Đáp án:
\(\begin{array}{l}
a)\\
{m_{C{H_3}COONa}} = 24,6g\\
b)\\
{C_\% }C{H_3}COOH = 3,61\% \\
{C_\% }C{H_3}COONa = 14,78\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{C{H_3}COOH}} = \dfrac{{120 \times 20\% }}{{60}} = 0,4\,mol\\
{n_{N{a_2}C{O_3}}} = \dfrac{{53 \times 30\% }}{{106}} = 0,15\,mol\\
N{a_2}C{O_3} + 2C{H_3}COOH \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
\text{ Lập tỉ lệ }:\dfrac{{{n_{N{a_2}C{O_3}}}}}{1} < \dfrac{{{n_{C{H_3}COOH}}}}{2}(0,15 < 0,2)\\
\Rightarrow \text{ $CH_3COOH$ dư } \\
{n_{C{H_3}COONa}} = 2{n_{N{a_2}C{O_3}}} = 0,3\,mol\\
{m_{C{H_3}COONa}} = 0,3 \times 82 = 24,6g\\
b)\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,15\,mol\\
{m_{{\rm{dd}}spu}} = 120 + 53 - 0,15 \times 44 = 166,4g\\
{n_{C{H_3}COOH}} \text{ dư }= 0,4 - 0,15 \times 2 = 0,1\,mol\\
{C_\% }C{H_3}COOH \text{ dư } = \dfrac{{0,1 \times 60}}{{166,4}} \times 100\% = 3,61\% \\
{C_\% }C{H_3}COONa = \dfrac{{24,6}}{{166,4}} \times 100\% = 14,78\%
\end{array}\)
