Theo đề, gọi \(a,b,c\) lần lượt là số mol của \(Al,Zn,Fe\) trong hỗn hợp.
\(\Rightarrow m_{hh}=27a+65b+56c=14,8\left(1\right)\)
Thí nghiệm 1: PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(a---------->\frac{3}{2}a\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(b--------->b\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(c--------->c\)
Theo ba phương trình trên: \(n_{H_2}=\frac{3}{2}a+b+c=\frac{0,7}{2}=0,35\left(mol\right)\left(2\right)\)
Thí nghiệm 2: PTHH:
\(2Al+3Cl_2\xrightarrow[]{t^0}2AlCl_3\)
\(a------>a\)
\(Zn+Cl_2\xrightarrow[]{t^0}ZnCl_2\)
\(b------>b\)
\(2Fe+3Cl_2\xrightarrow[]{t^0}2FeCl_3\)
\(c------>c\)
\(m\)muối\(=133,5a+136b+162,5c=43,2\left(g\right)\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta có hệ: \(\left\{{}\begin{matrix}27a+65b+56c=14,8\\\frac{3}{2}a+b+c=0,35\\133,5a+136b+162,5c=43,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=56c=56.0,1=5,6\left(g\right)\Rightarrow\%m_{Fe}=\frac{5,6}{14,8}.100\%=37,84\%\)