Câu 1:
\(a,PTHH:Fe+CuSO_4\to FeSO_4+Cu\\ b,\text{Đặt }n_{CuSO_4}=x(mol)\\ \Rightarrow n_{Fe}=n_{Cu}=x(mol)\\ \Rightarrow m_{Fe(tăng)}=m_{Cu(bám)}-m_{Fe(tan)}=64x-56x=1\\ \Rightarrow x=0,125(mol)\\ \Rightarrow m_{Fe}=56.0,125=7(g);m_{Cu}=64.0,125=8(g)\)