Đun nóng 0,1 mol hỗn hợp E chứa hai peptit X, Y bằng dung dịch NaOH (vừa đủ). Cô

$0,1\,mol\,E\,\left\{ \begin{array}{l}
{C_n}{H_{2n - 1}}NO\\
{H_2}O:\,0,1\,mol
\end{array} \right.\, + \,NaOH\, \to \,\left| \begin{array}{l}
m(g)\,muoi\,\underbrace {{C_n}{H_{2n}}N{O_2}Na\,}_{0,4\,mol} + \,{O_2} \to \left\{ \begin{array}{l}
{N_2}\\
C{O_2}(0,4n - 0,2)\,mol\\
N{a_2}C{O_3}:\,0,2mol
\end{array} \right.\\
{H_2}O
\end{array} \right.$
Ta có : ${m_{C{O_2}}}\, + {m_{{H_2}O}}\, = $ 44(0,4n – 0,2) + 18.0,4. N = 65,6 → n = 3
→ ${m_{h.hE(0,1\,mol\,E)}}\, = \,\,$ 0,4.(0,4n+29) + 0,1 . 18 = 30,2 (g)
m= 0,4. (14n+69) = 44,4
→ ${m_{h.h\,E(1,51\,mg\,E)}}\, = \,$ 1,51 . 44,4 = 67,044(g)
→ $\frac{{{m_{h.h\,E(1,51\,mg\,E)}}}}{{{m_{h.h\,E(0,1\,mol\,E)}}}} = \,\frac{{67,044}}{{30,2}} = 2,22(g)$
→ 1,51 m(g) E $\left\{ \begin{array}{l}
{C_n}{H_{2n - 1}}NO:\,0,4.2,22 = \,0,888\,mol\\
{H_2}O:\,0,1.2,22 = 0,222\,mol
\end{array} \right.\, + \,{O_2}\, \to \left\{ \begin{array}{l}
C{O_2}:\,0,88n\,mol\\
{H_2}O:\,0,88\,(n - 0,5)\, + \,0,222\\
{N_2}
\end{array} \right.$
Bảo toàn nguyên tố O → 0,888 + 0,222 + 2. ${n_{{O_2}}}$ = 2. 0,888n + 0,888(n-0,5)+0,222
→ ${n_{{O_2}}} = $ 3,33 mol → gần giá trị này nhất là 3,5