Thí nghiệm 1: \({{\rm{n}}_{{\rm{CaC}}{{\rm{O}}_3}}} = 0,05\;\)mol

\( \Rightarrow {n_{CO_3^{2 - }}} + {n_{HCO_3^ - }} = 0,05\,mol\)

Thí nghiệm 2: \({{\rm{n}}_{{\rm{C}}{{\rm{O}}_2}}} = 0,01\)mol; \({{\rm{n}}_{{\rm{HCl}}}} = 0,04\)mol

       \({{\rm{H}}^ + } + {\rm{CO}}_3^{2 - } \to {\rm{HCO}}_3^ - \)     

       \({\rm{HCO}}_3^ - + {{\rm{H}}^ + } \to {\rm{C}}{{\rm{O}}_2} + {{\rm{H}}_2}{\rm{O}}\)

\( \Rightarrow {n_{CO_3^{2 - }}} = {n_{HCl}} - {n_{C{O_2}}} = 0,04 - 0,01 = 0,03\,\,mol\)

\( \Rightarrow {n_{N{a_2}C{O_3}}} = {n_{CO_3^{2 - }}} = 0,03\,mol\)

Bảo toàn nguyên tố \[{\rm{C}}:{{\rm{n}}_{{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}}} + {{\rm{n}}_{{\rm{NaHC}}{{\rm{O}}_3}}} = {{\rm{n}}_{{\rm{CaC}}{{\rm{O}}_3}}} = 0,05 \Rightarrow {{\rm{n}}_{{\rm{NaHC}}{{\rm{O}}_3}}} = 0,02\]mol

\(\begin{array}{l}{{\rm{C}}_{{\rm{M}}\left( {{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}} \right)}} = \frac{{0,03}}{{0,02}} = 1,5{\rm{M}}\\{{\rm{C}}_{{\rm{M}}\left( {{\rm{NaHC}}{{\rm{O}}_3}} \right)}} = \frac{{0,02}}{{0,02}} = 1,0{\rm{M}}{\rm{.}}\end{array}\)

Chọn A.