\(n_{Cu} = a ; n_{Al} = b ; n_{Fe} = c(mol)\\ \Rightarrow 64a + 27b + 56c = 28,6(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\\ \text{Mặt khác} : n_{O_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3\\ \)
Ta có :
\(\dfrac{n_X}{n_{O_2}}=\dfrac{a+b+c}{0,5a +0,75b + 0,75c} = \dfrac{0,6}{0,4}(3)\\ (1)(2)(3)\Rightarrow a = \dfrac{317}{1460} ; b = \dfrac{121}{365}; c = \dfrac{15}{146}\\ \%m_{Cu} = \dfrac{\dfrac{317}{1460}.64}{28,6}.100\% = 48,59\%\\ \%m_{Al} = \dfrac{\dfrac{121}{365}.27}{28,6}.100\% = 31,3\%\\ \%m_{Fe} = 100\% - 41,59\% - 31,3\% = 27,11\%\)
