Lời giải:Chọn đáp án B
Giải thích:
\(\begin{array}{l}
+ \,\,\left\{ \begin{array}{l}
{n_E} = {n_X} + {n_Y} = 0,2\\
{n_{NaOH}} = 2{n_X} + {n_Y} = 0,25
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{n_X} = 0,05\\
{n_Y} = 0,15
\end{array} \right.;\,\,\left\{ \begin{array}{l}
E + NaOH:\,\,{n_{NaOH}} = {n_{{H_2}O}} = 0,25\\
BTKL:\,\,{m_E} = {m_{muoi}} + {m_{khi}} + {m_{{H_2}O}} - {m_{NaOH}} = 21,25
\end{array} \right.\\
+ \,\,\left\{ \begin{array}{l}
{\overline M _{khi}} = 27,5\,\, \Rightarrow \,N{H_3}\\
{n_E} = {n_{khi}} = 5,5:27,5 = 0,2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
X \to 0,05\,\,mol\,\,khi\,N{H_3}\\
Y \to 0,15\,\,mol\,\,khi\,\,C{H_3}N{H_2}
\end{array} \right.\,\,hoac\,\,\left\{ \begin{array}{l}
Y \to 0,15\,\,mol\,\,khi \,\,N{H_3}\\
X \to 0,05\,\,mol\,\,khi \,\,{C_3}{H_9}N
\end{array} \right.\\
+ \,\,\left\{ \begin{array}{l}
{C_n}{H_{2n + 4}}{O_4}{N_2} \to {C_n}{H_{2n}} + {H_4}{O_4}{N_2} \to C{H_2} + {H_4}{O_4}{N_2}\,\,({n_X} = {n_{{H_4}{O_4}{N_2}}})\\
{C_m}{H_{2m + 3}}{O_2}N \to {C_m}{H_{2m}} + {H_3}{O_2}N \to C{H_2} + {H_3}{O_2}N\,\,({n_Y} = {n_{{H_3}{O_2}N}})
\end{array} \right.\\
\Rightarrow E \to \left\{ \begin{array}{l}
{H_4}{O_4}{N_2}:\,\,0,05\,\,mol\\
{H_3}{O_2}N:\,\,0,15\,\,mol\\
C{H_2}:\,\,(21,25 - 0,05.96 - 0,15.49):14 = 0,65\,\,mol
\end{array} \right\}\\
+ \,\,Y:RCOO{H_3}NC{H_3}
\end{array}\)
Gọi số mol CH2 thêm vào X và Y lần lượt là a và b ta có:
\(\begin{array}{l}
0,05.a + 0,15.b = 0,65 \Rightarrow a = 4,\,b = 3\\
\Rightarrow X:\,{C_2}{H_4}{(COON{H_4})_2}\,\\
Y:\,C{H_3}COO{H_3}NC{H_3}\\
\Rightarrow {m_{X\,\,(\min )}} = 7,6\,\,gam\,\, \approx 8\,gam.
\end{array}\)
