Đáp án:
\( \% {m_{{C_3}{H_6}}} = 52,5\% ; \% {m_{C{H_4}}} = 47,5\% \)
\({m_{{C_2}{H_4}B{r_2}}} = 225,6{\text{ gam}}\)
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:
\({n_{B{r_2}}} = 0,6.2 = 1,2{\text{ mol = }}{{\text{n}}_{{C_2}{H_4}}} = {n_{{C_2}{H_4}B{r_2}}}\)
\( \to {m_{{C_2}{H_4}}} = 1,2.(12.2 + 4) = 33,6{\text{ gam}}\)
\( \to \% {m_{{C_3}{H_6}}} = \frac{{33,6}}{{64}} = 52,5\% \to \% {m_{C{H_4}}} = 47,5\% \)
\({m_{{C_2}{H_4}B{r_2}}} = 1,2(12.2 + 4 + 80.2) = 225,6{\text{ gam}}\)