`n_{hh khí}=\frac{4,48}{22,4}=0,2\ (mol)`
`n_{Br_2}=\frac{24}{160}=0,15\ (mol)`
Phương trình hóa học:
`C_2H_4 + Br_2 \to C_2H_4Br_2`
`n_{C_2H_4}=n_{Br_2}=0,15\ (mol)`
`⇒n_{CH_4}=n_{hh khí}-n_{C_2H_4}=0,2-0,15=0,05\ (mol)`
`m_{hh khí}=m_{CH_4}+m_{C_2H_4}=0,05×16+0,15×28=5\ (g)`
`⇒ \%m_{CH_4}=\frac{0,05×16}{5}.100\%=16\%`
`\%m_{C_2H_4}=\frac{0,15×28}{5}.100\%=84\%`
`\%V_{CH_4}=\frac{0,05×22,4}{4,48}.100\%=25\%`
`\%V_{C_2H_4}=\frac{0,15×22,4}{4,48}.100\%=75\%`