Đáp án:
\(C{\% _{{C_2}{H_4}B{r_2}}} = 22,7\% \)
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:
\({n_{{C_2}{H_4}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{B{r_2}}}\)
\( \to {m_{B{r_2}}} = 0,1.80.2 = 16{\text{ gam}}\)
\( \to {m_{dd{\text{ B}}{{\text{r}}_2}}} = \frac{{16}}{{20\% }} = 80{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{{C_2}{H_4}}} + {m_{dd{\text{ B}}{{\text{r}}_2}}} = 0,1.28 + 80 = 82,8{\text{ gam}}\)
\({n_{{C_2}{H_4}B{r_2}}} = {n_{{C_2}{H_4}}} = 0,1{\text{ mol}}\)
\( \to {m_{{C_2}{H_4}B{r_2}}} = 0,1.(12.2 + 4 + 80.2) = 18,8{\text{ gam}}\)
\( \to C{\% _{{C_2}{H_4}B{r_2}}} = \frac{{18,8}}{{82,8}} = 22,7\% \)
