Câu hỏi 24

E có công thức tổng quát là C_nH_2n+3N. 

\(\begin{array}{l}{C_n}{H_{2n + 3}}N + (1,5n + 0,75){O_2} \to nC{O_2} + (n + 1,5){H_2}O + 0,5{N_2}\\\,\,\,\,\,\,0,06\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,255\,\,\,\,\,\,\, \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,03\,(mol)\\ \Rightarrow 0,06.(1,5n + 0,75) = 0,255 \Rightarrow n = \frac{7}{3}\\m = (14.\frac{7}{3} + 3 + 14).0,06 = 2,98\,gam.\\{n_{HN{O_3}(pu)}} = {n_E} = 0,06\,mol \Rightarrow {m_{HN{O_3}(pu)}} = 3,78\,gam\\BTKL:{m_{muoi}} = m + {m_{HN{O_3}(pu)}} = 6,76\,gam\end{array}\)

Đáp án B