Đáp án:
3%
71,875 ml
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
nC{O_2} = \dfrac{{0,56}}{{22,4}} = 0,025\,mol\\
nC{H_3}COOH = 0,025 \times 2 = 0,05\,mol\\
C\% C{H_3}COOH = \dfrac{{0,05 \times 60}}{{100}} \times 100\% = 3\% \\
c)\\
2{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
n{C_2}{H_5}OH = 0,05 \times 2 = 0,1\,mol\\
V{C_2}{H_5}OH = \dfrac{{0,1 \times 46}}{{0,8}} = 5,75\,ml\\
Vr = \dfrac{{5,75}}{{10}} \times 100 = 57,5ml\\
Vr = \dfrac{{57,5}}{{80}} \times 100 = 71,875\,ml
\end{array}\)
