Question

Answer 1

a) Gọi \(\left\{{}\begin{matrix}n_{Cu}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow64a+27b=1,23\)  (1)

Bảo toàn electron: \(2n_{Cu}+3n_A=n_{NO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(\Rightarrow2a+3b=0,06\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,015\\b=0,01\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,015\cdot64}{1,23}\cdot100\%\approx78,05\%\\\%m_{Al}=21,95\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{CuO}=n_{Cu}=0,015\left(mol\right)\) \(\Rightarrow m_{CuO}=0,015\cdot80=1,2\left(g\right)\)