`n_{\text{hh khí}}=\frac{5,6}{22,4}=0,25(mol)`
`n_{Br_2}=\frac{4}{160}=0,025(mol)`
Phương trình:
`CH_2=CH_2+Br_2\to CH_2Br-CH_2Br`
Ta nhận thấy: `n_{Br_2}=n_{C_2H_4}=0,025(mol)`
`=> n_{CH_4}=0,25-0,025=0,225(mol)`
`C_2H_4+3O_2\overset{t^o}{\to}2CO_2+2H_2O`
`0,025` __ `0,075` (mol).
`CH_4+2O_2\overset{t^o}{\to}CO_2+2H_2O`
`0,225` __ `0,45` (mol).
Ta có: `n_{O_2}=0,075+0,45=0,525(mol)`
`=> V_{O_2}=0,525.22,4=11,76(l)`