Đáp án:
\(\% {V_{{C_2}{H_2}}} = 36,22\% ; \% {V_{C{H_4}}} = 63,78\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)
Ta có:
\({n_{{C_2}{H_2}B{r_4}}} = \frac{{18,8}}{{12.2 + 2 + 80.4}} = \frac{{47}}{{865}}{\text{ mol}}=n_{C_2H_2}\)
\( \to {V_{{C_2}{H_2}}} = \frac{{47}}{{865}}.22,4 = 1,217{\text{ lít}}\)
\( \to \% {V_{{C_2}{H_2}}} = \frac{{1,217}}{{3,36}} = 36,22\% \to \% {V_{C{H_4}}} = 100\% - 36,22\% = 63,78\% \)