a)PTHH: Fe + 2HCl -> FeCl2 + H2(1)

b)Theo bài ra: nFe=16.8 : 56= 0.3 (mol)

Theo PT(1): nFeCl2=nFe=0.3 (mol)

=>mFe= 0.3 × 127 = 38.1 (g)

c) PTHH: 3Fe + 2O2 -> Fe3O4 (2)

Theo PT(2): nFe3O4=0.5nO2=0.05 (mol)

=> mFe3O4= 0.05 × 232 = 11.6 (g)

a) CTHH: CaO, PTK: 40+16= 56 đvC

b)CTHH: NH3, PTK: 14+1×3=17 đvC

c)CTHH: CuSO4, PTK: 64+32+16×4=160 đvC