Đáp án:
a) 69,6 g
b) 13,2 g
Giải thích các bước giải:
\(\begin{array}{l}
P1:\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
P2:\\
C{H_3}COOH + KOH \to C{H_3}COOK + {H_2}O\\
nC{H_3}COOH = nKOH = 0,2 \times 1 = 0,2\,mol\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol \Rightarrow nhh = 0,15 \times 2 = 0,3\,mol\\
n{C_2}{H_5}OH = 0,3 - 0,2 = 0,1\,mol\\
a = 0,1 \times 2 \times 46 + 0,2 \times 2 \times 60 = 69,6g\\
b)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
\dfrac{{0,4}}{1} > \dfrac{{0,2}}{1} \Rightarrow\text{CH_3COOH dư} \\
nC{H_3}COO{C_2}{H_5} = n{C_2}{H_5}OH = 0,2\,mol\\
H = 75\% \Rightarrow m = 0,2 \times 75\% \times 88 = 13,2g
\end{array}\)
