Ta có: m bình tăng = mC2H4 = 2,8 (g)
\(\Rightarrow n_{C_2H_4}=\dfrac{2,8}{28}=0,1\left(mol\right)\)
\(\Rightarrow n_{C_2H_6}=\dfrac{12,395}{24,79}-0,1=0,4\left(mol\right)\)
⇒ mC2H6 = 0,4.30 = 12 (g)
\(\left\{{}\begin{matrix}\%V_{C_2H_6}=\dfrac{0,4.24,79}{12,395}.100\%=80\%\\\%V_{C_2H_4}=20\%\text{ }\end{matrix}\right.\)