Đáp án đúng: B
C3H8O2.
$\left\{ \begin{array}{l}\frac{{{n}_{{{H}_{2}}O}}}{{{n}_{C{{O}_{2}}}}}\,=\,\frac{4}{3}\\44{{n}_{C{{O}_{2}}}}\,+\,18{{n}_{{{H}_{2}}O}}\,=\,7,6\,+\,32.\frac{8,96}{22,4}\end{array} \right.\,\Rightarrow \,\left\{ \begin{array}{l}{{n}_{C{{O}_{2}}}}\,=\,0,3\\{{n}_{{{H}_{2}}O}}\,=\,0,4\end{array} \right.$
${{m}_{O}}\,=\,{{m}_{X}}\,-\,{{m}_{H}}\,-\,{{m}_{C}}\,=\,3,2\,gam\,\Rightarrow \,{{n}_{O}}\,=\,0,2\,mol$
${{n}_{C}}\,:\,{{n}_{H}}\,:\,{{n}_{O}}\,=\,3\,:\,8\,:\,2\,\Rightarrow \,{{C}_{3}}{{H}_{8}}{{O}_{2}}$
