\(n_{Fe}=\frac{2,24}{56}=0,04\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
______0,04------------------------->0,04_____(mol)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,04<---0,04_________________(mol)
=> \(m_{CuO}=0,04.80=3,2\left(g\right)\)