Đáp án:

1.B

2.D

3.C

4.A

5.C

6.C

7.C

Giải thích các bước giải:

1,

\(\begin{array}{l}
k = \dfrac{{3 \times 2 - 6 + 2}}{2} = 1\\
 \to HCOO{C_2}{H_5}\\
 \to C{H_3}COOC{H_3}\\
 \to {C_2}{H_5}COOH
\end{array}\)

2,

\(\begin{array}{l}
C{H_3}COO{C_2}{H_5} + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_3}COOH + {C_2}{H_5}OH\\
X:{C_2}{H_5}OH\\
Y:C{H_3}COOH\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O
\end{array}\)

3, \(C{H_2} = CHCOOH + C{H_3}OH \to C{H_2} = CHCOOC{H_3} + {H_2}O\)

4,

\(\begin{array}{l}
C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH\\
{n_{C{H_3}COO{C_2}{H_5}}} = 0,1mol\\
{n_{NaOH}} = 0,04mol\\
 \to {n_{C{H_3}COO{C_2}{H_5}}} > {n_{NaOH}}
\end{array}\)

Suy ra \(C{H_3}COO{C_2}{H_5}\) dư 

\(\begin{array}{l}
 \to {n_{C{H_3}COONa}} = {n_{NaOH}} = 0,04mol\\
 \to {m_{C{H_3}COONa}} = 3,28g
\end{array}\)

5,

\(\begin{array}{l}
{n_{C{O_2}}} = 0,26mol\\
{n_{{H_2}O}} = 0,26mol\\
 \to {n_C} = {n_{C{O_2}}} = 0,26mol\\
 \to {n_H} = 2{n_{{H_2}O}} = 0,52mol\\
 \to {m_C} + {m_H} = 3,64g\\
 \to {m_O} = 7,8 - 3,64 = 4,16g\\
 \to {n_O} = 0,26mol\\
 \to {n_C}:{n_H}:{n_O} = 0,26:0,52:0,26 = 1:2:1\\
 \to {(C{H_2}O)_n}\\
n = 2 \to {C_2}{H_4}{O_2}
\end{array}\)

6, \({(RCOO)_3}{C_3}{H_5} + 3NaOH \to 3RCOONa + {C_3}{H_5}{(OH)_3}\)

7, 

\(\begin{array}{l}
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = 0,2mol\\
 \to {n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,2mol\\
 \to {m_{C{H_3}COO{C_2}{H_5}}} = 17,6g\\
 \to H = \dfrac{{11}}{{17,6}} \times 100\%  = 62,5\% 
\end{array}\)