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?
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+.
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?
plzz reply
Yes, 1:1 means salicylate = Fe-salicylate complex.
(concn) = mols/L soln = mols/0.1 L = ?
so then would the concentration be 8.675x10^-5
That's what I calculated.
To determine the number of moles of Iron (III) Salicylate complex, we need to calculate the amount of Iron (III) Salicylate complex formed from the reaction between sodium salicylate and Iron (III) Chloride.
First, let's find the number of moles of sodium salicylate in the 1 ml of 8.675x10^-3 M solution:
Number of moles of sodium salicylate = concentration (M) x volume (L)
= 8.675x10^-3 M x 1x10^-3 L
= 8.675x10^-6 moles
When sodium salicylate reacts with Iron (III) Chloride, it forms one mole of Iron (III) Salicylate complex. So the number of moles of Iron (III) Salicylate complex formed will also be 8.675x10^-6 moles.
Hence, the number of moles of Iron (III) Salicylate complex = 8.675x10^-6 moles.