C_6H_5OH+3 Br_2-> C_6H_2Br_3OH+ 3HBr
n_{C_6H_5OH}={14,1}/{94}=0,15(mol) 
n_{C_6H_2Br_3OH}= n_{C_6H_5OH}= 0,15(mol)
m_{C_6H_2Br_3OH}= 331.0,15=49,65(g)
H=90%
-> m_{C_6H_2Br_3OH}= 49,65. 90%= 44,685(g)

n_{\text{phenol}}=(14,1)/(94)=0,15(mol)
PTHH:
C_6H_5OH+3Br_2\to C_6H_2Br_3OH\downarrow+3HBr
=>n_{C_6H_2Br_3OH}=0,15.90\%=0,135(mol)
=>m_{\text{kết tủa trắng}}=m_{C_6H_2Br_3OH}=0,135.331=44,685(g)