Cho 14,19 gam hỗn hợp gồm 3 amino axit vào dung dịch chứa 0,05

* Hướng dẫn giải

Đáp án C

$\begin{array}{l}+ Sơ đồ phản ứng:\\ \left\{\begin{array}{l}\underset{14,19 gam}{\underbrace{H_{2}HRCOOH}}\\ \underset{0,05 mol}{\underbrace{{\left(COOH\right)}_2}}\end{array}\right\}\underset{26,19 gam chấtắn Y}{\underbrace{\stackrel{0.3 mol NaOH}{\to }\left\{\begin{array}{l}{H}_{2}HRCOONa\\ (COONa{)}_2\\ ...\end{array}\right\}\stackrel{HCl dư}{\to }}}\underset{muối}{\underbrace{\left\{\begin{array}{l}Cl{H}_{3}HRCOOH\\ \underset{0,3 mol}{\underbrace{NaCl}}\end{array}\right\}}}\\ + \underset{18,69}{\underbrace{m_{chất \tan trong X}}}+\underset{12}{\underbrace{m_{NaOH}}}=\underset{26,19}{\underbrace{m_Y}}+18\underset{?}{\underbrace{n_{HOH}}}\Rightarrow n_{HOH}=0,25\Rightarrow NaOH dư.\\ + \underset{?}{\underbrace{n_{H^{+}/ a\min o axit}}}+\underset{0,1}{\underbrace{n_{H^{+}/{\left(COOH\right)}_2}}}=n_{HOH}=0,25\Rightarrow n_{H^{+}/ a\min o axit}=0,15.\\ \Rightarrow \left\{\begin{array}{l}{n}_{HCl pư viws a\min o axit}=n_{-N{H}_2}=n_{H^{+}/a\min o axit}=0,15\\ m_{muốii}=\underset{m_{NaCl}}{\underbrace{0,3.58,5}}+\underset{m_{a\min o axit}}{\underbrace{14,19}}+\underset{m_{HCl pư với aa}}{\underbrace{0,15.36,5}}=\end{array}\right.\end{array}$