Đáp án:
a. `C_2H_4+Br_2\to C_2H_4Br_2`
b. `V_{C_2H_4}=3,36\text{(l)}\qquad V_{CH_4}=1,12\text{(l)}`
c. `%V_{C_2H_4}=75%\qquad %V_{CH_4}=25%`
Giải thích các bước giải:
a. `C_2H_4+Br_2\to C_2H_4Br_2`
b.
`n_{Br_2}=\frac{24}{160}=0,15\text{(mol)}`
Theo PTHH: `n_{C_2H_4}=n_{Br_2}=0,15\text{(mol)}`
`\to V_{C_2H_4}=0,15\times 22,4=3,36\text{(l)}`
`\to V_{CH_4}=4,48-3,36=1,12\text{(l)}`
c. `%V_{C_2H_4}={3,36}/{4,48}\times 100%=75%`
`\to %V_{CH_4}=100%-75%=25%`