PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,1\left(mol\right)\Rightarrow V_{C_2H_4}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{CH_4}=6,72-2,24=4,48\left(l\right)\)