C6H5OH + 3Br2 → 2,4,6-Br3C6H2OH + 3HBr
\(n_{Br_2}=\frac{6,4}{160}=0,04\left(mol\right)\)
Theo PT : n 2,4,6-Br3C6H2OH = \(\frac{1}{3}n_{Br_2}=\frac{1}{75}\left(mol\right)\)
=> m 2,4,6-Br3C6H2OH =\(\frac{1}{75}.331=4,413\left(g\right)\)
C6H5OH + 3Br2 → 2,4,6-Br3C6H2OH + 3HBr
0,04-------0,0133
nBr2=6,4\160=0,04(mol)
=> m 2,4,6-Br3C6H2OH =0,0133.331=4,4(g)