Em tham khảo nha:

\(\begin{array}{l}
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{{C_2}{H_4}B{r_2}}} = \dfrac{{18,8}}{{188}} = 0,1\,mol\\
{n_{B{r_2}}} = {n_{{C_2}{H_4}B{r_2}}} = 0,1\,mol\\
{m_{B{r_2}}} = 0,1 \times 160 = 16g\\
b)\\
{n_{{C_2}{H_4}}} = {n_{{C_2}{H_4}B{r_2}}} = 0,1\,mol\\
{V_{{C_2}{H_4}}} = 0,1 \times 22,4 = 2,24l\\
{V_{C{H_4}}} = 4 - 2,24 = 1,76l\\
c)\\
\% {V_{{C_2}{H_4}}} = \dfrac{{2,24}}{4} \times 100\%  = 56\% \\
\% {V_{C{H_4}}} = 100 - 56 = 44\% 
\end{array}\)

Đáp án + Giải thích các bước giải:

$a\big)C_2H_4+Br_2\to C_2H_4Br_2$

Theo PT: $n_{Br_2}=n_{C_2H_4Br_2}=\frac{18,8}{188}=0,1(mol)$

$\to m_{Br_2}=0,1.160=16(g)$

$b\big)$ Theo PT: $n_{C_2H_4}=n_{Br_2}=0,1(mol)$

$\to V_{C_2H_4}=0,1.22,4=2,24(l)$

$\to V_{CH_4}=4-2,24=1,76(l)$

$c\big)\%V_{C_2H_4}=\frac{2,24}{4}.100\%=56\%$

$\to \%V_{CH_4}=100-56=44\%$