\(n_{SO_2} = \dfrac{7,28}{22,4} = 0,325(mol)\\ n_{Al} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 27a + 65b = 10,55(1)\\ \text{Bảo toàn electron :}\\ 3a + 2b = 0,325.2(2)\\ (1)(2) \Rightarrow a = 0,15 ; b = 0,1\\ \%m_{Al} = \dfrac{0,15.27}{10,55}.100\% = 38,39\%\)