Đáp án đúng: B
Giải chi tiết:X là tetrapeptit, Y là pentapeptit
nN2 = 7,392/22,4 = 0,33 => nN = 0,66 mol
\(\begin{gathered} \underbrace {E\left\{ \begin{gathered} CONH:0,66 \hfill \\ C{H_2}:x \hfill \\ {H_2}O:y \hfill \\ \end{gathered} \right.}_m\xrightarrow{{ + NaOH}}\underbrace {Muoi\left\{ \begin{gathered} {\text{COONa:0,66}} \hfill \\ N{H_2}:0,66 \hfill \\ C{H_2}:x \hfill \\ \end{gathered} \right.}_{m + 23,7}\xrightarrow{{ + O2}}\left\{ \begin{gathered} N{a_2}C{O_3}:0,33 \hfill \\ \xrightarrow{{BTNT:C}}C{O_2}:x + 0,33 \hfill \\ {H_2}O:x + 0,66 \hfill \\ \end{gathered} \right. \hfill \\ {m_{\,binh\,\tan g}} = {m_{C{O_2}}} + {m_{{H_2}O}} \to 44(x + 0,33) + 18(x + 0,66) = 84,06 \to x = 0,93mol \hfill \\ {m_{muoi}} = 0,66.67 + 0,66.16 + 0,93.14 = 67,8g \to m = 67,8 - 23,7 = 44,1g \hfill \\ {m_E} = 44,1 \to 0,66.43 + 14x + 18y = 44,1 \to y = 0,15 \hfill \\ Gly - Na:a \hfill \\ Ala - Na:b \hfill \\ \to \left\{ \begin{gathered} 2a + 3b = {n_C} = 0,66 + 0,93 \hfill \\ 97a + 111b = 67,8 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} a = 0,39 \hfill \\ b = 0,27 \hfill \\ \end{gathered} \right. \hfill \\ G/s:{n_X} = u,{n_Y} = v \hfill \\ \to \left\{ \begin{gathered} u + v = y = 0,15 \hfill \\ 4u + 5v = nN = 0,66 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} u = 0,09 \hfill \\ v = 0,06 \hfill \\ \end{gathered} \right. \hfill \\ X:Gl{y_n}Al{a_{4 - n}}(0,09) \hfill \\ Y:Gl{y_m}Al{a_{5 - m}}(0,06) \hfill \\ \to 0,09n + 0,06m = {n_{Gly}} = 0,39 \to 3n + 2m = 13(n < 4,m < 5) \hfill \\ \to \left\{ \begin{gathered} n = 3 \hfill \\ m = 2 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} X:Gl{y_3}Ala \hfill \\ Y:Gl{y_2}Al{a_3} \hfill \\ \end{gathered} \right. \hfill \\ \% mX = \frac{{0,09.260}}{{44,1}}.100\% = 53,1\% \hfill \\ \end{gathered} \)
Đáp án B
