\(a) C_6H_5OH + 3Br_2 \to C_6H_2Br_3OH + 3HBr\\ n_{C_6H_2Br_3OH} = \dfrac{49,65}{331} = 0,15(mol)\\ n_{Br_2} = 3n_{C_6H_2Br_3OH} = 0,45(mol)\\ m_{Br_2} = 0,45.160 = 72(gam)\\ b) n_{C_6H_5OH} = n_{C_6H_2Br_3OH} = 0,15(mol)\\ m_{C_6H_5OH} = 0,15.94 = 14,1(gam)\)