\(Phần\ 2 : \\ C_6H_5OH + 3Br_2 \to C_6H_2OHBr_3 + 3HBr\\ n_{C_6H_5OH} = n_{C_6H_2OHBr_3} = \dfrac{49,65}{331} = 0,15(mol)\\ Phần\ 1 : n_{H_2} = \dfrac{3,92}{22,4}=0,175(mol)\\ 2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2\\ 2C_6H_5OH + 2Na \to 2C_6H_5ONa + H_2\\ 2n_{H_2} = n_{C_2H_5OH} + n_{C_6H_5OH}\\ \Rightarrow n_{C_2H_5OH} = 0,175.2 - 0,15 = 0,2(mol)\\ \Rightarrow m = 2(0,2.46 + 0,15.94) = 46,6(gam)\)