\(n_{Cl_2} = a(mol) ; n_{O_2} = b(mol)\\ n_X = a + b = \dfrac{1,12}{22,4} = 0,05(mol)\\ m_X = 71a + 32b = 0,05.2.23,8 = 2,38(gam)\\ \Rightarrow a = 0,02 ; b = 0,03\)
Bảo toàn electron :
\(3n_{Fe} = 2n_{Cl_2} + 4n_{O_2}\\ \Rightarrow n_{Fe} = \dfrac{0,02.2 + 0,03.4}{3} = \dfrac{4}{75}(mol)\\ \Rightarrow m = \dfrac{4}{75}.160 = 8,53(gam)\)