nH2=0,1(mol)
PTHH: Fe+ H2SO4 -> FeSO4+ H2
-> nFe=nH2=0,1(mol) -> mFe=5,6(g)
=>%mFe=(5,6/12).100=46,667%
=>%mCu=53,333%
bạn tham khảo
https://qanda.ai/vi/solutions/H4NPGg8Pjh-Ho%E1%BA%A3-tan-gam-h%E1%BB%93n-h%E1%BB%A3p-Fe-v%C3%A0-trong-dung-d%E1%BB%8Bch-HCI-(du)-%C4%91%C6%B0%E1%BB%A3c-224-l%C3%ADtkhi-H-(%C4%91ktc)-Tinh