Hỗn hợp X gồm glyxin, alanin và

Chọn đáp án D

Giải thích:

$\left\{\begin{array}{l}{C}_2{H}_5{O}_{2}N\stackrel{cat}{\to }COO+C{H}_2+N{H}_3\\ C_3{H}_7{O}_{2}N\stackrel{cat}{\to }COO+C{H}_2+N{H}_3\\ C_5{H}_9{O}_{4}N\stackrel{cat}{\to }COO+C{H}_2+N{H}_3\\ Y\stackrel{\mathrm{0,08}mol{H}_2}{\to }Y\text{'}\stackrel{cat}{\to }COO+C{H}_2+H_2\end{array}\right\}\Rightarrow Z\stackrel{cat}{\to }\left\{\begin{array}{l}N{H}_3:xmol\\ H_2:ymol\\ C{H}_2:zmol\\ COO:t\end{array}\right\}\Rightarrow \left\{\begin{array}{l}x+y=\mathrm{0,28}\\ 3x+2y+6z=4.(\mathrm{7,11}+\mathrm{0,04})\\ (\mathrm{1,5}x+y+z)18=\mathrm{88,92}+\mathrm{0,08.18}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,2}\\ y=\mathrm{0,08}\\ z=\mathrm{4,64}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{C_5{H}_9{O}_{2}N}=u\\ \mathrm{0,2}+u+\mathrm{0,08.3}=t\\ \frac{147u}{\mathrm{17.0,2}+\mathrm{4,64.14}+44t}=\mathrm{15,957}\%\end{array}\right.\Rightarrow \left\{\begin{array}{l}t=\mathrm{0,54}\\ u=\mathrm{0,1}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{m}_{Z/\mathrm{0,28}mol}=\mathrm{92,12}gam\\ \Rightarrow m_{Z/\mathrm{0,14}mol}=\mathrm{46,06}gam.\end{array}\right.$