $n_{Mg} = a (mol); n_{R} = b (mol)$

- Xét TN1:

$\rightarrow 24a + M_R.b = 8$

$n_{H_2} = \dfrac{4,48}{22,4} = 0,2 (mol)$

PTHH:

$Mg + 2HCl \rightarrow MgCl_2 + H_2$

   a---------------------------------->a

$2R + 2nHCl \rightarrow 2RCl_n + nH_2$

b---------------------------------->0,5bn

$\rightarrow a + 0,5bn = 0,2 (1)$

Xét TN2:

$n_{Mg} = \dfrac{16}{8} . a = 2a (mol)$

$n_{R} = \dfrac{16}{8} . b = 2b (mol)$

$n_{Cl_2} = \dfrac{11,2}{22,4} = 0,5 (mol)$

PTHH:

$Mg + Cl_2 \xrightarrow{t^o} MgCl_2$

2a----->2a

$2R + nCl_2 \xrightarrow 2RCl_n$

2b--->bn

$\rightarrow 2a + bn = 0,5 (2)$

Từ $(1)(2) \rightarrow a = b = 0,1 (mol), n = 3$

$\rightarrow 0,1.24 + 0,1.M_R = 8(g)$

$\Leftrightarrow M_R = 56 (g/mol)$

$\Rightarrow$ R là Fe

$@An$