Đáp án:

\(\begin{array}{l}
a)\\
M:Natri(Na)\\
b)\\
\% {m_{N{a_2}C{O_3}}} = 42,74\% \\
\% {m_{N{a_2}S{O_4}}} = 57,26\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{P_1}:\\
{M_2}C{O_3} + {H_2}S{O_4} \to {M_2}S{O_4} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
 \Rightarrow {n_{{M_2}C{O_3}}} = {n_{C{O_2}}} = 0,15\,mol\\
{P_2}:\\
{M_2}C{O_3} + BaC{l_2} \to 2MCl + BaC{O_3}\\
{M_2}S{O_4} + BaC{l_2} \to 2MCl + BaS{O_4}\\
{n_{BaC{O_3}}} = {n_{{M_2}C{O_3}}} = 0,15\,mol\\
 \Rightarrow {m_{BaS{O_4}}} = 64,5 - 0,15 \times 197 = 34,95g\\
{n_{BaS{O_4}}} = \dfrac{m}{M} = \dfrac{{34,95}}{{233}} = 0,15\,mol\\
 \Rightarrow {n_{{M_2}S{O_4}}} = {n_{BaS{O_4}}} = 0,15\,mol\\
\text{ Ta có }\\
2 \times 0,15 \times (2M + 60) + 2 \times 0,15 \times (2M + 96) = 74,4\\
 \Leftrightarrow M = 23(g/mol) \Rightarrow M:Natri(Na)\\
b)\\
\% {m_{N{a_2}C{O_3}}} = \dfrac{{0,3 \times 106}}{{74,4}} \times 100\%  = 42,74\% \\
\% {m_{N{a_2}S{O_4}}} = 100 - 42,74 = 57,26\% 
\end{array}\)