Đáp án:
\({m_{{C_2}{H_4}B{r_2}}} = 9,4{\text{ gam}}\)
\(\% {V_{{C_2}{H_4}}} = 1\% ; \% {V_{C{H_4}}} = 99\% \)
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:
\({n_{B{r_2}}} = \frac{8}{{80.2}} = 0,05{\text{ mol = }}{{\text{n}}_{{C_2}{H_4}}} = {n_{{C_2}{H_4}B{r_2}}}\)
\( \to {m_{{C_2}{H_4}B{r_2}}} = 0,05.(12.2 + 4 + 80.2) = 9,4{\text{ gam}}\)
\({V_{{C_2}{H_4}}} = 0,05.22,4 = 1,12{\text{ lít}}\)
\( \to \% {V_{{C_2}{H_4}}} = \frac{{1,12}}{{112}} = 1\% \to \% {V_{C{H_4}}} = 99\% \)