`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`
`0,15` `0,15` `0,15` `(mol)`
`n_[H_2]=[3,7185]/[24,79]=0,15(mol)`
`a)%m_[Fe]=[0,15.56]/18 .100=46,67%`
`=>%m_[Cu]=100-46,67=53,33%`
`b)C_[M_[FeCl_2]]=[0,15]/[0,1]=1,5(M)`
`c)Cu + 2H_2 SO_[4(đ,n)] -> CuSO_4 + SO_2\uparrow + 2H_2 O`
`0,125` `0,25` `0,125` `(mol)`
`n_[Cu]=[18-0,15.56]/64=0,15(mol)`
`n_[H_2 SO_[4(đ,n)]]=[25.98]/[100.98]=0,25(mol)`
Có: `[0,15] > [0,25]/2->Cu` dư
`=>m_[CuSO_4]=0,125.160=20(g)`