Lời giải:Đáp án C
\(\begin{array}{l}
{n_{{O_2}}} = 0,48{\rm{ }}mol;{\rm{ }}{n_{NaOH}} = 0,3{\rm{ }}mol.\\
\to {m_{C{O_2}}} + {m_{{H_2}O}} = {m_E} + {m_{{O_2}}} = 17,28 + 0,48.32 = 32,64\\
\Rightarrow 44.{n_{C{O_2}}} + 18.{\rm{ }}{n_{{H_2}O}} = 32,64\left( I \right)
\end{array}\)
Dù là axit hay este khi tác dụng với NaOH ta luôn có sơ đồ:
–COO– + NaOH → –COONa + –OH
\( \Rightarrow {n_{--COO--}} = {n_{NaOH}} = 0,3{\rm{ }}mol \Rightarrow BTNT\left( O \right)\) ta có:
\(2.{n_{C{O_2}}} + {n_{{H_2}O}} = 2.0,3 + 2.0,48 = 1,56\,\,\left( {II} \right)\)
– Từ (I) và (II) \( \Rightarrow {n_{C{O_2}}} = 0,57;{\rm{ }}{n_{{H_2}O}} = 0,42.\)
\(\left\{ \begin{array}{l}
{n_E} = \frac{{{n_{ - COO - }}}}{2} = 0,15\\
{\overline C _E} = \frac{{0,57}}{{0,15}} = 3,8
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
X:C{H_2}{\left( {COOH} \right)_2}\\
Y:{C_2}{H_4}{\left( {COOH} \right)_2}\\
Z:{C_2}{H_6}{\left( {COO} \right)_2}\\
T:{C_3}{H_8}{\left( {COO} \right)_2}
\end{array} \right.\)
\(\left\{ \begin{array}{l}
Z:{C_2}{H_6}{\left( {COO} \right)_2}\\
T:{C_3}{H_8}{\left( {COO} \right)_2}
\end{array} \right. \to \left\{ \begin{array}{l}
3{\rm{ }}ancol{\rm{ }}\\
cu ng{\rm{ }}so{\rm{ }}mol
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
Z:HCOO - C{H_2} - C{H_2} - OOC - H:a\,\,mol\\
T:C{H_3} - OOC - COO - {C_2}{H_5}:a\,\,mol
\end{array} \right.\)
\(\begin{array}{l}
\Rightarrow {m_{3\,\,ancol}} = 62.a + 32.a + 46.a = 4,2 \Rightarrow a = 0,03\\
\left\{ \begin{array}{l}
{n_X} = x\\
{n_Y} = y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + y = {n_E} - {n_Z} - {n_T} = 0,09\\
{n_{C\left( {X,Y} \right)}} = 3x + 4y = {n_{C{O_2}}} - {n_{C\left( {z,T} \right)}} = 0,3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 0,06\\
y = 0,03
\end{array} \right.
\end{array}\)
Chọn C
