Đáp án:
44 A
45 C
Giải thích các bước giải:
\(\begin{array}{l}
44)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = \dfrac{{12}}{{60}} = 0,2\,mol\\
{n_{{C_2}{H_5}OH}} = \dfrac{{6,9}}{{46}} = 0,15\,mol\\
\dfrac{{0,2}}{1} > \dfrac{{0,15}}{1} \Rightarrow \text{ tính theo $C_2H_5OH$}\\
{n_{C{H_3}COO{C_2}{H_5}}} = \dfrac{{6,6}}{{88}} = 0,075\,mol\\
H = \frac{{0,075}}{{0,15}} \times 100\% = 50\% \\
45)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = \dfrac{6}{{60}} = 0,1\,mol\\
{n_{{C_2}{H_5}OH}} = \dfrac{6}{{46}} = 0,13\,mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,13}}{1} \Rightarrow \text{ tính theo $CH_3COOH$}\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,1\,mol\\
H = 50\% \Rightarrow {m_{C{H_3}COO{C_2}{H_5}}} = 0,1 \times 88 \times 50\% = 4,4g
\end{array}\)
