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Câu hỏi: Cho 9,97 gam hỗn hợp X gồm alanin và lysin tác dụng với dung dịch $\mathrm{NaOH}$ dư, thu được dung dịch chứa 11,73 gam muối. Mặc khác, 9,97 gam $X$ trên phản ứng hoàn toàn với dung dịch $\mathrm{HCl}$, thu được dung dịch Y chứa m gam muối. Giá trị của m là
A. 14,715.
B. 12,890.
C. 18,205.
D. 18,255.
A. 14,715.
B. 12,890.
C. 18,205.
D. 18,255.
$\underset{x}{\mathop{\text{N}{{\text{H}}_{2}}\text{CH}\left( \text{C}{{\text{H}}_{3}} \right)\text{COOH}}} +\underset{x}{\mathop{\text{NaOH}}} \to \text{N}{{\text{H}}_{2}}\text{CH}\left( \text{C}{{\text{H}}_{3}} \right)\text{COONa}+{{\text{H}}_{2}}\text{O} $
$\underset{y}{\mathop{\text{N}{{\text{H}}_{2}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{2}} \right)\text{COOH}}} +\underset{y}{\mathop{\text{NaOH}}} \to \text{N}{{\text{H}}_{2}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{2}} \right)\text{COONa}+{{\text{H}}_{2}}\text{O}$
$\underset{x}{\mathop{\text{N}{{\text{H}}_{2}}\text{CH}\left( \text{C}{{\text{H}}_{3}} \right)\text{COOH}}} +\underset{x}{\mathop{\text{HCl}}} \to \text{ClN}{{\text{H}}_{3}}\text{CH}\left( \text{C}{{\text{H}}_{3}} \right)\text{COOH}$
$\underset{y}{\mathop{\text{N}{{\text{H}}_{2}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{2}} \right)\text{COOH}}} +\underset{2y}{\mathop{2\text{HCl}}} \to \text{ClN}{{\text{H}}_{3}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{3}}\text{Cl} \right)\text{COOH}$
Ta có hệ phương trình:
$\left\{\begin{array}{l}89 x+146 y=9,97 \\ 111 x+168 y=11,73\end{array} \rightarrow\left\{\begin{array}{l}x=0,03 \\ y=0,05\end{array}\right.\right.$
${{n}_{\text{HCl}}}={{n}_{\text{alanin }}}+2{{n}_{\text{lysin }}}=0,13~ \text{mol}$
$\text{BTKL}:{{m}_{muối}}={{m}_{\text{aminoaxit }}}+{{m}_{\text{HCl}}}=9,97+0,13.36,5=14,715$ gam
$\underset{y}{\mathop{\text{N}{{\text{H}}_{2}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{2}} \right)\text{COOH}}} +\underset{y}{\mathop{\text{NaOH}}} \to \text{N}{{\text{H}}_{2}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{2}} \right)\text{COONa}+{{\text{H}}_{2}}\text{O}$
$\underset{x}{\mathop{\text{N}{{\text{H}}_{2}}\text{CH}\left( \text{C}{{\text{H}}_{3}} \right)\text{COOH}}} +\underset{x}{\mathop{\text{HCl}}} \to \text{ClN}{{\text{H}}_{3}}\text{CH}\left( \text{C}{{\text{H}}_{3}} \right)\text{COOH}$
$\underset{y}{\mathop{\text{N}{{\text{H}}_{2}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{2}} \right)\text{COOH}}} +\underset{2y}{\mathop{2\text{HCl}}} \to \text{ClN}{{\text{H}}_{3}}{{\left[ \text{C}{{\text{H}}_{2}} \right]}_{4}}\text{CH}\left( \text{N}{{\text{H}}_{3}}\text{Cl} \right)\text{COOH}$
Ta có hệ phương trình:
$\left\{\begin{array}{l}89 x+146 y=9,97 \\ 111 x+168 y=11,73\end{array} \rightarrow\left\{\begin{array}{l}x=0,03 \\ y=0,05\end{array}\right.\right.$
${{n}_{\text{HCl}}}={{n}_{\text{alanin }}}+2{{n}_{\text{lysin }}}=0,13~ \text{mol}$
$\text{BTKL}:{{m}_{muối}}={{m}_{\text{aminoaxit }}}+{{m}_{\text{HCl}}}=9,97+0,13.36,5=14,715$ gam
Đáp án A.