Hỗn hợp X gồm glyxin, axit glutamic

Chọn đáp án C.

Khi cho X tác dụng với NaOH thì:$n_{H_{2}O}=n_{NaOH}=n_{COOH}=\mathrm{0,5}mol\stackrel{BTKL}{\to }{m}_X=\mathrm{30,05}\left(g\right)$

Đặt$\left\{\begin{array}{l}C{O}_2:xmol\\ H_{2}O:ymol\\ N_2:zmol\end{array}\right.\to \left\{\begin{array}{l}x+t=\mathrm{0,925}\\ \stackrel{BT:O}{\to }2x+y=\mathrm{2,575}\\ \stackrel{BTKL}{\to }44x+18y+28z=\mathrm{55,25}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,85}\\ y=\mathrm{0,875}\\ z=\mathrm{0,075}\end{array}\right.$

Đặt$\left\{\begin{array}{l}Gly:amol\\ Glu:bmol\\ C_n{H}_{2n}{O}_2:cmol\end{array}\right.\to \left\{\begin{array}{l}a+b=2n_{N_2}=\mathrm{0,15}\\ -\mathrm{0,5}a+\mathrm{0,5}b=n_{C{O}_2}-n_{H_{2}O}=-\mathrm{0,025}\\ a+2b+c=\mathrm{0,5}\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\mathrm{0,1}\\ b=\mathrm{0,05}\\ c=\mathrm{0,3}\end{array}\right.\stackrel{BT:C}{\to }n=\frac{4}{3}$

$\Rightarrow$ Axit cacboxylic nhỏ hơn trong X là HCOOH: 0,2 mol $\Rightarrow$%m = 30,62%.