Đáp án:
\({m_{{C_2}{H_4}B{r_2}}} = 4,7{\text{ gam}}\)
\({\text{\% }}{{\text{V}}_{{C_2}{H_4}}} = 20\% ;\% {n_{C{H_4}}} = 80\% \)
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:
\({n_{B{r_2}}} = \frac{4}{{80.2}} = 0,025{\text{ mol = }}{{\text{n}}_{{C_2}{H_4}}} = {n_{{C_2}{H_4}B{r_2}}} \to {m_{{C_2}{H_4}B{r_2}}} = 0,025.(12.2 + 4 + 80.2) = 4,7{\text{ gam}}\)
\({V_{{C_2}{H_4}}} = 0,025.22,4 = 0,56{\text{ lít}} \to {\text{\% }}{{\text{V}}_{{C_2}{H_4}}} = \frac{{0,56}}{{2,8}} = 20\% \to \% {n_{C{H_4}}} = 80\% \)