nFe=m:M=2,8:56=0,05(mol)
nCuO=m:M=8:80=0,1(mol)
nCu=m:M=6,4:64=0,1(mol)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(n_{CuO}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\)
\(n_{Cu}=\dfrac{m}{M}=\dfrac{6,4}{64}=0,1\left(mol\right)\)