FeCO3 is a substance which is only partially soluble in water. The equilibrium reaction in water can be expressed as below.

FeCO3(s) -----------> Fe2+(aq) + CO3(2-)(aq)

For this reaction, the equilibrium constant,Ksp is given by,

Ksp = [Fe2+(aq)][CO3(2-)(aq)]

To solve this problem, you need to find the solubility product or Ksp for FeCO3. Assuming you are refering to a soultion at ambient temperature (25 C), I found the value of the Ksp as,

Ksp = 3.13 x 10^(-11)

Now in a pure FeCO3 solution, according to stoichiometry,

[Fe2+(aq)] = [CO3(2-)(aq)]

Ksp = [Fe2+(aq)][CO3(2-)(aq)]

But, since [Fe2+(aq)] = [CO3(2-)(aq)]

3.13 x 10^(-11) = ([Fe2+(aq)])^2

Therefore,

[Fe2+(aq)] = 5.595 x 10^(-06) mol/L

Therefore number of moles of Fe2+ dissolved in solution is 5.595 x 10^(-06) mol/L. Again according to stoichiometry amount of FeCO3 dissolved is 5.595 x 10^(-06) mol/L.

Therefore molar solubility of FeCO3 is 5.595 x 10^(-06) mol/L.

Consider 15 L of this FeCO3 solution,

Number of FeCO3 mols in 15 L,

        = 5.595 x 10^(-06) mol/L x 15 L

        = 8.3925 x 10^(-5) mol

The molar mass of FeCO3 is,

        = (55.8+12+3x16)

        = 115.8 g/mol

Therefore FeCO3 grams dissolved in 15 L is,
       

        = 116 g/mol x 8.3925 x 10^(-5) mol

        = 9.719 x 10^(-3) g.

Therefore FeCO3 grams dissolved in 15 L is 9.719 mg.