n(Fe2+)=0.2*0.5=0.1(mol)
因为酸亚铁溶液中铁的含量少于6g,所以沉淀中含有Fe与Cu
设沉淀中n(Cu)=x,n(Fe)=y
64x+56y=5.2
2x+2*3*(6/56-y-x)=0.1*2-------电荷守恒
x=33/700mol
y=0.04mol
m(Cu)=0.04*64=2.56g
n(Fe2+)=0.2*0.5=0.1(mol)
因为酸亚铁溶液中铁的含量少于6g,所以沉淀中含有Fe与Cu
设沉淀中n(Cu)=x,n(Fe)=y
64x+56y=5.2
2x+2*3*(6/56-y-x)=0.1*2-------电荷守恒
x=33/700mol
y=0.04mol
m(Cu)=0.04*64=2.56g