\(n_{Fe}=\dfrac{3,92}{56}=0,07\left(mol\right)\\ a,Fe+CuSO_4\rightarrow FeSO_4+Cu\\ b,n_{Cu}=n_{FeSO_4}=n_{Fe}=0,07\left(mol\right)\\ b,m_{FeSO_4}+m_{Cu}=0,07.\left(152+64\right)=0,07.216=15,12\left(g\right)\\ c,C_{MddCuSO_4}=\dfrac{0,07}{0,1}=0,7\left(M\right)\)