Đáp án đúng: B
Giải chi tiết:
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 B
