Đáp án đúng: B
Giải chi tiết:\(\eqalign{ & \left\{ \matrix{ X(8C) \hfill \cr Y(9C) \hfill \cr Z(11C) \hfill \cr T:{C_n}{H_{2n}}{O_2} \hfill \cr} \right.\buildrel {quy\,doi} \over \longrightarrow \underbrace {\left\{ \matrix{ C{H_2}:x \hfill \cr CONH:y \hfill \cr {H_2}O:z \hfill \cr {O_2}:T \hfill \cr} \right.}_{249,56(g)}\left| \matrix{ P1\buildrel { + {O_2}} \over \longrightarrow \left\{ \matrix{ C{O_2}:amol \hfill \cr {H_2}O:a - 0,11\,mol \hfill \cr} \right. \hfill \cr P2\buildrel { + NaOH:(y + t)\,mol} \over \longrightarrow {H_2}O:z\,mol + \,{C_2}{H_5}OH:t\,mol + G\left\{ \matrix{ {C_n}{H_{2n}}{O_2}NNa \hfill \cr {C_m}{H_{2m - 1}}{O_2}Na \hfill \cr} \right. \hfill \cr} \right. \cr & G\left\{ \matrix{ {C_n}{H_{2n}}{O_2}NNa \hfill \cr {C_m}{H_{2m - 1}}{O_2}Na \hfill \cr} \right.\buildrel { + {O_2}:3,385mol} \over \longrightarrow \left\{ \matrix{ N{a_2}C{O_3}:0,5y + 0,5t \hfill \cr C{O_2}:a + 0,5y - 2,5t \hfill \cr {H_2}O:x + y - 2,5t \hfill \cr} \right. \cr} \)
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
\(\eqalign{ & \left\{ \matrix{ {n_{peptit}} = 0,1{\rm{ }}mol \to 0,8 < n < 1,1 \hfill \cr {n_{C(hh)}} = 4,98 + 0,42 = 5,4 \hfill \cr} \right. \to 4,3 < {n_{C(este)}} < 4,6 \to {{4,3} \over {1,1}} < C(este) < {{4,6} \over {1,1}} \cr & \to Este:{C_4}{H_8}{O_2} \cr & \to \left\{ \matrix{ {n_{C(peptit)}} = 5,4 - 4,4 = 1 \hfill \cr {n_{peptit}} = z = 0,1 \hfill \cr {n_{CONH}} = 0,42 \to so\,mat\,xich\,tb = 4,2 \hfill \cr} \right. \to \left\{ \matrix{ Z:Gl{y_4}Ala(m) \hfill \cr Y:Gl{y_3}Ala(n) \hfill \cr X:ValAla(p) \hfill \cr} \right. \to \left\{ \matrix{ m + n + p = 0,1 \hfill \cr 11m + 9n + 8p = 1 \hfill \cr 5m + 4n + 2p = 0,42 \hfill \cr} \right. \to \left\{ \matrix{ m = 0,06 \hfill \cr n = 0,02 \hfill \cr z = 0,02 \hfill \cr} \right. \cr & \to \% {m_Y} = {{0,02.(75.3 + 89 - 18.3)} \over {124,78}}.100\% = 4,17\% \cr} \)
Đáp án B
