a, \(C_2H_4+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{Br2}=\frac{64}{160}=0,4\left(mol\right)\)
\(\rightarrow n_{C2H2}=0,2\left(mol\right)\)
\(V_{C2H2}=0,2.22,4=4,48\left(l\right)\)
\(\rightarrow\%_{C2H2}=\frac{4,48}{6,72}.100\%=66,67\%\)
\(\%_{CH4}=100\%-66,67\%=33,33\%\)
b, \(CH_4+Cl_2\rightarrow CH_3Cl+HCl\)
\(n_{CH4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(\rightarrow n_{Cl2}=n_{CH4}=0,1\left(mol\right)\)
\(\rightarrow V_{CL2}=0,1.22,4=2,24\left(l\right)\)