Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{MgC{O_3}}} = 45,65\% \\
\% {m_{CaC{O_3}}} = 54,35\% \\
b)\\
{m_{BaC{O_3}}} = 19,7g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
MgC{O_3} + 2HCl \to MgC{l_2} + C{O_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{MgC{O_3}}} = {n_{CaC{O_3}}} = a\,mol\\
{m_{hh}} = 18,4g \Leftrightarrow 84a + 100a = 18,4 \Rightarrow a = 0,1\,mol\\
\% {m_{MgC{O_3}}} = \dfrac{{0,1 \times 84}}{{18,4}} \times 100\%  = 45,65\% \\
\% {m_{CaC{O_3}}} = 100 - 45,65 = 54,35\% \\
b)\\
{n_{C{O_2}}} = 0,1 + 0,1 = 0,2\,mol\\
{n_{NaOH}} = 0,2 \times 1,5 = 0,3\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,3}}{{0,2}} = 1,5\\
1 < T < 2 \Rightarrow \text{ Tạo cả 2 muối } \\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O(1)\\
N{a_2}C{O_3} + C{O_2} + {H_2}O \to 2NaHC{O_3}(2)\\
{n_{N{a_2}C{O_3}(1)}} = {n_{C{O_2}(1)}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{n_{N{a_2}C{O_3}(2)}} = {n_{C{O_2}(2)}} = 0,2 - 0,15 = 0,05\,mol\\
{n_{N{a_2}C{O_3}}} = 0,15 - 0,05 = 0,1\,mol\\
N{a_2}C{O_3} + BaC{l_2} \to 2NaCl + BaC{O_3}\\
{n_{BaC{O_3}}} = {n_{N{a_2}C{O_3}}} = 0,1\,mol\\
{m_{BaC{O_3}}} = 0,1 \times 197 = 19,7g
\end{array}\)