Đáp án:
\(\% {m_{C{H_4}}}= 16\% ; \% {m_{{C_2}{H_4}}} = 84\% \)
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:
\({n_{B{r_2}}} = \frac{{24}}{{80.2}} = 0,15{\text{ mol = }}{{\text{n}}_{{C_2}{H_4}}}\)
\({n_{hh}} = {n_{C{H_4}}} + {n_{{C_2}{H_4}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)
\( \to {n_{C{H_4}}} = 0,2 - 0,15 = 0,05{\text{ mol}}\)
\( \to {m_{C{H_4}}} = 0,05.16 = 0,8{\text{ gam;}}{{\text{m}}_{{C_2}{H_4}}} = 0,15.(12.2 + 4) = 4,2{\text{ gam}}\)
\( \to \% {m_{C{H_4}}} = \frac{{0,8}}{{0,8 + 4,2}} = 16\% \to \% {m_{{C_2}{H_4}}} = 84\% \)