Câu hỏi 19

T gồm các amin no, đơn chức, mạch hở nên gọi công thức chung cho các chất trong T là C_nH_2n+3N.

Theo đề bài, có: n_O2 = 0,15 mol; n_CO2 = 0,075 mol.

Cách 1: 

\(\begin{array}{l}{C_n}{H_{2n + 3}}N + (1,5n + 0,75){O_2} \to nC{O_2} + (n + 1,5){H_2}O + 0,5{N_2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,075\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol)\\ \Rightarrow 0,15n = 0,075.(1,5n + 0,75) \Rightarrow n = 1,5 \Rightarrow T:\left\{ \begin{array}{l}C{H_3}N{H_2}\\{C_2}{H_5}N{H_2}\end{array} \right.\end{array}\)

Cách 2: 

\(\begin{array}{l}BTNT\,oxi:\,2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} \Rightarrow {n_{{H_2}O}} = 2.0,15 - 2.0,075 = 0,15\,mol.\\\frac{{{n_H}}}{{{n_C}}} = \frac{{0,15.2}}{{0,075}} = \frac{{2n + 3}}{n} \Rightarrow n = 1,5 \Rightarrow T:\left\{ \begin{array}{l}C{H_3}N{H_2}\\{C_2}{H_5}N{H_2}\end{array} \right.\end{array}\)

Đáp án B