\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(m_{HCl}=245.16,7\%=40,915\left(g\right)\Rightarrow n_{HCl}=\dfrac{40,915}{36,5}=1,121\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{1,121}{2}\), 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 = 1,121 - 0,2 = 0,921 (mol)
Ta có: m dd sau pư = 5,6 +245 - 0,1.2 = 250,4 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127}{250,4}.100\%\approx5,07\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,921.36,5}{250,4}.100\%\approx13,43\%\end{matrix}\right.\)
Đúng 1
Bình luận (0)
