What volume of 0.400 M Hg(NO3)2 is needed to precipitate all the I- ions from 250.0 mL of 0.300 M FeI3?
Is the answer 93.8mL??
I don't think so.
Hg^2+ + 2I^- ==> HgI2
mols I^- = M x L x 3 = 0.225
So you need 1/2 that mols Hg^2+.
M = mols/L or L = mols/M
0.1125/0.400 = 0.281L.