$Na+{{H}_{2}}O\to NaOH+\frac{1}{2}{{H}_{2}}$

$Ca+2{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}+{{H}_{2}}$

$O{{H}^{-}}+HCO_{3}^{-}\to CO_{3}^{2-}+{{H}_{2}}O$

$C{{a}^{2+}}+CO3_{3}^{2-}\to CaC{{O}_{3}}$

Vì ${{n}_{CaC{{O}_{3}}}}=0,07\,mol<{{n}_{HC{{O}_{3}}}}=0,08\,mol$

Có ${{n}_{OH}}=2{{n}_{{{H}_{2}}}}=0,08\,mol\Rightarrow $ Chứng tỏ $C{{a}^{2+}}$ thiếu, $CO_{3}^{2-}$ dư

$\Rightarrow {{n}_{C{{a}^{2+}}}}=0,07mol={{n}_{CaC{{O}_{3}}}}\Rightarrow {{n}_{Ca}}=0,07-0,04=0,03\,mol$

Có: ${{n}_{Na}}+2{{n}_{Ca}}=0,08\,mol\Rightarrow {{n}_{Na}}=0,02\,mol$

$\Rightarrow m=1,66g$

Đáp án A