Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_5}OH + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_5}OH}} = 0,1mol\\
\to {n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = 0,3mol\\
\to {V_{{O_2}}} = 6,72l\\
\to {V_{KK}} = 5{V_{{O_2}}} = 33,6l\\
{V_{{C_2}{H_5}OH}} = \dfrac{{4,6}}{{0,8}} = 5,75ml\\
\to {V_{{C_2}{H_5}OH}}{\rm{dd}} = \dfrac{{5,75 \times 100}}{8} = 71,875ml\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
{n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 0,1mol\\
\to {m_{C{H_3}COOH}} = 6g\\
\to {m_{C{H_3}COOH}}thực= \dfrac{{6 \times 80}}{{100}} = 4,8g
\end{array}\)