Đáp án:
\(\begin{array}{l}
b)\\
{V_{{C_2}{H_5}OH}} = 115\,ml\\
{m_{{C_2}{H_5}OH}} = 92g\\
c)\\
{m_{C{H_3}COOH}} = 108g\\
d)\\
{m_{C{H_3}COONa}} = 82g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_5}OH + {O_2} \xrightarrow{\text{ lên men }} C{H_3}COOH + {H_2}O\\
b)\\
{V_{{C_2}{H_5}OH}} = 575 \times \dfrac{{20}}{{100}} = 115\,ml\\
{m_{{C_2}{H_5}OH}} = 115 \times 0,8 = 92g\\
c)\\
{n_{{C_2}{H_5}OH}} = \dfrac{{92}}{{46}} = 2\,mol\\
{n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 2\,mol\\
H = 90\% \Rightarrow {m_{C{H_3}COOH}} = 2 \times 90\% \times 60 = 108g\\
d)\\
{n_{NaOH}} = 0,5 \times 2 = 1\,mol\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{C{H_3}COOH}} < {n_{NaOH}}(1,8 > 1) \Rightarrow \text{ $CH_3COOH$ dư } \\
{n_{C{H_3}COONa}} = {n_{NaOH}} = 1\,mol\\
{m_{C{H_3}COONa}} = 1 \times 82 = 82g
\end{array}\)