Gọi $x$, $y$ là số mol $C_2H_2$, $C_2H_4$
$n_{hh}=\dfrac{5,6}{22,4}=0,25(mol)$
$\Rightarrow x+y=0,25$ $(1)$
$m_{\text{tăng}}=6,8g=m_{hh}$
$\Rightarrow 26x+28y=6,8$ $(2)$
$(1)(2)\Rightarrow x=0,1; y=0,15$
$V_{C_2H_2}=0,1.22,4=2,24l$
$V_{C_2H_4}=0,15.22,4=3,36l$