Đáp án đúng: A
Giải chi tiết:\(\begin{gathered} \left\{ \begin{gathered} X(8C) \hfill \\ Y(9C) \hfill \\ Z(11C) \hfill \\ T:{C_n}{H_{2n}}{O_2} \hfill \\ \end{gathered} \right.\xrightarrow{{quy\,doi}}\underbrace {\left\{ \begin{gathered} C{H_2}:x \hfill \\ CONH:y \hfill \\ {H_2}O:z \hfill \\ {O_2}:T \hfill \\ \end{gathered} \right.}_{249,56(g)}\left| \begin{gathered} P1\xrightarrow{{ + {O_2}}}\left\{ \begin{gathered} C{O_2}:amol \hfill \\ {H_2}O:a - 0,11\,mol \hfill \\ \end{gathered} \right. \hfill \\ P2\xrightarrow{{ + NaOH:(y + t)\,mol}}{H_2}O:z\,mol + \,{C_2}{H_5}OH:t\,mol + G\left\{ \begin{gathered} {C_n}{H_{2n}}{O_2}NNa \hfill \\ {C_m}{H_{2m - 1}}{O_2}Na \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \hfill \\ G\left\{ \begin{gathered} {C_n}{H_{2n}}{O_2}NNa \hfill \\ {C_m}{H_{2m - 1}}{O_2}Na \hfill \\ \end{gathered} \right.\xrightarrow{{ + {O_2}:3,385mol}}\left\{ \begin{gathered} N{a_2}C{O_3}:0,5y + 0,5t \hfill \\ C{O_2}:a + 0,5y - 2,5t \hfill \\ {H_2}O:x + y - 2,5t \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \)
P1: nCO2 –nH2O = 0,11 => 0,5y – z = 0,11 (1)
P2: nNaOH = y + t => nNa2CO3 = 0,5y + 0,5t
BT “C”: x + y -2t = 0,5y + 0,5t + nCO2
=> nCO2 = x + 0,5y – 2,5t
BT “H”: (2x + y + 2z)+(y + t) -2z - 6t = 2nH2O
=> nH2O = x + y – 2,5t
BT “O”: nO(muối) + 3,385.2 = 3(0,5y + 0,5t) + 2(x + 0,5y – 2,5t) + (x + y – 2,5t)
=> 3x + 1,5y – 8t = 6,77 (2)
Phương trình về KL: 14x + 43y + 18z + 32t = 249,56:2 (3)
BTKL: 124,78 + 4(y+t) = 133,18 + 18z + 46t
=> 40y – 18z – 6t = 8,4 (4)
(1) (2) (3) (4) => x = 4,98; y = 0,42; z = 0,1; t = 1,1
\(\begin{gathered} \left\{ \begin{gathered} {n_{peptit}} = 0,1{\text{ }}mol \to 0,8 < n < 1,1 \hfill \\ {n_{C(hh)}} = 4,98 + 0,42 = 5,4 \hfill \\ \end{gathered} \right. \to 4,3 < {n_{C(este)}} < 4,6 \to \frac{{4,3}}{{1,1}} < C(este) < \frac{{4,6}}{{1,1}} \hfill \\ \to Este:{C_4}{H_8}{O_2} \hfill \\ \to \left\{ \begin{gathered} {n_{C(peptit)}} = 5,4 - 4,4 = 1 \hfill \\ {n_{peptit}} = z = 0,1 \hfill \\ {n_{CONH}} = 0,42 \to so\,mat\,xich\,tb = 4,2 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} Z:Gl{y_4}Ala(m) \hfill \\ Y:Gl{y_3}Ala(n) \hfill \\ X:ValAla(p) \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} m + n + p = 0,1 \hfill \\ 11m + 9n + 8p = 1 \hfill \\ 5m + 4n + 2p = 0,42 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} m = 0,06 \hfill \\ n = 0,02 \hfill \\ z = 0,02 \hfill \\ \end{gathered} \right. \hfill \\ \to \% {m_Y} = \frac{{0,02.(75.3 + 89 - 18.3)}}{{124,78}}.100\% = 4,17\% \hfill \\ \end{gathered} \)
Đáp án A
