[an_ncka430_smnc]

The solubility of FeSO4 at 60 Celsius is of 48 g/ 100 g H2O, and at 10 Celsius is of 22 g/100 g H2O. We have 468,66 saturated solution of FeSO4 at 60 Celsius. Calculate the amount of FeSO4*7H2O which shall damp after cooling the solution at 10 Celsius.

[an_ncka430_smnc]

Please help me, cause I have no idea how it is done 

[an_ncka430_smnc]

This is my attempt...


FeSO4*7H20=A(Fe)+A(S)+4*A(O)+7*(A(H)*2+A(O))=56+32+64+7*18=278 g*mol-1


148 g...48 g FeSO4....100 g water
468,66...a g FeSO4.....b g water

=> a=468,66*48/148=152 g FeSO4
     b=46866/148=316,66

122 g...22 g FeSO4....100 g water
468,66 g....x g FeSO4.....y g water

=> x= 468,66*22/122=84,51 g FeSo4
    y=46866/122=384,14 g water

122 g...22 g FeSO4....100 g water
                  z g FeSO4 . 366,16 g water

=> z = 366,16*22/100=80,55 g FeSO4

84,51-80,55=3,96

I'm confused

[Borek]

The solubility of FeSO4 at 60 Celsius is of 48 g/ 100 g H2O

That means that every 148g of saturated solution contains 48 g of the sulfate. Can you calculate total amount of sulfate in 468,66 g if solution?

[an_ncka430_smnc]

So I should calculate the FeSO4 from the solution?

[an_ncka430_smnc]

278 g solution.....152 g FeSO4...126 g water
468,66 g solution...x.....y

x=468,66*152/278=256,24 g FeSO4
y=468,66*126/278=212,41 g water

[an_ncka430_smnc]

I'm lost 

[Borek]

At 60 deg C:

148 g solution.. 48 g FeSO₄ . 100 g H₂O
468.66 g solution . x g ...y g

At 40 deg C mass of water won't change, but mass of solid dissolved will be smaller.

Edit: typo (1000 instead of 100) corrected, sorry about confusion.

[an_ncka430_smnc]

Why is it 1000 g H2O???

[sjb]

Just a typo, I think, take 100g, instead

[Borek]

Yes, that was a typo, sorry about that.