1/ PTHH:
$CH_2=CH-CH=CH_2+ 2Br_2 \to CH_2Br-CHBr-CHBr-CH_2Br$
2/
Coi T gồm C (a mol), H (b mol)
=> $12a+b=14,9$ (1)
$n_{CO_2}= n_C= a$
$n_{H_2O}= 0,5n_H= 0,5b$
$m_{tăng}= m_{CO_2}+ m_{H_2O}$
=> $44a+ 18.0,5b= 66,9$ (2)
(1)(2) => $a=1,05; b=2,3$
T gồm $C_3H_8, C_3H_4, C_3H_6$
Bảo toàn C: $n_T= \frac{1}{3}n_{CO_2}= 0,35 mol$
=> $\overline{M}_T= \frac{14,9}{0,35}= \frac{298}{7}$
=> $d_{T/H_2}= \frac{149}{7}$