Đáp án:
a. `C_2H_4+Br_2\to C_2H_4Br_2`
b. `%m_{CH_2}=66,67%\qquad %m_{C_2H_4}=33,33%`
Giải thích các bước giải:
a. PTHH: `C_2H_4+Br_2\to C_2H_4Br_2`
b. `n_{C_2H_4Br_2}=\frac{4,7}{188}=0,025\text{(mol)}`
Theo PTHH: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,025\text{(mol)}\)
`\to m_{C_2H_4}=0,025\times 28=0,7\text{(g)}`
`n_{hh}=n_{CH_2}+n_{C_2H_4}=\frac{2,8}{22,4}=0,125\text{(mol)}`
`\to n_{CH_2}=0,125-0,025=0,1\text{(mol)}`
`\to m_{CH_2}=0,1\times 14=1,4\text{(g)}`
`\to %m_{CH_2}={1,4}/{1,4+0,7}\times 100%=66,67%`
`\to %m_{C_2H_4}=100%-66,67%=33,33%`