Đáp án đúng: A
18,8
${{n}_{Ca{{(OH)}_{2}}}}\,=\,0,1\,mol;\,\,{{n}_{CaC{{O}_{3}}}}\,=\,0,05\,mol$
Lọc kết tủa, đun nóng dung dịch lại thấy có kết tủa ⟹ có$Ca{{(HC{{O}_{3}})}_{2}}$
${{n}_{Ca{{(HC{{O}_{3}})}_{2}}}}\,=\,{{n}_{Ca{{(OH)}_{2}}}}\,-\,{{n}_{CaC{{O}_{3}}}}\,=\,0,05\,mol$
$\Rightarrow \,{{n}_{C{{O}_{2}}}}\,=\,{{n}_{CaC{{O}_{3}}}}\,+\,2{{n}_{Ca{{(HC{{O}_{3}})}_{2}}}}\,=\,0,15\,mol$
$\Rightarrow \,{{n}_{{{N}_{2}}}}\,=\,0,1\,mol$$\Rightarrow {{M}_{X}}\,=\,\frac{{{m}_{{{N}_{2}}}}\,+\,{{m}_{C{{O}_{2}}}}}{{{n}_{X}}}\,=\,37,6\,\Rightarrow \,{{d}_{X/{{H}_{2}}}}\,=\,18,8$