Trung hòa hoàn toàn 14,16 gam một amin X (bậc 1) bằng axit HCl

* Hướng dẫn giải

Đáp án C

$\begin{array}{l}+ \left\{\begin{array}{l}BTKL: \underset{14,16}{\underbrace{m_{a\min }}}+36,5\underset{?}{\underbrace{n_{HCl}}}=\underset{22,92}{\underbrace{m_{muối}}}\\ n_{a\min }=n_{HCl}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{a\min }=n_{HCl}=0,24\\ M_{a\min }=\frac{14,16}{0,24}=59\end{array}\right.\\ \Rightarrow a\min X là C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2\end{array}$