\(n_{NO}=0,28mol\)
\(\left\{{}\begin{matrix}BTe:3n_{Al}+2n_{Cu}=3n_{NO}=0,84\\27n_{Al}+64n_{Cu}=15,84\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,16mol\\n_{Cu}=0,18mol\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,16\cdot27}{15,84}\cdot100=27,27\%\)
\(\Rightarrow\%m_{Cu}=100\%-27,27\%=72,73\%\)