Chem 101 help
Suppose 0.691 grams of ammonium acetate is dissolved in 0.250mL of a 0.30M aqueous solution of sodium chromate.
Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the ammonium acetate is dissolved in it.
I've been trying to solve similar problems for almost 2 hours now, but I simply can't get the correct answer (sig figs included). The pset is called "limiting reactants in solution." Thanks for your help!
October 5, 2014
5 Comments • Newest first
Ok... Let's look whats given.
1) Equation
Na2CrO4 + 2C2H3O2NH4 -> 2C2H3O2- + CrO4^2- + 2Na+ + NH4+ (Thx u @dorks)
2) Grams of NH4C2H3O2
0.691g of NH4C2H3O2
3) Volume
0.00025 L
4) mols of Na2CrO4
0.30 M
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First off, let's find the mols of NH4C2H3O2..
mols = mass / molar mass
so..
0.691g of NH4C2H3O2 / 77.0828 g/mol = 0.00896 mol of NH4C2H3O2
Second step is to find molarity of NH4C2H3O2.
We can do this by using the equation c = n / v
0.00896 mols of NH4C2H3O2 / 0.00025 L = 35.84 mols/L
Last step is to find the molarity of C2H3O2.
We can do this by using the molarity of NH4C2H3O2 which we got from the second step.
C2H3O2- mol/L = 35.84 mol/L of NH4C2H3O2 x 1 mol of C2H3O2- / 2 mol of NH4C2H3O2
And the answer is 17.92 mol/L
(I didnt use significant digits at all, but i'm sure you can do that on your own.)
I'm learning this stuff in AP Chem and I think it's rather confusing sometimes. I'm guessing the sodium chromate is just there to throw you off thinking you need to use it but ya don't?
Here's a similar question I did in class, hope it helps. [url=http://imgur.com/14RVKCI]![/url]
also. If you need anymore of these written out examples of stoic including all this jazz I'll post some
easiest way to not get overwhelmed is to write out the equation first so you know the mol ratios and label the info given
Na2CrO4 + 2C2H3O2NH4 -> 2C2H3O2- + CrO4^2- + 2Na+ + NH4+
i haven't done basic chem in a while tho sorry
What grade?
What unit?
Write the chemical equation down for me
0.691g NH4C2H3O2 x (1 mol / Molar mass of sodium acetate) = # number of moles
0.250mL = 2.50 x 10^-4 L
# of moles / (2.50 x 10^-4 L) = molarity
(Molarity is moles per liter)
You probably also need the molar ration like Dorks said