Giải bài 7 trang 55 - SGK môn Hóa học lớp 11 nâng cao

Gọi x, y lần lượt là số mol của $NO;\,{{N}_{2}}O$ trong hỗn hợp

 $\begin{aligned} & {{n}_{Al}}=0,5\,mol \\ & \overset{0}{\mathop{Al}}\,-3e\to \overset{+3}{\mathop{Al}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overset{+5}{\mathop{N}}\,+3e\to \overset{+2}{\mathop{N}}\,O \\ & 0,5\,\,\,\,\,1,5\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x\,\,\,\,\,\,\,\,\,\,x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overset{+5}{\mathop{2N}}\,+8e\to {{\overset{+1}{\mathop{N}}\,}_{2}}O \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8y\,\,\,\,\,\,\,\,\,\,\,y \\ & \left\{ \begin{aligned} & {{n}_{e\text{ nhường }\!\!}}=1,5\,mol \\ & {{d}_{hh/{{H}_{2}}}}=19,2 \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & 3x+8y=1,5 \\ & \dfrac{30x+44y}{x+y}=19,2.2 \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & x=0,1 \\ & y=0,15 \\ \end{aligned} \right. \\ & \Rightarrow {{n}_{HN{{O}_{3}}}}={{n}_{e\text{ nhường }}}+{{n}_{NO}}+2{{n}_{{{N}_{2}}O}}=1,9\,mol \\ & \Rightarrow {{C}_{M\,(HN{{O}_{3}})}}\approx 0,86M \\ \end{aligned}$

Ghi nhớ:

Tổng số mol e nhường = tổng số mol e nhận