Đáp án:

$m_{Br_2}=60(g)$

Giải thích các bước giải:

Quy đổi hỗn hợp X gồm: $\begin{cases} CH_4\\C_2H_4\\C_3H_4\\C_4H_4\\H_2 \end{cases}$ thành: $C_nH_4$ và $H_2.$ 

$\underbrace{C_nH_4}_{X}+H_2\xrightarrow{t^o,\ Ni}\underbrace{C_nH_m}_{Y}$

$n_Y=\dfrac{5,6}{22,4}=0,25(mol)\\n_{Br_2}=\dfrac{72}{160}=0,45(mol)$

Y có dạng: $C_nH_{2n+2-2k}$

$k=\dfrac{n_{Br_2}}{n_Y}=\dfrac{0,45}{0,25}=1,8$

$\to Y:\ C_nH_{2n-1,6}$

Mà: $M_Y=1×46=46(g/mol)$

$\to 14n-1,6=46\\\to n=3,4$

Vây X là $C_{3,4}H_4$ và Y là $C_{3,4}H_{5,2}$

$C_{3,4}H_4+0,6H_2\xrightarrow{t^o,\ Ni}C_{3,4}H_{5,2}$

Đặt: $\begin{cases} n_{C_{3,4}H_4}=x\ (mol)\\n_{H_2}=0,6x\ (mol) \end{cases}$

$n_X=0,25=x+0,6x\\\to x=0,15625(mol)\\k'=\dfrac{3,4×2+2-4}{2}=2,4\\n_{Br_2}=x×k'=0,375(mol)\\m_{Br_2}=0,375×160=60(g)$

\(hh{\text{ X gom`: CH}}_4^{};C_2^{}H_4^{};C_3^{}H_4^{};C_4^{}H_4^{};H_2^{}\)

\({\text{cac hidrocacbon trong X deu co 4 hidro}}\)

\( \to {\text{C}}_n^{}H_4^{};H_2^{}\)

\(n_Y^{} = \frac{{5,6}}{{22,4}} = 0,25\;{\text{mol; M}}_Y^{} = 46g/mol\)

\({\text{C}}_n^{}H_4^{}(X) + Br_2^{} \to C_n^{}H_a^{}(Y)\)

\(14n + a = 46\)

\(\Delta  = \frac{{2C - H + 2}}{2}.n_Y^{} = 0,45\)

\( \to \frac{{2.n - a + 2}}{2}.0,25 = 0,45\)

\( \to n = 3,4;a = 5,2\)

\( \to C_{3,4}^{}H_{5,2}^{}(Y)\)

\(C_{3,4}^{}H_4^{} + 2,4Br_2^{} \to C_{3,4}^{}H_4^{}Br_{4,8}^{}(HCHC\;{\text{no)}}\)

\({\text{0}}{\text{,25}}\,{\text{      0}}{\text{,6}}\)

\({\text{m}}_{Br_2^{}}^{} = 0,6.160 = 96g\)