Đáp án:
\({m_{B{r_2}}} = 80{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)
Khí thoát ra là \(CH_4\)
\( \to {V_{C{H_4}}} = 4,48{\text{ lít}} \to {{\text{V}}_{{C_2}{H_2}}} = 10,08 - 4,48 = 5,6{\text{ lít}}\)
\( \to \% {V_{C{H_4}}} = \frac{{4,48}}{{10,08}} = 44,44\% \to \% {V_{{C_2}{H_2}}} = 55,56\% \)
\( \to {n_{{C_2}{H_2}}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}} \to {{\text{n}}_{B{r_2}}} = 2{n_{{C_2}{H_2}}} = 0,5{\text{ mol}}\)
\( \to {m_{B{r_2}}} = 0,5.(80.2) = 80{\text{ gam}}\)