Đáp án:

\({V_{{C_2}{H_4}}} = 4,48{\text{ lít;}}{{\text{V}}_{C{H_4}}} = 2,24{\text{ lít}} \to {\text{\% }}{{\text{V}}_{{C_2}{H_4}}} = \frac{{4,48}}{{6,72}} = 66,67\% ;\% {V_{C{H_4}}} = 33,33\% \)

\({{\text{m}}_{{C_2}{H_4}B{r_2}}} = 37,6{\text{ gam}}\)

Giải thích các bước giải:

 Ta có: \({n_{hh}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)

Cho hỗn hợp qua brom thì chỉ có etilen phản ứng.

\({C_2}{H_4} + B{r_2}\xrightarrow{{}}{C_2}{H_4}B{r_2}\)

Ta có: 

\({n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,2.1 = 0,2{\text{ mol}} \to {n_{C{H_4}}} = 0,1{\text{ mol}}\)

\(\to {V_{{C_2}{H_4}}} = 0,2.22,4 = 4,48{\text{ lít}} \to {{\text{V}}_{C{H_4}}} = 2,24{\text{ lít}} \to {\text{\% }}{{\text{V}}_{{C_2}{H_4}}} = \frac{{4,48}}{{6,72}} = 66,67\% ;\% {V_{C{H_4}}} = 33,33\% \)

\({n_{{C_2}{H_4}B{r_2}}} = {n_{B{r_2}}} = 0,2{\text{ mol}} \to {{\text{m}}_{{C_2}{H_4}B{r_2}}} = 0,2.(12.2 + 4 + 80.2) = 37,6{\text{ gam}}\)

a,

$n_{Br_2}=0,2(mol)$

$C_2H_4+Br_2\to C_2H_4Br_2$

$\Rightarrow n_{C_2H_4}=n_{C_2H_4Br_2}=0,2(mol)$

$V_{C_2H_4}=0,2.22,4=4,48l$

$V_{CH_4}=6,72-4,48=2,24l$

b,

$\%V_{CH_4}=\dfrac{2,24.100}{6,72}=33,33\%$

$\%V_{C_2H_4}=66,67\%$

c,

$m_{C_2H_4Br_2}=0,2.188=37,6g$