a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(m_{HCl}=43,8.10\%=4,38\left(g\right)\Rightarrow n_{HCl}=\dfrac{4,38}{36,5}=0,12\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,12}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,12-0,1=0,02\left(mol\right)\)
Ta có: m dd sau pư = 2,8 + 43,8 - 0,05.2 = 46,5 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,05.127}{46,5}.100\%\approx13,66\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,02.36,5}{46,5}.100\%\approx1,57\%\end{matrix}\right.\)
