Đáp án:
\(\% {V_{{C_2}{H_4}}} = \% {V_{{C_2}{H_6}}} = 30\% ;\% {V_{{C_2}{H_2}}} = 40\% \)
\(\% {m_{{C_2}{H_4}}} 30,2\% ;\% {m_{{C_2}{H_6}}} = 32,4\% \)
\( \% {m_{{C_2}{H_2}}} = 37,4\% \)
Giải thích các bước giải:
Dẫn hỗn hợp \(X\) qua dung dịch \(Br_2\) dư
\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)
\({C_2}{H_4} + B{r_2}\xrightarrow{{}}{C_2}{H_4}B{r_2}\)
Khí thoát ra là \(C_2H_6\)
\({n_{{C_2}{H_6}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol;}}{{\text{n}}_X} = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol}}\)
Cho hỗn hợp khí trên qua \(AgNO_3/NH_3\)
\({C_2}{H_2} + 2AgN{O_3} + 2N{H_3}\xrightarrow{{}}A{g_2}{C_2} + 2N{H_4}N{O_3}\)
\( \to {n_{A{g_2}{C_2}}} = \frac{{36}}{{240}} = 0,15{\text{ mol = }}{{\text{n}}_{{C_2}{H_2}}}\)
\( \to {n_{{C_2}{H_4}}} = 0,5 - 0,15 - 0,2 = 0,15{\text{ mol}}\)
Vì % số mol=% thể tích
\( \to \% {V_{{C_2}{H_4}}} = \% {V_{{C_2}{H_6}}} = \frac{{0,15}}{{0,5}} = 30\% \to \% {V_{{C_2}{H_2}}} = 40\% \)
\({m_{{C_2}{H_4}}} = 0,15.28 = 4,2{\text{ gam;}}{{\text{m}}_{{C_2}{H_6}}} = 0,15.30 = 4,5{\text{ gam;}}{{\text{m}}_{{C_2}{H_2}}} = 0,2.26 = 5,2{\text{ gam}}\)
\( \to \% {m_{{C_2}{H_4}}} = \frac{{4,2}}{{4,2 + 4,5 + 5,2}} = 30,2\% ;\% {m_{{C_2}{H_6}}} = \frac{{4,5}}{{4,2 + 4,5 + 5,2}} = 32,4\% \)
\( \to \% {m_{{C_2}{H_2}}} = 37,4\% \)
