(1)
m₁(CuSO₄)=200×75.4/175.4=85.97g
m₁(H₂O)=200-85.97=114.03g
设0℃时析出胆矾(CuSO₄•5H₂O)中含硫酸铜为x,含水为y
则CuSO₄~ 5H₂O
160 90
x y=90x/160=9x/16
m₂(H₂O)=114.03-50-y=64.03-9x/16
(85.97-x):(64.03-9x/16)=14.3:100
解得x=83.53g,y=46.99g
∴析出胆矾=x+y=130.52g
(2)
设100℃时析出胆矾(CuSO₄•5H₂O)中含硫酸铜为x,含水为y
则y=9x/16
溶液中m(CuSO₄)=86+16-x=102-x
溶液中m(H₂O)=114-y=114-9x/16
∴(102-x):(114-9x/16)=75.4:100
解得x=27.86,y=15.67
析出胆矾=x+y=43.53g