Đáp án đúng: D
C3H8.
Khối lượng dung dịch sau phản ứng đúng bằng khối lượng dung dịch trước phản ứng nên${{m}_{C{{O}_{2}}}}\,+\,{{m}_{{{H}_{2}}O}}\,=\,{{m}_{CaC{{O}_{3}}}}\,=\,20,4\,gam$
TH1: không tạo Ca(HCO3)2
$\Rightarrow \,{{n}_{C{{O}_{2}}}}\,=\,{{n}_{CaC{{O}_{3}}}}\,=\,0,204\,mol$
$\Rightarrow \,{{m}_{{{H}_{2}}O}}\,=\,20,4\,-\,44.0,204\,=\,11,424\,gam$
$\Rightarrow \,{{n}_{{{H}_{2}}O}}\,=\,\frac{238}{375}\,mol\,\Rightarrow \,{{n}_{C}}\,:\,{{n}_{H}}\,=\,{{n}_{C{{O}_{2}}}}\,:\,(2{{n}_{{{H}_{2}}O}})\,=\,9\,:\,56$
Không có HC nào thỏa mãn.
TH2: tạo muối Ca(HCO3)2
${{n}_{Ca{{(HC{{O}_{3}})}_{2}}}}\,=\,{{n}_{Ca{{(OH)}_{2}}}}\,-\,{{n}_{CaC{{O}_{3}}}}\,=\,0,048\,mol$
$\Rightarrow \,{{n}_{C{{O}_{2}}}}\,=\,{{n}_{CaC{{O}_{3}}}}\,+\,2{{n}_{Ca{{(HC{{O}_{3}})}_{2}}}}\,=\,0,3\,mol$
$\Rightarrow \,{{m}_{{{H}_{2}}O}}\,=\,20,4\,-\,44.0,3\,=\,7,2\,gam$
$\Rightarrow \,{{n}_{{{H}_{2}}O}}\,=\,0,4\,mol\,\Rightarrow \,{{n}_{C}}\,:\,{{n}_{H}}\,=\,{{n}_{C{{O}_{2}}}}\,:\,(2{{n}_{{{H}_{2}}O}})\,=\,3\,:\,8$
$\Rightarrow \,{{C}_{3}}{{H}_{8}}$
