Đáp án:
\( \% {n_{C{H_4}}} = 25\% ;\% {n_{{C_2}{H_2}}} = 75\% \)
\( {m_{C{H_4}}} = 1,6{\text{ gam;}}{{\text{m}}_{{C_2}{H_2}}} = 7,8{\text{ gam}}\)
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:
\({n_{B{r_2}}} = 0,2.0,75 = 0,15{\text{ mol}}\)
\( \to {n_{{C_2}{H_2}}} = \frac{1}{2}{n_{B{r_2}}} = \frac{1}{2}.0,15 = 0,075{\text{ mol}}\)
\({n_X} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {{\text{n}}_{C{H_4}}} = 0,1 - 0,075 = 0,025{\text{ mol}}\)
\( \to \% {n_{C{H_4}}} = \frac{{0,025}}{{0,1}} = 25\% \to \% {n_{{C_2}{H_2}}} = 75\% \)
Trong 8,96 lít hỗn hợp
\({n_{C{H_4}}} = 0,025.\frac{{8,96}}{{2,24}} = 0,1{\text{ mol;}}{{\text{n}}_{{C_2}{H_2}}} = 0,075.\frac{{8,96}}{{22,4}} = 0,3{\text{ mol}}\)
\( \to {m_{C{H_4}}} = 0,1.16 = 1,6{\text{ gam;}}{{\text{m}}_{{C_2}{H_2}}} = 0,3.26 = 7,8{\text{ gam}}\)
