Amino axit X có công thức H2NCxHy(COOH)2. Cho 0,1 mol X vào 0,2 lít

* Hướng dẫn giải

Chọn đáp án C.

Quy luật phản ứng :

$\begin{array}{l}{H}^{+}+O{H}^{-}\to H_{2}O\\ -C\text{OO}H+O{H}^{-}\to -C{\text{OO}}^{-}+H_{2}O\end{array}$

Theo giả thiết: $\left\{\begin{array}{l}{n}_{(KOH,NaOH)}=n_{O{H}^{-}}=\underset{0,2}{\underbrace{n_{H^{+}}}}+\underset{0,2}{\underbrace{n_{-C\text{OOH}}}}=0,4\\ n_{NaOH}:n_{KOH}=1:3\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NaOH}=0,1\\ n_{KOH}=0,3\end{array}\right.$

Theo bảo toàn khối lượng, ta có :

$\underset{?}{\underbrace{m_{H_{2}NR{\left(C{\text{OO}}^{\text{-}}\right)}_2}}}+\underset{0,3.39}{\underbrace{m_{K^{+}}}}+\underset{0,1.23}{\underbrace{m_{N{a}^{+}}}}+\underset{0,1.96}{\underbrace{m_{S{O_4}^{2-}}}}=36,7$

$$\begin{gathered} \Rightarrow m_{H_{2}NR{\left(C{\text{OO}}^{\text{-}}\right)}_2}=13,1\Rightarrow M_{H_{2}NR{\left(C{\text{OO}}^{\text{-}}\right)}_2}=\frac{13,1}{0,1}=131 \\ \Rightarrow M_{H_{2}NR{\left(C\text{OOH}\right)}_2}=133\Rightarrow \%m_{NtrongX}=\frac{14}{133}.100\%=\boxed{10,526\%} \end{gathered}$$