Cho 0,2 mol hỗn hợp E gồm chất X

Giải thích:

$\left\{\begin{array}{l}{n}_E=n_X+n_Y=\mathrm{0,2}\\ n_{NaOH}=2n_X+n_Y=\mathrm{0,25}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_X=\mathrm{0,05}\\ n_Y=\mathrm{0,15}\end{array}\right.;\left\{\begin{array}{l}E+NaOH:n_{NaOH}=n_{H_{2}O}=\mathrm{0,25}\\ BTKL:m_E=m_{muoái}+m_{khí}+m_{H_{2}O}-m_{NaOH}=\mathrm{21,25}\end{array}\right.$

$\left\{\begin{array}{l}{\overline{M}}_{khí}=\mathrm{27,5}\Rightarrow N{H}_3\\ n_E=n_{khí}=\mathrm{5,5}:\mathrm{27,5}=\mathrm{0,2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}X\stackrel{}{\to }\mathrm{0,05}molkhíN{H}_3\\ Y\stackrel{}{\to }\mathrm{0,15}molkhíC{H}_{3}N{H}_2\end{array}\right.hoac\left\{\begin{array}{l}Y\stackrel{}{\to }\mathrm{0,15}molkhíN{H}_3\\ X\stackrel{}{\to }\mathrm{0,05}molkhíC_3{H}_{9}N\end{array}\right.$

$\left\{\begin{array}{l}{C}_n{H}_{2n+4}{O}_4{N}_2\stackrel{cat}{\to }{C}_n{H}_{2n}+H_4{O}_4{N}_2\stackrel{cat}{\to }C{H}_2+H_4{O}_4{N}_2(n_X=n_{H_4{O}_4{N}_2})\\ C_m{H}_{2m+3}{O}_{2}N\stackrel{cat}{\to }{C}_m{H}_{2m}+H_3{O}_{2}N\stackrel{cat}{\to }C{H}_2+H_3{O}_{2}N(n_Y=n_{H_3{O}_{2}N})\end{array}\right.$

$\Rightarrow E\stackrel{cat}{\to }\left\{\begin{array}{l}{H}_4{O}_4{N}_2:\mathrm{0,05}mol\\ H_3{O}_{2}N:\mathrm{0,15}mol\\ C{H}_2:(\mathrm{21,25}-\mathrm{0,05.96}-\mathrm{0,15.49}):14=\mathrm{0,65}mol\end{array}\right\}$