Câu 1:
Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}n_{CuO}=n_{Cu}=b\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}a\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow102\cdot\dfrac{1}{2}a+80b=21,1\) (1)
Ta có: \(n_{O_2}=\dfrac{3,92}{22,4}=0,175\left(mol\right)\)
Bảo toàn electron: \(3a+2b=0,7\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,1\cdot27=2,7\left(g\right)\\m_{Cu}=0,2\cdot64=12,8\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{2,7+12,8}\cdot100\%\approx17,42\%\\\%m_{Cu}=82,58\%\end{matrix}\right.\)