Đáp án:
\(\begin{array}{l}
{V_{kk}} = 100,8l\\
{V_{C{O_2}}} = 13,44l\\
{m_{C{H_3}COOH}} = 14,4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{P_1}:\\
{C_2}{H_5}OH + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 3{H_2}O\\
{m_{{C_2}{H_5}OH}} = \dfrac{{27,6}}{2} = 13,8g\\
{n_{{C_2}{H_5}OH}} = \dfrac{{13,8}}{{46}} = 0,3\,mol\\
{n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = 0,9\,mol\\
{V_{kk}} = 0,9 \times 5 \times 22,4 = 100,8l\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_5}OH}} = 0,6\,mol\\
{V_{C{O_2}}} = 0,6 \times 22,4 = 13,44l\\
{P_2}:\\
{C_2}{H_5}OH + {O_2} \xrightarrow{\text{ lên men }} C{H_3}COOH + {H_2}O\\
{n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 0,3\,mol\\
H = 80\% \Rightarrow {m_{C{H_3}COOH}} = 0,3 \times 60 \times 80\% = 14,4g
\end{array}\)