PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=0,3\cdot1=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) \(\Rightarrow\) HCl còn dư
\(\Rightarrow n_{FeCl_2}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\) \(\Rightarrow C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,3}\approx0,33\left(M\right)=C_{M_{FeCl_2}}\)
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