Đáp án đúng: D
C2H4O2.
Bình 1:
${{m}_{{{H}_{2}}S{{O}_{4}}}}\,=\,100.96,48{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}\,=\,96,48\,gam$$\Rightarrow {{m}_{{{H}_{2}}O}}\,=\,3,52\,gam$
${{m}_{\text{dd}\,{{H}_{2}}S{{O}_{4}}\,sau}}\,=\,\frac{96,48}{90}.100\,=\,107,2\,gam$$\Rightarrow {{m}_{{{H}_{2}}O\,\,(X)}}\,=\,107,2\,-\,96,48\,-\,3,52\,=\,7,2\,gam$
$\Rightarrow \,{{n}_{{{H}_{2}}O}}\,=\,0,4\,mol$
Bình 2:
${{n}_{{{K}_{2}}C{{O}_{3}}}}\,=\,\frac{55,2}{138}\,=\,0,4\,mol$$\Rightarrow {{n}_{C{{O}_{2}}}}\,=\,0,4\,mol$
Gọi CTPT của X là CxHyO2
$\Rightarrow \,x\,=\,\frac{{{n}_{C{{O}_{2}}}}}{{{n}_{X}}}\,=\,2;\,\,y\,=\,\frac{2{{n}_{{{H}_{2}}O}}}{{{n}_{X}}}\,=\,4$ ⟹ C2H4O2