Hỗn hợp X gồm glyxin, analin và axit glutami...

- Hướng dẫn giải

$m_O =0,412m \Rightarrow n_O =\frac{0,412\text{m}}{16} \Rightarrow n_{-COOH} = \frac{1}{2}{n}_O = \frac{0,412\text{m}}{32}$

−COOH    +    NaOH    ⟶    −COONa    +    H₂O

 →  $\frac{0,412\text{m}}{32}\to \frac{0,412\text{m}}{32}$           $\frac{0,412\text{m}}{32}$

$\left\{\begin{array}{c}{\text{H}}_{2}N-{\text{CH}}_2-COOH\\ {\text{H}}_{2}NCH\left({\text{CH}}_3\right)-COOH\\ {\text{H}}_{2}N{\text{C}}_3{\text{H}}_5{\left(\text{COOH}\right)}_2\end{array}\right. + NaOH \left\{\begin{array}{c}{\text{H}}_{2}N-{\text{CH}}_2-COONa\\ {\text{H}}_{2}NCH\left({\text{CH}}_3\right)-COONa\\ {\text{H}}_{2}N{\text{C}}_3{\text{H}}_5{\left(\text{COONa}\right)}_2\end{array}\right. + H_{2}O$

   m(gam)                           $0,412m/32$  (mol)           20,532(gam)              $0,412m/32$(mol)

Bảo toàn khối lượng ta có:

$m+ \frac{0,412\text{m}}{32}.40=20,532+\frac{0,412\text{m}}{32} .18$

⇒ 1,28325m = 20,532

⇒ m = 16

Đáp án cần chọn là: C