PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, Giả sử: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
⇒ 27x + 56y = 5,5 (1)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}x+y\left(mol\right)\Rightarrow\dfrac{3}{2}x+y=0,2\left(2\right)\)
Từ (1) và (2) ⇒ x = 0,1 (mol), y = 0,05 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,5}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)
b, Theo PT: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,05}=8M\)
c, Theo p/a, ta có: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\end{matrix}\right.\)
PT: \(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_{3\downarrow}+3NaCl\)
_____0,1_______________0,1 (mol)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_{2\downarrow}+2NaCl\)
0,05________________0,05 (mol)
⇒ m kết tủa = mAl(OH)3 + mFe(OH)2 = 0,1.78 + 0,05.90 = 12,3 (g)
Bạn tham khảo nhé!
