a, PT: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
b, Ta có: \(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02\left(mol\right)\)
\(m_{HCl}=400.10\%=40\left(g\right)\Rightarrow n_{HCl}=\dfrac{40}{36,5}=\dfrac{80}{73}\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{\dfrac{80}{73}}{6}\), ta được HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_3}=2n_{Fe_2O_3}=0,04\left(mol\right)\\n_{HCl\left(pư\right)}=6n_{Fe_2O_3}=0,12\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=\dfrac{80}{73}-0,12=\dfrac{1781}{1825}\left(mol\right)\)
Ta có: m dd sau pư = 3,2 + 400 = 403,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_3}=\dfrac{0,04.162,5}{403,2}.100\%\approx1,61\%\\C\%_{HCl\left(dư\right)}=\dfrac{\dfrac{1781}{1825}.36,5}{403,2}.100\%\approx8,83\%\end{matrix}\right.\)
Bạn tham khảo nhé!
