CTTQ 2 ankin là $C_nH_{2n-2}$
$n_{Br_2}=0,15 mol$
Bảo toàn $\pi$, $n_{\text{ankin}}=\dfrac{0,15}{2}=0,075 mol$
$\Rightarrow \overline{M}_{\text{ankin}}=\dfrac{3,7}{0,075}=49,37=14n-2$
$\Leftrightarrow n=3,6$
Vậy 2 ankin là $C_3H_4 (a mol), C_4H_6$ (b mol)
$\Rightarrow 40a+54b=3,7$ (1)
$n_{Br_2}=0,15 mol \Rightarrow 2a+2b=0,15$ (2)
(1)(2)$\Rightarrow a=0,025; b=0,05$
$m_{C_3H_4}=0,025.40=1g$
$\Rightarrow m_{C_4H_6}=3,7-1=2,7g$