Hỗn hợp E gồm hai amin X (CnHmN); Y (CnHm+1N2 với n ≥ 2 ) và hai

* Hướng dẫn giải

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{A} \\ \left\{\begin{array}{l}\mathrm{X}:{\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}}\mathrm{N}\\ \mathrm{Y}:{\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}+1}{\mathrm{N}}_2\\ \mathrm{Anken}:{\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}\end{array}\right.\underset{}{\overset{+{\mathrm{O}}_2}{\to }}\left\{\begin{array}{l}{\mathrm{CO}}_2\text{ 0,22}\\ {\mathrm{H}}_2\mathrm{O}\text{ 0,3}\\ {\mathrm{N}}_2\text{ 0,03}\end{array}\right. \\ \mathrm{Từ} \mathrm{CTPT} \to \mathrm{X},\mathrm{Y} \mathrm{cùng} \mathrm{số} \mathrm{liên} \mathrm{kết} \\ \to {\text{n}}_{\mathrm{\pi }}=0,22+0,03-0,3+0,08=0,03<0,08 \to \mathrm{X}, \mathrm{Y} \mathrm{cùng} \mathrm{no} \\ {\mathrm{n}}_{\mathrm{anken}}={\text{n}}_{\mathrm{\pi }}=0,03\underset{}{\overset{}{\to }}{\mathrm{n}}_{\mathrm{X},\mathrm{Y}}=0,05\underset{}{\overset{}{\to }}\left\{\begin{array}{l}{\mathrm{n}}_{\mathrm{Y}}=0,03.2-0,05=0,01\\ {\mathrm{n}}_{\mathrm{X}}=0,04\end{array}\right. \\ \mathrm{BT}\left(\mathrm{C}\right):0,04\mathrm{n}+0,01\mathrm{n}+0,03\mathrm{x}=0,22\underset{}{\overset{}{\to }}\left\{\begin{array}{l}\mathrm{n}=3\\ \mathrm{x}=7/3\end{array}\right. \\ \underset{}{\overset{}{\to }}\mathrm{Y}:{\mathrm{C}}_3{\mathrm{H}}_{10}{\mathrm{N}}_2\text{ 0,01}\underset{}{\overset{}{\to }}\mathrm{\%}{\mathrm{m}}_{\mathrm{Y}}=18,14\mathrm{\%} \end{gathered}$$