\(n_{H_2}=\dfrac{3,7185}{24,79}=0,15(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,4}=0,75M\)
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