Question

取一定量的CuO粉末与0.5L稀H2SO4,充分反应后将一根50g的铁棒插入上述溶液中至铁棒质量不再时

Answer 1

Fe+H2SO4==FeSO4+H2,生成224mL标准状况下的氢气,n(H2)=0.01mol,溶解的n(Fe)=0.01mol,

m(Fe)=0.56g.

CuO+H2SO4=CuSO4+H2O

Fe+CuSO4=FeSO4+Cu △m

56g 64g 8g

m(Cu) (0.24+0.56)g

m(Cu) =6.4g,n(CuO)=n(Cu)=0.1mol