Đáp án đúng: B
Giải chi tiết:\(\underbrace {\underbrace {{C_2}{H_4}}_{X\,\,\,mol}\underbrace {CH \equiv C - C{H_3}}_{y\,mol}}_{a\,mol\,}\left\langle \begin{gathered} \xrightarrow{{ + AgN{O_3}/N{H_3}}}0,12\,mol\,\,CAg \equiv C - C{H_3}\, = > \,y = \,0,12\,mol \hfill \\ \xrightarrow{{ + 0,\,34\,mol\,{H_2}}}{n_{{C_2}{H_4}}} = \,n{\,_{{H_2}}}\, - 2{n_{{C_3}{H_4}}} = 0,34 - 0,12.2 = 0,1 \hfill \\ \end{gathered} \right. = > \,a\, = \,0,12\, + 0,1\, = 0,22\,mol\)
Đáp án B