a) Gọi số mol C2H4, C2H2 là a, b (mol)
=> \(a+b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) (1)
mtăng = mC2H4 + mC2H2
=> 28a + 26b = 5,4 (2)
(1)(2) => a = 0,1 (mol); b = 0,1 (mol)
PTHH: C2H4 + Br2 --> C2H4Br2
0,1--->0,1
C2H2 + 2Br2 --> C2H2Br4
0,1--->0,2
=> \(m_{Br_2}=\left(0,1+0,2\right).160=48\left(g\right)\)
b)
\(\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,1}{0,2}.100\%=50\%\)