Đáp án:
\(\begin{array}{l}
a)\\
{m_{C{H_3}COOH}} = 18g\\
{m_{{C_2}{H_5}OH}} = 4,6g\\
b)\\
{m_{Na}} = 9,2g\\
c)\\
{V_{{H_2}}} = 4,48l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
C{H_3}COOH + KOH \to C{H_3}COOK + {H_2}O\\
{n_{KOH}} = 0,3 \times 1 = 0,3\,mol\\
{n_{C{H_3}COOH}} = {n_{KOH}} = 0,3\,mol\\
{m_{C{H_3}COOH}} = 0,3 \times 60 = 18g\\
{m_{{C_2}{H_5}OH}} = 22,6 - 18 = 4,6g\\
b)\\
{n_{{C_2}{H_5}OH}} = \dfrac{{4,6}}{{46}} = 0,1\,mol\\
{n_{Na}} = 0,3 + 0,1 = 0,4\,mol\\
{m_{Na}} = 0,4 \times 23 = 9,2g\\
c)\\
{n_{{H_2}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l
\end{array}\)