Đáp án:
Em tham khảo nha
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
a)\\
V{C_2}{H_5}OH = 1,5 \times \dfrac{{20}}{{100}} = 0,3l = 300ml\\
b)\\
m{C_2}{H_5}OH = 300 \times 0,8 = 240g\\
n{C_2}{H_5}OH = \dfrac{{240}}{{46}} = \dfrac{{120}}{{23}}\,mol\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
nC{H_3}COOH = n{C_2}{H_5}OH = \dfrac{{120}}{{23}}\,mol\\
mC{H_3}COOH = \dfrac{{120}}{{23}} \times 60 = 313g\\
2)\\
a)\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
b)\\
nC{O_2} = nCaC{O_3} = \dfrac{{10}}{{100}} = 0,1\,mol\\
nC{H_3}COOH = 0,1 \times 2 = 0,2\,mol\\
n{H_2} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol \Rightarrow nhhA = 0,2 \times 2 = 0,4\,mol\\
n{C_2}{H_5}OH = 0,4 - 0,2 = 0,2\,mol\\
\% mC{H_3}COOH = \dfrac{{0,2 \times 60}}{{0,2 \times 46 + 0,2 \times 60}} \times 100\% = 56,6\% \\
\% m{C_2}{H_5}OH = 100 - 56,6 = 43,4\%
\end{array}\)
