Đáp án:

\(\begin{array}{l}
\% {m_{N{a_2}C{O_3}}} = 33,87\% \\
\% {m_{{K_2}C{O_3}}} = 66,13\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O(1)\\
{K_2}C{O_3} + 2HCl \to 2KCl + C{O_2} + {H_2}O(2)\\
{n_{C{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{n_{HCl}} \text{ tham gia }= 2{n_{C{O_2}}} = 0,5\,mol\\
{n_{HCl}} = 0,4 \times 1,5 = 0,6\,mol\\
{n_{HCl}} \text{ dư }= 0,6 - 0,5 = 0,1\,mol\\
NaOH + HCl \to NaCl + {H_2}O(3)\\
{n_{NaCl(3)}} = {n_{HCl}}  \text{ dư }= 0,1\,mol\\
 \Rightarrow {m_{NaCl(1)}} + {m_{KCl}} = 39,9 - 0,1 \times 58,5 = 34,05g\\
hh:N{a_2}C{O_3}(a\,mol),{K_2}C{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
2a \times 58,5 + 2b \times 74,5 = 34,05\\
a + b = 0,25
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,15\\
\% {m_{N{a_2}C{O_3}}} = \dfrac{{0,1 \times 106}}{{0,1 \times 106 + 0,15 \times 138}} \times 100\%  = 33,87\% \\
\% {m_{{K_2}C{O_3}}} = 100 - 33,87 = 66,13\% 
\end{array}\)