Đáp án đúng: B
CH2(NH2)COOH; CH2(NH2)COOCH3.
${{M}_{A}}\,=\,44,5.2\,=\,89$
${{n}_{A}}\,=\,8,9/89\,=\,0,1\,mol$
${{n}_{C{{O}_{2}}}}\,=\,\frac{13,2}{44}\,=\,0,3$;${{n}_{{{H}_{2}}O}}\,=\,\frac{6,3}{18}\,=\,0,35$;${{n}_{{{N}_{2}}}}\,=\,\frac{1,12}{22,4}\,=\,0,05$
${{C}_{x}}{{H}_{y}}{{O}_{z}}{{N}_{t}}\,\xrightarrow{{{O}_{2}},\,{{t}^{o}}}\,xC{{O}_{2}}\,+\,\frac{y}{2}{{H}_{2}}O\,+\,\frac{1}{2}{{N}_{2}}$
Bảo toàn nguyên tố có
$\left\{ \begin{array}{l}0,1x\,=\,0,3\\0,1y\,=\,0,35.2\\0,1t\,=\,0,05.2\end{array} \right.$
⇒
$\left\{ \begin{array}{l}x\,=\,3\\y\,=7\\t\,=\,1\end{array} \right.$
⇒
$z\,=\,\frac{89-3.12-7-14}{16}\,=\,2$
Vậy A là${{H}_{2}}N-C{{H}_{2}}-COOC{{H}_{3}}$ và${{H}_{2}}N-C{{H}_{2}}-COOH$