\(n_{Br_2}=\dfrac{1,92}{160}=0,012\left(mol\right)\)
Gọi số mol C2H4, C2H2 là a, b (mol)
=> \(a+b=\dfrac{0,224}{22,4}=0,01\) (1)
PTHH: C2H4 + Br2 --> C2H4Br2
a---->a
C2H2 + 2Br2 --> C2H2Br4
b----->2b
=> a + 2b = 0,012 (2)
(1)(2) => a = 0,008 (mol); b = 0,002 (mol)
=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,008}{0,01}.100\%=80\%\\\%V_{C_2H_2}=100\%-80\%=20\%\end{matrix}\right.\)