a/
\(2Al\left(x\right)+6HCl\left(3x\right)\rightarrow2AlCl_3\left(x\right)+3H_2\left(1,5x\right)\)
\(Fe\left(y\right)+2HCl\left(2y\right)\rightarrow FeCl_2\left(y\right)+H_2\left(y\right)\)
Gọi số mol của Al và Fe lần lược là x, y thì ta có:
\(27x+56y=11\left(1\right)\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow1,5x+y=0,4\left(2\right)\)
Từ (1) và (2) ta có hệ: \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%Al=\dfrac{5,4}{11}.100\%=49,09\%\\\%Fe=100\%-49,09\%=50,91\%\end{matrix}\right.\)
b/ \(n_{HCl}=3x+2y=3.0,2+2.0,1=0,8\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,8}{0,5}=1,6\left(l\right)=1600\left(ml\right)\)
c/ \(\Rightarrow m_{ddHCl}=1,2.1600=1920\left(g\right)\)
\(m_{H_2}=0,4.2=0,8\left(g\right)\)
\(\Rightarrow m_{ddspu}=11+1920-0,8=1930,2\left(g\right)\)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\m_{FeCl_2}=0,1.127=12,7\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{26,7}{1930,2}.100\%=1,38\%\\C\%\left(FeCl_2\right)=\dfrac{12,7}{1930,2}.100\%=0,66\%\end{matrix}\right.\)