Đáp án đúng: D
Ta có
$\left\{ \begin{array}{l}{{n}_{Z}}\,=\,{{n}_{{{N}_{2}}}}\,+\,{{n}_{C{{O}_{2}}}}\,=\,\frac{0,56}{22,4}\,=\,0,025\\{{m}_{Z}}\,=\,28{{n}_{{{N}_{2}}}}\,+\,44{{n}_{C{{O}_{2}}}}\,=\,0,025.20,4.2\,=\,1,02\end{array} \right.$ ⇒$\left\{ \begin{array}{l}{{n}_{{{N}_{2}}}}\,=\,0,005\\{{n}_{C{{O}_{2}}}}\,=\,0,02\end{array} \right.$
${{C}_{x}}{{H}_{y}}{{O}_{z}}{{N}_{t}}$
+
$(x+\frac{y}{4}-\frac{z}{2}){{O}_{2}}$
$\xrightarrow{{{t}^{o}}}$
$xC{{O}_{2}}$
+
$\frac{y}{2}{{H}_{2}}O$
+
$\frac{t}{2}{{N}_{2}}$
$\displaystyle 0,01\text{ }mol$
→
$0,01(x+\frac{y}{4}-\frac{z}{2})$
→
$0,01x$
→
$0,01.\frac{y}{2}$
→
$0,01.\frac{t}{2}$
Do đó ta có hệ phương trình
$\left\{ \begin{array}{l}0,01(x+\frac{y}{4}-\frac{z}{2})\,=\,0,0275\\0,01x\,=\,0,02\\0,01\frac{t}{2}\,=\,0,005\end{array} \right.$
⇒
$\left\{ \begin{array}{l}y-2z=3\\x=2\\t=1\end{array} \right.$
⇒
$\left\{ \begin{array}{l}z=1\\y=5\\x=2\\t=1\end{array} \right.$
hoặc
$\left\{ \begin{array}{l}z=2\\y=7\\x=2\\t=1\end{array} \right.$
Vậy A là${{C}_{2}}{{H}_{5}}ON$ hoặc${{C}_{2}}{{H}_{7}}{{O}_{2}}N$
