1 ml of 8.675x10^-3 M sodium salicylate was dispensed in a 100 ml volumetric flask, It was diluted to 100ml mark with acidified 0.02 M Iron (III) Chloride solution. What is the number of moles of Iron (III) Salicylate complex?

Answer 1

The way I see it is the salicylate is the limiting reagent and you have mols = M x L = 8.675E-3M x 0.001L = ?
The iron complex is
Fe^3+ + sal^- ==> Fe(sal)²⁺.

Answer 2

So does that mean that the salicylate and the iron 3 complex will have an equal amount of mol since they are reacting in a 1:1 mol ratio. If this is true than how will i calculate the concentration of the iron 3 salicylate complex?

Answer 3

plzz reply

Answer 4

Yes, 1:1 means salicylate = Fe-salicylate complex.
(concn) = mols/L soln = mols/0.1 L = ?

Answer 5

so then would the concentration be 8.675x10^-5

Answer 6

That's what I calculated.