$2M + 3Cl_2 \to 2MCl_3$
Ta có: $n_{M}=\dfrac{5,6}{M_M}\,(\text{mol})\\n_{MCl_3}=\dfrac{16,25}{M_M+106,5}\,(\text{mol})\\n_M=n_{MCl_3}\to \dfrac{5,6}{M_M}=\dfrac{16,25}{M_M+106,5}\\\to 5,6M_M + 596,4=16,25M_M\\\to 10,65M_M=596,4\\\to M_M=56\,(\text{g/mol})$$\to$ M là $Fe \to A$