Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{500.14,6\%}{36,5}=2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{2}{2}\), ta được HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)
⇒ nHCl (dư) = 2 - 0,2 = 1,8 (mol)
a, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
c, Ta có: m dd sau pư = 5,6 + 500 - 0,1.2 = 505,8 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{12,7}{505,8}.100\%\approx2,51\%\\C\%_{HCl\left(dư\right)}=\dfrac{1,8.36,5}{505,8}.100\%\approx12,99\%\end{matrix}\right.\)
Bạn tham khảo nhé!