`n_{hh\ khí}=\frac{6,72}{22,4}=0,3(mol)`
`n_{Br_2}=\frac{64}{160}=0,4(mol)`
`a)` `C_2H_2+2Br_2\to C_2H_2Br_4`
`=> n_{C_2H_2}=0,5n_{Br_2}=0,2(mol)`
`=>%V_{C_2H_2}=\frac{0,2.100%}{0,3}\approx 66,7%`
`%V_{CH_4}=\frac{0,1,100%}{0,3}\approx33,3%`
`b)` `n_{CH_4}=0,1(mol)`
`CH_4+Cl_2\to CH_3Cl+HCl`
`n_{Cl_2}=n_{CH_4}`
`=> V_{Cl_2}=0,1.22,4=2,24(l)`