2Al + 6HCl → 2AlCl3 + 3H2↑ (1)
Fe + 2HCl → FeCl2 + H2↑ (2)
\(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
a) Gọi x,y lần lượt là số mol của Al, 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 hệ: \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0.2\\y=0,1\end{matrix}\right.\)
Vậy \(n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)
\(n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1\times56=5,6\left(g\right)\)
b) Theo PT1,2: \(\Sigma n_{HCl}=2\Sigma n_{H_2}=2\times0,4=0,8\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=0,8\times36,5=29,2\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{29,2}{14,6\%}=200\left(g\right)\)
