Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 2,8l\\
b)\\
{V_{{\rm{dd}}{C_2}{H_5}OH}} = 119,79ml
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
{n_{C{H_3}COOH}} = \dfrac{{300 \times 5\% }}{{60}} = 0,25\,mol\\
{n_{{H_2}}} = \dfrac{{0,25}}{2} = 0,125\,mol\\
{V_{{H_2}}} = 0,125 \times 22,4 = 2,8l\\
b)\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
{n_{{C_2}{H_5}OH}} = {n_{C{H_3}COOH}} = 0,25\,mol\\
H = 80\% \Rightarrow {m_{{C_2}{H_5}OH}} = \dfrac{{0,25 \times 46}}{{80\% }} = 14,375g\\
{V_{{C_2}{H_5}OH}} = \dfrac{{14,375}}{{0,8}} = 17,96875\,ml\\
{V_{{\rm{dd}}{C_2}{H_5}OH}} = \dfrac{{17,96875 \times 100}}{{15}} = 119,79ml
\end{array}\)
