Đáp án:
\(\% {n_{C{H_4}}} = 50\% \)
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:
\({n_{B{r_2}}} = \frac{{32}}{{80.2}} = 0,2{\text{ mol}}\)
\( \to {n_{{C_2}{H_2}}} = \frac{1}{2}{n_{B{r_2}}} = 0,1{\text{ mol}}\)
\( \to {m_{{C_2}{H_2}}} = 0,1.(12.2 + 2) = 2,6{\text{ gam}} \to {{\text{m}}_{C{H_4}}} = 1,6{\text{ gam}}\)
\( \to {n_{C{H_4}}} = \frac{{1,6}}{{12 + 4}} = 0,1{\text{ mol}}\)
\( \to \% {n_{C{H_4}}} = \frac{{0,1}}{{0,1 + 0,1}} = 50\% \)