\(\begin{array}{l}
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\

b)\\
{n_{hh}} = \dfrac{{0，56}}{{22，4}} = 0，025mol\\
{n_{B{r_2}}} = \dfrac{{5，6}}{{160}} = 0，035mol\\
hh:{C_2}{H_4}(a,mol)，{C_2}{H_2}(b,mol)\\
a + b = 0，025(1)\\
a + 2b = 0，035(2)\\
 \text{Từ (1) ѵà (2)}\Rightarrow a = 0，015mol;b = 0，01mol\\
\% {V_{{C_2}{H_4}}} = \dfrac{{0，015 \times 22，4}}{{0，56}} \times 100\%  = 60\% \\
\% {V_{{C_2}{H_2}}} = 100 – 60 = 40\% 
\end{array}\)

Đáp án + Giải thích các bước giải:

$a\big)$

$C_2H_4+Br_2\to C_2H_4Br_2$

$C_2H_2+2Br_2\to C_2H_2Br_4$

$b\big)$

Đặt $n_{C_2H_4}=x(mol);n_{C_2H_2}=y(mol)$

$\to x+y=\frac{0,56}{22,4}=0,025(1)$

Theo PT: $x+2y=n_{Br_2}=\frac{5,6}{160}=0,035(2)$

Từ $(1)(2)\to x=0,015(mol);y=0,01(mol)$

$\to \%V_{C_2H_4}=\frac{0,015}{0,025}.100\%=60\%$

$\to \%V_{C_2H_2}=100-60=40\%$