Đáp án đúng: C
Giải chi tiết:\(\begin{gathered} {n_{N{H_3}}} = \frac{{33,6}}{{22,4}} = 1,5{\text{ }}mol \hfill \\ {N_2}\,\,\, + \,\,\,\,\,3{H_2}\,\,\,\overset {{t^o},xt} \leftrightarrows \,\,\,\,2N{H_3} \hfill \\ 0,75\,\,\,\,\,\,\,2,25\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,5 \hfill \\ \to {n_{{N_2}(LT)}} = 0,75.\frac{{100}}{{60}} = 1,25\,mol \to {V_{{N_2}(LT)}} = 28\,(l). \hfill \\ \to {n_{{H_2}(LT)}} = 2,25.\frac{{100}}{{60}} = 3,75\,mol \to {V_{{N_2}(LT)}} = 84\,(l). \hfill \\ \hfill \\ \end{gathered} \)
Đáp án: C