Gọi số mol C2H4, C2H2 là a, b (mol)
=> a + b = \(\dfrac{13,44}{22,4}=0,6\) (1)
nBr2 = 0,8.1 = 0,8 (mol)
PTHH: C2H4 + Br2 --> C2H4Br2
a----->a
C2H2 + 2Br2 --> C2H2Br4
b----->2b
=> a + 2b = 0,8 (2)
(1)(2) => a = 0,4 (mol); b = 0,2 (mol)
=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,4}{0,6}.100\%=66,67\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,6}.100\%=33,33\%\end{matrix}\right.\)