Đáp án:

\(\begin{array}{l}
a)\\
{m_{{\rm{dd}}C{H_3}COOH}} = 90g\\
b)\\
{m_{C{H_3}COOK}} = 29,4g\\
{V_{C{O_2}}} = 3,36l\\
c)\\
{C_\% }C{H_3}COOK = 28,24\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2C{H_3}COOH + {K_2}C{O_3} \to 2C{H_3}COOK + C{O_2} + {H_2}O\\
{n_{{K_2}C{O_3}}} = \dfrac{{20,7}}{{138}} = 0,15\,mol\\
{n_{C{H_3}COOH}} = 2{n_{{K_2}C{O_3}}} = 0,3\,mol\\
{m_{{\rm{dd}}C{H_3}COOH}} = \dfrac{{0,3 \times 60}}{{20\% }} = 90g\\
b)\\
{n_{C{H_3}COOK}} = 2{n_{{K_2}C{O_3}}} = 0,3\,mol\\
{m_{C{H_3}COOK}} = 0,3 \times 98 = 29,4g\\
{n_{C{O_2}}} = {n_{{K_2}C{O_3}}} = 0,15\,mol\\
{V_{C{O_2}}} = 0,15 \times 22,4 = 3,36l\\
c)\\
{m_{{\rm{dd}}spu}} = 20,7 + 90 - 0,15 \times 44 = 104,1g\\
{C_\% }C{H_3}COOK = \dfrac{{29,4}}{{104,1}} \times 100\%  = 28,24\% 
\end{array}\)