1, 2 (hình)
3.
Y: $C_2H_2$, $H_2$, $C_2H_4$, $C_2H_6$
Z: $C_2H_6$, $H_2$
$m_{\text{hh bđ}}=0,06.26+0,04.2=1,64g$
Bảo toàn khối lượng: $m_Y=m_{\text{hh bđ}}=1,64g$
$m_{\text{tăng}}=m_{C_2H_2}+m_{C_2H_4}=1,32g$
$\Rightarrow m_Z=1,64-1,32=0,32g$
Gọi $x$, $y$ là số mol $C_2H_6$, $H_2$
$\Rightarrow 30x+2y=0,32$
$\overline{M}_Z=0,5.32=16$
$\Rightarrow x+y=\dfrac{0,32}{16}=0,02$
Giải hệ: $x=y=0,01$
$\to V=(0,01+0,01).22,4=0,448l$
$\%V_{C_2H_6}=\%V_{H_2}=50\%$