\(n_{Al} = a ; n_{Fe} = b\Rightarrow 27a + 56b = 27,6(1)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,5a(mol)\\ n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = \dfrac{b}{3}(mol)\\ \Rightarrow 0,5a.102 + \dfrac{b}{3}232 = 43,6(2)\\ (1)(2) \Rightarrow a = 0,4 ; b = 0,3\\ m_{Al} = 0,4.27 = 10,8(gam) ; m_{Fe} = 0,3.56 = 16,8(gam)\)