Question

okay so in lab we made up 2 solutions of FeCl3 and KI. We varied the volumes...I was just wondering how to calculate a new concentration for those solutions. We weighed out .825g of the FeCl3 and 1.35 g of the KI. For trial one we used 20 mL of both. For trial 2 we used 20.0 mL Fe and 10.0 mL KI, skip to trial 4, 15 mL Fe and 10 mL KI. The reaction eq = 2I+2Fe3+-->I2 + 2Fe2+.

I know that M=mol/L but the listed concentrations in the answers were .020 M Fe and .020 M KI (Trial 1), .020 Fe, .010M KI (trial 2), trial 4 i don't remember.

Answer 1

You don't give enough information to calculate it but here is how you do it.
mols FeCl3 = 0.825/162.2 = 0.00509 mols.
How what was the volume you diluted this to? I will call it 250 mL but use what you did in the experiment.
So for Fe in trial 1, you take 20 mL of the 250. You have removed 0.00509 x 20/250 = 4.07E-4 mols. You're placing that with 20 mL of the KI solution which gives 40 mL total. So (Fe) in trial 1 is 4.07E-4 mol/0.040 L = ?
KI is done similarly.

Answer 2

ahh thank you!