Hòa tan hoàn toàn 43,56 gam hỗn hợp X gồm Mg, CuO, Fe3O4 và FeCO3 (trong đó nguyên tố oxi chiếm 22,04% về khối lượng) vào 960 ml dung dịch HNO3 2M,

Ta có: $n_{O\left(X\right)}=\frac{43,56.0,2204}{16}=0,6\left(mol\right)$$Z\left\{\begin{array}{l}{N}_{2}O,C{O}_2:a\\ NO:b\end{array}\right.\to \left\{\begin{array}{l}a+b=0,12\\ 44a+30b=0,\mathrm{12.2.18},5\end{array}\right.\to \left\{\begin{array}{l}a=0,06\\ b=0,06\end{array}\right.$

Quy đổi hỗn hợp X, ta có sơ đồ phản ứng sau:$X\left\{\begin{array}{l}Mg,Fe,Cu\\ O\\ C{O}_2:0,06-x\end{array}\right.\stackrel{HN{O}_3:1,92}{\to }Y\left\{\begin{array}{l}K{L}^{m+}\\ N{H}_4^{+}\\ N{O}_3^{-}:1,77\end{array}\right.+Z\left\{\begin{array}{l}{N}_{2}O:x\\ C{O}_2:0,06-x\\ NO:0,06\end{array}\right.$$\stackrel{BTNT\left(O\right)}{\to }{n}_O=0,6-2.\left(0,06-x\right)=2x+0,48$$\stackrel{BTNT\left(N\right)}{\to }{n}_{N{H}_4^{+}}=1,92-1,77-2x-0,06=0,09-2x$$n_{HN{O}_3}=10.\left(0,09-2x\right)+10x+4.0,06+2.\left(2x+0,48\right)=1,92\to x=0,03$$\to n_{C{O}_2}=0,06-0,03=0,03$$X\underset{43,56gam}{\underbrace{\left\{\begin{array}{l}Mg:y\\ CuO:z\\ F{e}_3{O}_4:t\\ FeC{O}_3:0,03\end{array}\right.}}\stackrel{HN{O}_3}{\to }Y\stackrel{NaOH}{\to }T\stackrel{t°}{\to }\underset{48gam}{\underbrace{\left\{\begin{array}{l}MgO:y\\ CuO:z\\ F{e}_2{O}_3:1,5t+0,015\end{array}\right.}}$$\left\{\begin{array}{l}{m}_x=24y+80z+232t+116.0,03=43,56\\ m_{oxit}=40y+80z+160.\left(1,5t+0,015\right)=48\\ n_{O\left(X\right)}=z+4t+3.0,03=0,6\end{array}\right.\to \left\{\begin{array}{l}y=0,3\\ z=0,15\\ t=0,09\end{array}\right.$$\to \%m_{CuO}=\frac{80.0,15}{43,56}.100\%=27,55\%$

Chọn đáp án D.