\(Fe+2HCl\rightarrow FeCl_2+H_2\\ 0,2mol\text{:}0,4mol\rightarrow0,2mol\text{:}0,2mol\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=0,5.1=0,5\left(mol\right)\)
Ta có \(0,2< \dfrac{0,5}{2}\) nên HCl dư.
\(n_{HCldu}=0,5-0,4=0,1\left(mol\right)\)
\(m_{HCldu}=0,1.36,5=3,65\left(g\right)\)
\(n_{Fe}=11,2:56=0,2(mol)\\n_{HCl}=0,5.1=0,5(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ Ta\ có\ \dfrac{n_{Fe}}{1}<\dfrac{n_{HCl}}{2}\\ \Rightarrow HCl\ dư\\ n_{HCl\ dư}=0,5-0,4=0,1(mol)\\ \Rightarrow m_{Hcl\ dư}=0,1.36,5=3,65(g)\)