Đáp án:
b) 0,4l
c) 14,08g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
b)\\
nN{a_2}C{O_3} = \dfrac{m}{M} = \dfrac{{10,6}}{{106}} = 0,1\,mol\\
nC{H_3}COOH = 2N{a_2}C{O_3} = 0,2\,mol\\
VC{H_3}COOH = \dfrac{n}{{{C_M}}} = \dfrac{{0,2}}{{0,5}} = 0,4l\\
c)\\
n{C_2}{H_5}OH = \dfrac{{11,5}}{{46}} = 0,25\,mol\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
nC{H_3}COO{C_2}{H_5} = nC{H_3}COOH = 0,2\,mol\\
H = 80\% \Rightarrow mC{H_3}COO{C_2}{H_5} = 0,2 \times 88 \times 80\% = 14,08g
\end{array}\)