Đáp án:
\({m_{B{r_2}}}= 1,6{\text{ gam}}\)
\(\% {V_{{C_2}{H_4}}} = 7,467\%; \% {B_{C{H_4}}} = 92,533\% \)
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:
\({n_{{C_2}{H_4}B{r_2}}} = \frac{{1,88}}{{12.2 + 4 + 80.2}} = 0,01{\text{ mol = }}{{\text{n}}_{{C_2}{H_4}}}\)
\( \to {n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,01{\text{ mol}}\)
\( \to {m_{B{r_2}}} = 0,01.(80.2) = 1,6{\text{ gam}}\)
\({V_{{C_2}{H_4}}} = 0,01.22,4 = 0,224{\text{ lít}}\)
\( \to \% {V_{{C_2}{H_4}}} = \frac{{0,224}}{3} = 7,467\% \to \% {B_{C{H_4}}} = 92,533\% \)