Đáp án:
\({m_{{C_2}{H_2}B{r_4}}}= 43,25{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{hh}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)
\( \to {n_{{C_3}{H_8}}} = {n_{{C_2}{H_2}}} = \frac{{0,3}}{2} = 0,15{\text{ mol}}\)
Dẫn hỗn hợp qua \(Br_2\)
\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)
Ta có:
\({n_{B{r_2}}} = 0,5.0,5 = 0,25{\text{ mol < 2}}{{\text{n}}_{{C_2}{H_2}}}\)
Vậy \(C_2H_2\) dư
\( \to {n_{{C_2}{H_2}B{r_4}}} = \frac{1}{2}{n_{B{r_2}}} = 0,125{\text{ mol}}\)
\( \to {m_{{C_2}{H_2}B{r_4}}} = 0,125.(12.2 + 2 + 80.4) = 43,25{\text{ gam}}\)