$n_{Br_2} = \dfrac{m}{M} = \dfrac{160}{160} = 1 (mol) \\ C_2H_4 + Br_2 \xrightarrow\ C_2H_4Br_2 \\ n_{C_2H_4} = n_{Br_2} = n_{C_2H_4Br_2 } = 1 (mol) \\ 1) V_{C_2H_4} = n . 22,4 = 1 . 22,4 = 22,4 (l) \\ V_{CH_4} = V_{hh} - V_{C_2H_4} = 67,2 - 22,4 = 44,8 (l) \\ 2) m_{C_2H_4Br_2} = n . M = 1 . 188 = 188 (g) $

Đáp án:

\({V_{{C_2}{H_4}}}  = 22,4{\text{ lít}}; {{\text{V}}_{C{H_4}}} = 44,8{\text{ lít}}\)

\({m_{{C_2}{H_4}B{r_2}}} = 188{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

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

Ta có:

\({n_{B{r_2}}} = \frac{{160}}{{80.2}} = 1{\text{ mol}}\)

\( \to {n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 1{\text{ mol}}\)

\( \to {V_{{C_2}{H_4}}} = 1.22,4 = 22,4{\text{ lít}} \to {{\text{V}}_{C{H_4}}} = 67,2 - 22,4 = 44,8{\text{ lít}}\)

Ta có:

\({n_{{C_2}{H_4}B{r_2}}} = {n_{B{r_2}}} = 1{\text{ mol}}\)

\( \to {m_{{C_2}{H_4}B{r_2}}} = 1.(12.2 + 4 + 80.2) = 188{\text{ gam}}\)