\(\begin{array}{l} Fe + 2HCl \to FeC{l_2} + {H_2}\\ 2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\ a)\\ hh:Fe(a\,mol),Al(b\,mol)\\ n{H_2} = \dfrac{{0,4}}{2} = 0,2\,mol\\ 56a + 27b = 5,5\\ a + 1,5b = 0,2\\ = > a = 0,05;\,b = 0,1\\ \% mFe = \dfrac{{0,05 \times 56}}{{5,5}} \times 100\% = 50,9\% \\ \% mAl = 100 - 50,9 = 49,1\% \\ b)\\ nHCl = 0,2 \times 2,1 = 0,42\,mol\\ nHCl\,pu = 2n{H_2} = 0,4mol\\ CMHCl = \dfrac{{0,02}}{{0,2}} = 0,1M\\ CMFeC{l_2} = \dfrac{{0,05}}{{0,2}} = 0,25M\\ CMAlC{l_3} = \dfrac{{0,1}}{{0,2}} = 0,5M\\ c)\\ NaOH + HCl \to NaCl + {H_2}O\\ nNaOH = nHCl = 0,02\,mol\\ VNaOH = \dfrac{{0,02}}{2} = 0,01l = 10ml \end{array}\)