Nung m gam hỗn hợp gồm Mg và Cu(NO3)2 trong điều kiện không có không khí

* Hướng dẫn giải

Đáp án B

$\begin{array}{l}+ Trong Z: \left\{\begin{array}{l}{n}_{H_2}+n_{N_2}=0,05\\ 2n_{H_2}+28n_{N_2}=0,05.5,7.4=1,14\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_2}=0,01\\ n_{N_2}=0,04\end{array}\right.\\ + Sơ đồ phản ứng:\\ \underset{m \left(g\right)}{\underbrace{\left\{\begin{array}{l}Mg: x mol\\ Cu(N{O}_3{)}_2: y mol\end{array}\right\}}}\underset{X}{\underbrace{\left\{\begin{array}{l}Mg, MgO\\ CuO\\ Cu(N{O}_3{)}_2\end{array}\right\}}}+\underset{0,45 mol}{\underbrace{\left\{\begin{array}{l}N{O}_2\uparrow \\ O_2\uparrow \end{array}\right\}}}\\ \underset{X}{\underbrace{\left\{\begin{array}{l}Mg, MgO\\ CuO\\ Cu(N{O}_3{)}_2\end{array}\right\}}}\underset{Y, m_{muối}=71,87 \left(g\right)}{\underbrace{\left\{\begin{array}{l}MgC{l}_2: x\\ CuC{l}_2: y\\ N{H}_{4}Cl\end{array}\right\}}}+\underset{Z}{\underbrace{\left\{\begin{array}{l}{N}_2: 0,04\\ H_2: 0,01\end{array}\right\}}}+H_{2}O\\ + \left\{\begin{array}{l}BTNT O: n_{O/X}=n_{H_{2}O}=6n_{Cu{\left(N{O}_3\right)}_2}-2n_{(N{O}_2, O_2)}=6y-0,9\\ BTNT H: n_{N{H}_{4}Cl}=\frac{1,3-0,01.2-2(6y-0,9)}{4}=(0,77-3y)\end{array}\right.\\ + \left\{\begin{array}{l}{m}_{muối}=95x+135y+53,5(0,77-3y)=71,87\\ BTNT Cl: 2x+2y+(0,77-3y)=1,3\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,39\\ y=0,25\end{array}\right.\\ \Rightarrow m=56,36 \end{array}$

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