Cho m gam Mg tác dụng với dung dịch hỗn hợp Cu(NO3)2 và H2SO4 đun nóng

* Hướng dẫn giải

Đáp án A

$\begin{array}{l}+ \left\{\begin{array}{l}{\overline{M}}_Y=16\\ Y chứa NO (hóa nâu)\end{array}\right.\Rightarrow Y có \left\{\begin{array}{l}NO: x\\ H_2: y\end{array}\right.\Rightarrow \left\{\begin{array}{l}x+y=0,04\\ 30x+2y=0,04.16\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,02\\ y=0,02\end{array}\right.\\ + Kim loại gồm Mg dư, Cu, n_{Mg dư}=n_{Cu}=\frac{1,76}{64+24}=0,02 mol.\\ + Sơ đồ phản ứng:\\ \\ \underset{m \left(g\right)}{\underbrace{Mg}}+ \left\{\begin{array}{l}{H}_{2}S{O}_4\\ Cu(N{O}_3{)}_2\end{array}\right\} \underset{ }{\to }\left\{\begin{array}{l}\left\{\begin{array}{l}NO: 0,02\\ H_2: 0,02\end{array}\right\}\\ \left\{\begin{array}{l}M{g}^{2+}\\ S{O_4}^{2-}\\ N{H_4}^{+}\end{array}\right\}\\ \left\{\begin{array}{l}Cu: 0,02\\ Mg dư: 0,02\end{array}\right\}\end{array}\right. \\ \\ + \left\{\begin{array}{l}BTNT Cu: n_{Cu{\left(N{O}_3\right)}_2}=n_{Cu}=0,02\\ BTNT N: n_{N{H_4}^{+}}=2n_{Cu{\left(N{O}_3\right)}_2}-n_{NO}=0,02\\ BTE: 2n_{Mg pư}=3n_{NO}+2n_{H_2}+8n_{N{H_4}^{+}}+2n_{C{u}^{2+}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Mg pư}=0,15\\ m=4,08\end{array}\right.\end{array}$