Пусть в смеси х моль C6H5NH2 и y моль C6H4(CH3)OH;
1)
C6H5NH2 + HBr = [C6H5NH3]+Br- и
2) C6H4(CH3)OH + NaOH = C6H4(CH3)ONa + H2O;
n HBr = n C6H5NH2, а n NaOH = n C6H4(CH3)OH; т.к. по условию m HBr = m NaOH, то 81*х = 40*y, y = 2.025*x;
3) C6H5NH2 + 3Br2 = C6H2(Br3)NH2 + 3HBr и
4) C6H4(CH3)OH + 2Br2 = C6H2(Br2)(CH3)OH + 2HBr;
из х моль C6H5NH2 пол-я 330*х г C6H2(Br3)NH2, из 2.025*x моль C6H4(CH3)OH - 266*2.025*x = 538.65*х г C6H2(Br2)(CH3)OH;
по условию, 330*х+538.65*х = 52.12, откуда х = 0.06 моль;
m смеси C6H5NH2 и C6H4(CH3)OH = 93*0.06+108*0.06*2.025 = 18.702 г, а м.д. соотв-о равны (93*0.06)/18.702 = 0.2984 (29.84%) и (108*0.06*2.025)/18.702 = 0.7016 (70.16%).