\(n_S = \dfrac{3,2}{32} = 0,1(mol)\ ; n_{H_2S} = \dfrac{0,056}{22,4} = 0,0025(mol) S^{+6} + 6e \to S\\ S^{+6} + 8e \to S^{-2}\\ \text{Bảo toàn electron :}\\ n_{Al} = \dfrac{0,1.6 + 0,0025.8}{3} = \dfrac{31}{150}(mol)\\ m = \dfrac{31}{150}.27 = 5,58(gam)\)