X, Y (MX < MY) là hai pep...

A

- Hướng dẫn giải

Phương pháp giải:

Quy đổi hỗn hợp ban đầu thành CONH, CH₂, H₂O, C₂H₅OH (với số mol H₂O là số mol peptit)

Giải chi tiết:

Do tạo ancol etylic nên este là Ala-C₂H₅

\(\begin{array}{l}\left\{ \begin{array}{l}X,Y\\Ala - {C_2}{H_5}:0,025\end{array} \right. + NaOH:0,25 \to \left\{ \begin{array}{l}Cn{H_{2n}}{O_2}NNa:0,25\\{C_2}{H_5}OH:0,025\end{array} \right.\\{C_{\overline n }}{H_{2\overline n }}{O_2}NNa + \frac{{3\overline n  - 1,5}}{2}{O_2} \to (\overline n  - 0,5)C{O_2} + \overline n {H_2}O + 0,5{N_2} + 0,5N{a_2}C{O_3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,795\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,125\,mol\\ \to 0,125.\frac{{3\overline n  - 1,5}}{2} = 0,795.\frac{1}{2}\\ \to 2(Gly) < \overline n  = 2,62 < 3(Ala)\\\left\{ \begin{array}{l}Gly - Na:x\\Ala - Na:y\end{array} \right. \to \left\{ \begin{array}{l}x + y = 0,25\\\frac{{2x + 3y}}{{0,25}} = 2,62\end{array} \right. \to \left\{ \begin{array}{l}x = 0,095\\y = 0,155\end{array} \right. \to {m_{muoi}} = 0,095.97 + 0,155.111 = 26,42(g)\end{array}\)

\(\begin{array}{l}Quy\,doi\,E\,thanh\\18,29(g)\left\{ \begin{array}{l}CONH:0,25\\C{H_2}:a\\{H_2}O:b\\{C_2}{H_5}OH:0,025\end{array} \right. + NaOH \to 26,47(g)\left\{ \begin{array}{l}COONa:0,25\\N{H_2}:0,25\\C{H_2}:a\end{array} \right. \to a = 0,405 \to b = 0,04\\{n_{N(peptit)}} = 0,25 - {n_{N({\rm{es}}te)}} = 0,25 - 0,025 = 0,225\,mol\\5(pentapeptit) < \overline N  = \frac{{0,225}}{{0,04}} = 5,625 < 6(h{\rm{ex}}apeptit)\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}X:u\\Y:v\end{array} \right. \to \left\{ \begin{array}{l}u + v = 0,04\\5u + 6v = {n_{N(peptit)}} = 0,225\end{array} \right. \to \left\{ \begin{array}{l}u = 0,015\\v = 0,025\end{array} \right. \to \left\{ \begin{array}{l}Gl{y_n}Al{a_{5 - n}}:0,015\\Gl{y_m}Al{a_{6 - m}}:0,025\end{array} \right.\\ \to {n_{Gly}} = 0,015n + 0,025m = 0,095 \to 3n + 5m = 19 \to n = 3;m = 2\\\left\{ \begin{array}{l}Gl{y_3}Al{a_2}:0,015\\Gl{y_2}Al{a_4}:0,025\end{array} \right. \to \% {m_Y} = \frac{{0,025.416}}{{18,29}}.100\%  = 56,86\% \end{array}\)

Đáp án A