Finding limited reactant (Chem 101)

1. Sep 11, 2010

legendarium

1. The problem statement, all variables and given/known data

Find the limited reactant: Silver Nitrate and Potassium Chromate mix in a double replacement.

Silver Nitrate: 2.15 mL, 0.153 M ; Potassium Chromate: 1.08 mL, 0.197 M.

3. The attempt at a solution

I balanced and have a chemical reaction of:
2AgNO3 + K2Cro4 -> Ag2Cro4 + 2KNO3

I can solve this problem if i was given the grams of each molecule. But im not sure how to start the problem with Molarity and Volume.

I think i would need to do this:

0.153 mol AgNO3 /1L (0.00215L AgN03) (2mol KNO3/2mol AgNo3) (101.1 g KNO3/1mol KNO3) = 0.033gAgNo3

So, how does this look? This is only one of the two equations but i can figure the other one out if if this is right.

Last edited: Sep 11, 2010
2. Sep 12, 2010

Staff: Mentor

Do you know how to read chemical reaction equation? It is given in terms of moles, so while it can be easily converted to masses, there is no need for that.

By definition

$$C = \frac n V$$

Solve for n.

3. Sep 12, 2010

legendarium

Yes i figured that out last night...

So, is this right?

4. Sep 12, 2010

Staff: Mentor

As I told you - don't calculate mass of the reactant, calculate number of moles.

And I have no idea what you did - KNO3 doesn't play any role here and is not involved.

Last edited by a moderator: Aug 13, 2013
5. Sep 12, 2010

legendarium

im assuming my original post is not clear.

i think i did exactly that.

Thank you.

6. Sep 18, 2010

~christina~

What you need to find is the mass of the products formed using the amounts of each reactant present at the begining of the rxn.

The concept is correct but you are mixing up the reactants and the products.

Technically it would be: (using AgNO3 as the reactant to find product)

(0.00215 L AgNO3 * 0.153 mol/L AgNO3)*(1 mol Ag2CrO4 / 2 mol AgNO3)*(331.73g / 1 mol Ag2CrO4) = 0.0546 g Ag2CrO4

You can figure out the grams of Ag2CrO4 produced using the amount of the potassium chromate given. That should tell you which is the limiting reagent.

Hope this helps.

Last edited: Sep 18, 2010
7. Sep 25, 2010

legendarium

It did, thank you.