\(n_{CO_2}=0,75\ mol;\ n_{N_2}=0,125\ mol;\ n_{H_2O}=1,125\ mol;\)
\(BTC:\ n_{C\ (X)}=n_{CO_2}=0,75\ mol;\ \)
\(BTH:\ n_{H\ (X)}=2n_{H_2O}=2,25\ mol;\)
\(BTN:\ n_{N\ (X)}=2n_{N_2}=0,25\ mol;\)
có: \(m_X=m_C+m_H+m_O+m_N\)
\(\Rightarrow m_O=m_X-(m_C+m_H+m_N)=14,75-(12.0,75+2,25+14.0,25)=0\)
\(\Rightarrow\) Trong X không có O
\(n_C:n_H:n_N=0,75:2,25:0,25=3:9:1\)
\(\Rightarrow CTĐGN:\ C_3H_9N\)
Chúc bạn học tốt!