Đáp án:
\(B\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Ca{C_2} + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {C_2}{H_2}\)
Ta có:
\({n_{Ca{C_2}}} = \frac{{12,8}}{{40 + 12.2}} = 0,2{\text{ mol = }}{{\text{n}}_{{C_2}{H_2}}}\)
\({C_2}{H_2} + {H_2}\xrightarrow{{{t^o}}}{C_2}{H_4}\)
\({C_2}{H_2} + 2{H_2}\xrightarrow{{Ni,{t^o}}}{C_2}{H_6}\)
Khí \(B\) gồm \(C_2H_6;C_2H_4;C_2H_2\)
Cho \(B\) qua \(AgNO_3\)
\({C_2}{H_2} + 2AgN{O_3} + 2N{H_3}\xrightarrow{{}}A{g_2}{C_2} + 2N{H_4}N{O_3}\)
Ta có:
\({n_{A{g_2}{C_2}}} = \frac{{14,4}}{{240}} = 0,06{\text{ mol = }}{{\text{n}}_{{C_2}{H_2}}}\)
Khí thoát ra là \(C_2H_4;C_2H_6\)
\(3{C_2}{H_4} + 2KMn{O_4} + 4{H_2}O\xrightarrow{{}}3{C_2}{H_4}{(OH)_2} + 2KOH + 2Mn{O_2}\)
Ta có:
\({n_{KMn{O_4}}} = \frac{{12,64}}{{39 + 55 + 16.4}} = 0,08{\text{ mol}} \to {{\text{n}}_{{C_2}{H_4}}} = \frac{3}{2}{n_{KMn{O_4}}} = 0,12{\text{ mol}}\)
\( \to {n_{{C_2}{H_6}}} = 0,2 - 0,06 - 0,12 = 0,02{\text{ mol}}\)
\( \to {m_B} = 0,02.30 + 0,12.28 + 0,06.26 = 5,52{\text{ gam}}\)
\({n_B} = 0,2{\text{ mol}} \to {{\text{M}}_B} = \frac{{5,52}}{{0,2}} = 27,6\)
\( \to {d_{B/{H_2}}} = \frac{{27,6}}{2} = 13,8\)
