$ n_{Fe} = \dfrac{m}{M} = \dfrac{28}{56} = 0,5 (mol) $
$ Fe + 2HCl \rightarrow FeCl_2 + H_2 ↑ $
$ n_{HCl} = 2.n_{Fe} = 2.0,5 = 1 (mol) $
$ m_{ctHCl} = n . M = 1 . 36,5 = 36,5 (g) $
$ m_{ddHCl} = \dfrac{36,5}{18,25}.100 = 200 (g) $
$ n_{H_2} = n_{Fe} = 0,5 (mol) $
$ m_{H_2} = n . M = 0,5 . 2 = 1 (g) $
$ m_{dd spứ} = m_{Fe} + m_{ddHCl} - m_{H_2} = 28 + 200 - 1 = 227 (g) $
$ m_{ctFeCl_2} = n . M = 0,5 . 127 = 63,5 (g) $
$ \rightarrow C\%_{FeCl_2} = \dfrac{m_{ct}}{m_{dd}}.100\% = \dfrac{63,5}{227}.100 = 27,97\% $
