Đáp án đúng: A
CH4 và C2H6
Khí ra khỏi bình là oxi dư ${{n}_{{{O}_{2}}\,du}}\,=\,\frac{PV}{RT}\,=\,\frac{0,4.11,2}{0,082.273}\,=\,0,2\,mol$$\Rightarrow {{n}_{{{O}_{2}}\,pu}}\,=\,2\,-\,0,2\,=\,1,8\,mol$
${{n}_{C{{O}_{2}}}}\,=\,{{n}_{CaC{{O}_{3}}}}\,=\,1\,mol\,\Rightarrow \,{{n}_{{{H}_{2}}O}}\,=\,2{{n}_{{{O}_{2}}\,pu}}\,-\,2{{n}_{C{{O}_{2}}}}\,=\,1,6\,mol$
${{n}_{{{H}_{2}}O}}\,>\,{{n}_{C{{O}_{2}}}}\,\Rightarrow \,ankan\,\Rightarrow \,{{n}_{ankan}}\,=\,{{n}_{{{H}_{2}}O}}\,-\,{{n}_{C{{O}_{2}}}}\,=\,0,6\,mol$
Số C =$\frac{{{n}_{C{{O}_{2}}}}}{{{n}_{ankan}}}$ = 1,67 ⟹ CH4 và C2H6