Đáp án đúng: B
Giải chi tiết:TN1:
nO2 = 0,36 mol
nCO2 = 0,32 mol => nC = 0,32 mol
nH2O = 0,16 mol => nH = 0,32 mol
BTNT O: nO = 2nCO2 + nH2O – 2nO2 = 0,32.2 + 0,32 – 0,36.2 = 0,08 mol
=> C:H:O = 0,32:0,32:0,08 = 8:8:2 => C8H8O2
TN2: nNaOH = 0,07 mol
neste = 0,5nO = 0,04 mol
nNaOH/neste = 0,07/0,04 = 1,75 => 1 este của phenol
\(\begin{gathered} E\left\{ \begin{gathered} A(este\,cua\,phenol):x \hfill \\ B:y \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} x + y = {n_E} = 0,04 \hfill \\ 2x + y = {n_{NaOH}} = 0,07 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} x = 0,03 \hfill \\ y = 0,01 \hfill \\ \end{gathered} \right. \hfill \\ \to {n_{{H_2}O}} = x = 0,03\,mol \hfill \\ A(0,03\,mol)\left[ \begin{gathered} HCOO{C_6}{H_4}C{H_3} \hfill \\ C{H_3}COO{C_6}{H_5} \hfill \\ \end{gathered} \right.;\,B(0,01\,mol)\left[ \begin{gathered} HCOOC{H_2}{C_6}{H_5} \hfill \\ {C_6}{H_5}COOC{H_3} \hfill \\ \end{gathered} \right. \hfill \\ TH1:HCOO{C_6}{H_4}C{H_3}(A)\,va\,{C_6}{H_5}COOC{H_3}(B) \to {m_{muoi}} = 7,38gam(loai) \hfill \\ TH2:C{H_3}COO{C_6}{H_5}(A)\,va\,{C_6}{H_5}COOC{H_3}(B) \to {m_{muoi}} = 7,38gam(loai) \hfill \\ TH3:C{H_3}COO{C_6}{H_5}(A)\,va\,HCOOC{H_2}{C_6}{H_5}(B) \to {m_{muoi}} = 6,62gam(thỏa mãn) \hfill \\ \to {m_{muoi\,axit\,cacboxylic}} = {m_{CH3COONa}} + {m_{HCOONa}} = 0,03.82 + 0,01.68 = 3,14\,gam \hfill \\ \end{gathered} \)
Đáp án B
