解析:
1.C(CuSO4)=n/V=2.0*V/2V=1.0mol/L
C(H2SO4)=n/V=1.0*V/2V=0.5mol/L
2.C(H+)=2C(H2SO4)=1.0mol/L
C(Cu2+)= C(CuSO4)=1.0mol/L
C(SO42-)= n/V= (2.0V+1.0*V)/(V+V)=1.5mol/L
3.因为Fe过量,所以Cu2+与H+全反应完全.
Fe + Cu2+ = Fe2+ + Cu
1 1
.1.0*2V n=2V
Fe + 2H+ = Fe2+ + H2(^)
.2 1
1.0*2V n=V
C(Fe2+)= n/V= (2V+V)/2V=1.5mol/L 答:略