Đáp án + Giải thích các bước giải:

$a\big)$

$C_2H_4+Br_2\to C_2H_4Br_2$

Theo PT: $n_{C_2H_4}=n_{Br_2}=\frac{32}{160}=0,2(mol)$

$\to \%V_{C_2H_4}=\frac{0,2.22,4}{6,72}.100\%=66,7\%$

$\to \%V_{CH_4}=100-66,7=33,3\%$

$b\big)$

$C_2H_4+H_2O\xrightarrow{t^o,axit}C_2H_5OH$

    $0,2$            $←$            $0,2(mol)$

Mà $H=80\%$

$\to m_{C_2H_5OH}=0,2.80\%.46=7,36(g)$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{B{r_2}}} = \dfrac{{32}}{{160}} = 0,2\,mol\\
{n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,2\,mol\\
\% {V_{{C_2}{H_4}}} = \dfrac{{0,2 \times 22,4}}{{6,72}} \times 100\%  = 66,67\% \\
\% {V_{C{H_4}}} = 100 - 66,67 = 33,33\% \\
b)\\
{C_2}{H_4} + {H_2}O \to {C_2}{H_5}OH\\
{n_{{C_2}{H_5}OH}} = {n_{{C_2}{H_4}}} = 0,2\,mol\\
H = 80\%  \Rightarrow {m_{{C_2}{H_5}OH}} = 0,2 \times 46 \times 80\%  = 7,36g
\end{array}\)