a) 2Al + 6HCl → 2AlCl3 + 3H2 (1)
Fe + 2HCl → FeCl2 + H2 (2)
b) \(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
Gọi x,y lần lượt là số mol của Al và Fe
Ta có: \(27x+56y=11\) (*)
Theo pT1: \(n_{H_2}=\frac{3}{2}n_{Al}=1,5x\left(mol\right)\)
Theo Pt2: \(n_{H_2}=n_{Fe}=y\left(mol\right)\)
ta có: \(1,5x+y=0,4\) (**)
Từ (*)(**) ta có: \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
Vậy \(n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)
\(n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1\times56=5,6\left(g\right)\)
\(\%m_{Al}=\frac{5,4}{11}\times100\%=49,09\%\)
\(\%m_{Fe}=\frac{5,6}{11}\times100\%=50,91\%\)
c) Theo Pt1: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)
Theo pt2: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,1\times127=12,7\left(g\right)\)
\(\Rightarrow m_{muối}=26,7+12,7=39,4\left(g\right)\)
\(n_{Al}=x;n_{Fe}=y\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow hpt:\left\{{}\begin{matrix}27x+56y=11\\22,4\left(\frac{3x}{2}+y\right)=8,96\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\rightarrow\%m_{Al}=\frac{27.0,2}{11}.100\%=49,09\left(\%\right)\\n_{Fe}=0,1\left(mol\right)\rightarrow\%m_{Fe}=100-49,09=50,91\left(\%\right)\end{matrix}\right.\\ m_M=m_{AlCl_3}+m_{FeCl_2}=0,2.133,5+0,1.127=39,4\left(g\right)\)