`n_{\text{hh khí}}=\frac{3,36}{22,4}=0,15(mol)`
`n_{CH_4}=\frac{2,24}{22,4}=0,1(mol)`
`=> n_{C_2H_2}=0,15-0,1=0,05(mol)`
`a)` `C_2H_2+2Br_2\to C_2H_2Br_4`
`b)` `%V_{CH_4}=\frac{0,1.100%}{0,15}\approx 66,7%`
`%V_{C_2H_2}=\frac{0,05.100%}{0,15}\approx 33,3%`
`c)` `CH_4+2O_2\overset{t^o}{\to}CO_2+2H_2O`
`C_2H_2+\frac{5}{2}O_2\overset{t^o}{\to}2CO_2+H_2O`
`n_{O_2}=2n_{CH_4}+2,5n_{O_2}`
`=> n_{O_2}=2.0,1+2,5.0,05=0,325(mol)`
Do `V_{(kk)}=5.V_{O_2}`
`=>V_{(kk)}=5.0,325.22,4=36,4(l)`