Hòa tan hoàn toàn 23,46 gam hỗn hợp X gồm Mg, Cu(OH)2 và Fe(NO3)2 trong dung

Đáp án A

BTKL: $m_{H_{2}O}=8,1\to n_{H_{2}O}=0,45$

$$\begin{gathered} Y\left\{\begin{array}{l}M{g}^{2+}:a\\ C{u}^{2+}:b\\ F{e}^{2+}:c\\ F{e}^{3+}:d\\ N{a}^{+}:0,92\\ N{H}_4^{+}:e\\ S{O}_4^{2-}:0,92\end{array}\right.\underset{}{\overset{Ba{\left(OH\right)}_2}{\to }}\left\{\begin{array}{l}\downarrow \\ N{H}_3\\ H_{2}O\end{array}\right.\left\{\begin{array}{l}2a+2b+2c+3d+e+0,92=0,92.2\left(BTĐT\right)\\ 2a+2b+2c+3d+e=n_{O{H}^{-}\left(Ba\left(O{H}_2\right)\right)}\end{array}\right. \\ \to n_{O{H}^{-}\left(Ba\left(O{H}_2\right)\right)}=0,92 \\ {n}_{S{O}_4^{2-}}=0,92\to n_{Ba{\left(OH\right)}_2}=0,92 \\ {n}_{O{H}^{-}\left(\downarrow \right)}=0,92-e \\ \to m_{\downarrow }=m_{KL}+m_{O{H}^{-}\left(\downarrow \right)}\to m_{KL}=29,23-17.\left(0,92-e\right) \\ {m}_m=m_{KL}+m_{S{O}_4^{2-}}+m_{N{a}^{+}}+m_{N{H}_4^{+}}=123,42 \\ \to 29,23-17\left(0,92-e\right)+0,92.96+0,92.23+18e=123,42\to e=0,01 \\ \left\{\begin{array}{l}{N}_2:x\\ NO:z\\ H_2:y\end{array}\right. \end{gathered}$$

BT H: $n_{Cu{\left(OH\right)}_2}=\frac{0,01.4+2y+0,45.2-0,92}{2}=y+0,01$

$$\begin{gathered} \to \left\{\begin{array}{l}x+y+z=0,1\\ 28x+30z+2y=0,1.5,85.4\\ 2.\left(y+0,01\right)+12x+2y+4z+0,01.10=0,92\left(n_{H^{+}}\right)\end{array}\right.\to \left\{\begin{array}{l}x=0,05\\ y=0,02\\ z=0,03\end{array}\right. \\ \to n_{Cu{\left(OH\right)}_2}=0,03 \end{gathered}$$

BTNT N: $n_{Fe{\left(N{O}_3\right)}_2}=\frac{0,05.2+0,03+0,01}{2}=0,07\to \%m_{Mg}=33,76\%$ .