Hỗn hợp E gồm chất X (C3H10N2O4) và chất

* Hướng dẫn giải

Chọn đáp án D

Từ giả thiết suy ra $Xla{H}_{4}NOOC-COO{H}_{3}NC{H}_3.$

$\left\{\begin{array}{l}\underset{\mathrm{0,05}mol}{\underbrace{MuoiX}}+\underset{\mathrm{0,1}mol}{\underbrace{2HCl}}\stackrel{}{\to }\underset{\mathrm{0,05}mol}{\underbrace{HOOC-COOH}}+\underset{muoivoco}{\underbrace{N{H}_{4}Cl}}+\underset{\mathrm{0,05}mol}{\underbrace{C{H}_{3}N{H}_{3}Cl}}\\ \underset{\mathrm{0,1}mol}{\underbrace{TripeptitY}}+\underset{\mathrm{0,2}mol}{\underbrace{2H_{2}O}}+\underset{\mathrm{0,3}mol}{\underbrace{3HCl}}\stackrel{}{\to }muoicloruacuaa\min oaxit\end{array}\right.$

$\Rightarrow m_{chathuuco}=\underset{m_{{\left(COOH\right)}_2}}{\underbrace{\mathrm{0,05.90}}}+\underset{m_{C{H}_{3}N{H}_{3}Cl}}{\underbrace{\mathrm{0,05.67,5}}}+\underset{muoicloruacuaa\min oaxit}{\underbrace{\mathrm{20,3}+\mathrm{0,2.18}+\mathrm{0,3.36,5}}}=\mathrm{42,725}gam.$