Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2CuO + C \to 2Cu + C{O_2}\\
C{O_2} + 2KOH \to {K_2}C{O_3} + {H_2}O(1)\\
C{O_2} + KOH \to KHC{O_3}(2)\\
2KHC{O_3} + 2NaOH \to {K_2}C{O_3} + N{a_2}C{O_3} + 2{H_2}O(3)\\
b)\\
{n_{CuO}} = 1mol\\
{n_{NaOH}} = 0,2mol\\
 \to {n_{KHC{O_3}}} = {n_{NaOH}} = 0,2mol\\
 \to {n_{KOH(2)}} = {n_{C{O_2}(2)}} = {n_{KHC{O_3}}} = 0,2mol\\
 \to {n_{C{O_2}(1)}} = 0,3mol \to {n_{KOH(1)}} = 2{n_{C{O_2}(1)}} = 0,6mol\\
 \to C{M_{KOH}} = \dfrac{{0,6 + 0,2}}{{0,2}} = 4M\\
c)\\
{n_{N{a_2}C{O_3}(3)}} = {n_{{K_2}C{O_3}(3)}} = \dfrac{1}{2}{n_{NaOH}} = 0,1mol\\
 \to {m_{N{a_2}C{O_3}(3)}} = 10,6g\\
 \to {n_{{K_2}C{O_3}(3)}} + {n_{{K_2}C{O_3}(1)}} = 0,4mol\\
 \to {m_{{K_2}C{O_3}}} = 55,2g\\
 \to \% {m_{N{a_2}C{O_3}(3)}} = \dfrac{{10,6}}{{55,2 + 10,6}} \times 100\%  = 16,11\% \\
 \to \% {m_{{K_2}C{O_3}}} = \dfrac{{55,2}}{{55,2 + 10,6}} \times 100\%  = 83,89\% 
\end{array}\)