a) Fe + 2HCl → FeCl2 + H2
b) \(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
\(m_{HCl}=400\times7,3\%=29,2\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{29,2}{36,5}=0,8\left(mol\right)\)
Theo pT: \(n_{Fe}=\frac{1}{2}n_{HCl}\)
Theo bài: \(n_{Fe}=\frac{1}{4}n_{HCl}\)
Vì \(\frac{1}{4}< \frac{1}{2}\) ⇒ HCl dư
Dung dịch A gồm: HCl dư và FeCl2
Theo pT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,2\times2=0,4\left(g\right)\)
Ta có: \(m_{dd}saupư=11,2+400-0,4=410,8\left(g\right)\)
Theo pT: \(n_{HCl}pư=2n_{Fe}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=0,8-0,4=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=0,4\times36,5=14,6\left(g\right)\)
\(\Rightarrow C\%_{HCl}dư=\frac{14,6}{410,8}\times100\%=3,55\%\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,2\times127=25,4\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\frac{25,4}{410,8}\times100\%=6,18\%\)
