\(\begin{array}{l} 1)\\ 2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\ Fe + 2HCl \to FeC{l_2} + {H_2}\\ n{H_2} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\ hh:Al(a\,mol),Fe(b\,mol)\\ 27a + 56b = 11,1\\ 1,5a + b = 0,3\\ \Rightarrow a = 0,1;b = 0,15\\ \% mAl = \dfrac{{0,1 \times 27}}{{11,1}} \times 100\% = 24,32\% \\ \% mFe = 100 - 24,32 = 75,68\% \\ 2)\\ 2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\ C{l_2} + 2Na \to 2NaCl\\ 2NaCl _{tinh\, \, thể}+ {H_2}S{O_4} _{đặc} \to N{a_2}S{O_4} + 2HCl\\ HCl + AgN{O_3} \to AgCl + HN{O_3}\\ C{l_2} + Ca{(OH)_2} \to CaOC{l_2} + {H_2}O\\ CaOC{l_2} + 2HCl \to CaC{l_2} + C{l_2} + {H_2}O\\ C{l_2} + 2NaBr \to 2NaCl + B{r_2}\\ 3B{r_2} + 2Al \to 2AlB{r_3} \end{array}\)
