B gồm $C_2H_4, C_2H_2, C_2H_6, H_2$
$m_{\text{brom tăng}}= m_{C_2H_4}+ m_{C_2H_2}= 1,64g$
$m_X= 0,06.30+0,02.2= 1,84g$
Bảo toàn khối lượng: $m_A= m_B= 1,64+1,84= 3,48g$
Ta có: $26x+0,18.2=3,48$
$\Leftrightarrow x= 0,12$
$\%V_{C_2H_2}= \frac{0,12.100}{0,12+0,18}= 40\%$