Calculate the molar solubility of fe(oh)2 in pure water. (the value of ksp for fe(oh)2 is 4.87×10−17.)

When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2  + 2OH-

when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
 when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17 
[Fe+2]= X
[OH-] = 2X
so by substitution 
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !