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[smanx]

This is going to be kind of long but I'm stuck right now. Not completely lost though because I know the what formulaes to use but I just don't know when to use them, so I need to be prodded to the right direction.

We were given this:

2 KMnO₄ + 5 H₂C₂O₄* 2H₂O + 3 H₂SO₄ --> 3 MnSO₄ + K₂SO₄ + 10 CO₂ + 18 waters

This was a titration lab and on the first run we got:

.22g oxalic acid dihydrate
15 mL  3 M H₂SO₄
9.90 mL KMnO<sub>4/sub]

Calculate:

1) # of moles of oxalic acid dissolved (Note: oxalic acid was added to the 3 M H2SO4 solution)

So far I've just found the gmm of oxalic acid and divided the mass used by that to get the moles. And I got .2200g/124.04956g/mol = .00177348 mol

What I am unsure of here is if the other solution should be included somehow.

2) # of moles KMnO4 consumed(this was the solution added from the buret)

Not entirely sure what to do in this part.

3) molar concentration of the KMnO4 solution

I'm pretty sure I just find the molarity of the solution for this using the concentration formula but I need the first two to get that.

So just wondering if you guys can tell me what steps to take and I can surely slave through the calculations myself.</sub>

[Chemistry Tutor]

ok do you have the molarity of the KMnO4

[smanx]

I'm quite sure it was a 2M solution.

[Chemistry Tutor]

just 2 M, there should be more significant numbers for the concentration

[smanx]

Oh okay I think you're asking me to calculate for step 3. I'm pretty sure I can't accurately do that yet without doing step 2 first, which is what I am stuck on. Am I supposed to figure out the number of moles consumed by determining the excess? Then just use M= n/V

[Chemistry Tutor]

sorry didn't read that far, my bad

ok if you have the moles of oxalic acid then you just use a molar ratio from the balanced chemical equation to find moles of permanganate

.00177348 mol oxalic acid  x 2 mole KMnO4  /  5 mole oxalic acid

[smanx]

Yes, thanks that's the step I forgot. Okay I think I should be fine for now. I'll do the calculations and hopefully not get stuck again... but chances are I will. Will post my calculations later.

[smanx]

Here's what I got for my run 1:

1) oxalic acid:

gmm: 124.04956 g/mol
m: .22000 g

n= m/gmm
  = .00177348 mol

* I'm not sure if the 15mL 3M H₂SO₄ this was dissolved in should be included in the calculations.

2) KMnO₄:

molar ratio:

2 mol KMnO₄ =  x mol of KMnO₄
5 mol oxalic acid           =  .00177348 mol of oxalic acid

x= .000709392 mol

3) molarity of KMnO₄:

given: V= .00990 L
         n= .000709392

M=n/V

=.071655758 mol/L

Would like to know if I did these right before I go into the other runs and the Part B(which I may also need help on).