Cho 0,2 mol hỗn hợp E gồm chất X (CnH2n+4O4N2) và chất Y (CmH2m+3O2N) đều mạch hở tác dụng vừa đủ với 0,25 mol NaOH

* Hướng dẫn giải

$\begin{array}{l}+\left\{\begin{array}{l}{n}_E=n_X+n_Y=0,2\\ n_{NaOH}=2n_X+n_Y=0,25\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_X=0,05\\ n_Y=0,15\end{array}\right.;\left\{\begin{array}{l}E+NaOH:n_{NaOH}=n_{H_{2}O}=0,25\\ BTKL:m_E=m_{muối}+m_{khí}+m_{H_{2}O}-m_{NaOH}=21,25\end{array}\right.\\ +\left\{\begin{array}{l}{\overline{M}}_{khí}=27,5\Rightarrow N{H}_3\\ n_E=n_{khí}=5,5:27,5=0,2\end{array}\right.\Rightarrow \left\{\begin{array}{l}X\stackrel{}{\to }0,05molkhíN{H}_3\\ Y\stackrel{}{\to }0,15molkhíC{H}_{3}N{H}_2\end{array}\right.hoặc\left\{\begin{array}{l}Y\stackrel{}{\to }0,15molkhíN{H}_3\\ X\stackrel{}{\to }0,05molkhíC_3{H}_{9}N\end{array}\right.\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}{C}_n{H}_{2n+4}{O}_4{N}_2\stackrel{cắt}{\to }{C}_n{H}_{2n}+H_4{O}_4{N}_2\stackrel{cắt}{\to }C{H}_2+H_4{O}_4{N}_2(n_X=n_{H_4{O}_4{N}_2})\\ C_m{H}_{2m+3}{O}_{2}N\stackrel{cắt}{\to }{C}_m{H}_{2m}+H_3{O}_{2}N\stackrel{cắt}{\to }C{H}_2+H_3{O}_{2}N(n_Y=n_{H_3{O}_{2}N})\end{array}\right.\\ \Rightarrow E\stackrel{cắt}{\to }\left\{\begin{array}{l}{H}_4{O}_4{N}_2:0,05mol\\ H_3{O}_{2}N:0,15mol\\ C{H}_2:(21,25-0,05.96-0,15.49):14=0,65mol\end{array}\right\}\\ +\left\{\begin{array}{l}{m}_{X}nhỏnhấtkhiYlàH\mathrm{...}COO{H}_{3}NC{H}_3\\ Xcótốithiểu2Ckhilà{\text{\hspace{0.17em}(COONH}}_3{\text{)}}_2\end{array}\right.\Rightarrow \left\{\begin{array}{l}\Delta C{H}_2=0,65-0,05.2-0,15.2=0,25=0,05.2+0,15\\ XlàC_2{H}_4{\left(COON{H}_4\right)}_{2}vàYlàC{H}_{3}COO{H}_{3}NC{H}_3\end{array}\right.\\ \Rightarrow m_{X\left(\min \right)}=7,6gam\boxed{gầnnhấtvới8gam}\end{array}$

Chọn B