C2H4+Br2-to>C2H4Br2
x--------x mol
C2H2+2Br2-to>C2H2Br4
y----------2y mol
ta có :
\(\left\{{}\begin{matrix}x+y=0,3\\160x+320y=80\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=>%nC2H2=\(\dfrac{0,1}{0,3}.100=33,3\%\)
=>%nC2H4=100-33,3=66,7%