[jonjon1324]

2. If 100. mL of 0.186 M Na2C2O4 is reacted with a excess of Ca(NO3)2 (aq), how much precipitate forms assuming a 76.1% yield?

I have no idea how to do this..Can anyone ...point me in the right direction?

I know what the product is, and the precipitate is Ca2C2O4...But I don't know what to do from there...And I don't know what units I'm supposed to find this in..grams? liters? Is it supposed to tell me or am I just supposed to assume?

[enahs]

Start with a balanced equation.
Then simple Stoichiometry.

http://www.shodor.org/UNChem/basic/stoic/index.html

[jonjon1324]

Start with a balanced equation.
Then simple Stoichiometry.

http://www.shodor.org/UNChem/basic/stoic/index.html

I don't even know how to start...=/
The balanced equation is (NO3)2 (aq) + Na2C2O4 (aq) → CaC2O4 (s) + 2NaNO3 (aq)
I don't know where to go from there..

[Arkcon]

You start with a certain volume of a certain molarity of Ca(NO₃)₂.  That's some mLs of some grams/mL, which is ... grams.  You get a solid, you can measure it in liters if you want to, or molarity because some tiny amount dissolves, but really, the units should be grams of dry powdered precipitate.

[jonjon1324]

You start with a certain volume of a certain molarity of Ca(NO₃)₂.  That's some mLs of some grams/mL, which is ... grams.  You get a solid, you can measure it in liters if you want to, or molarity because some tiny amount dissolves, but really, the units should be grams of dry powdered precipitate.

...Okay...so.....I don't know.
Do I have to find the amount of moles, and then convert that into grams? Because Molarity = mol/mL, and I know what Molarity and mL are, so is finding the amount of moles involved in this?

[enahs]

Yes. Err.

First, write a proper balanced equation.

Now, how many mols of Sodium Oxalate do you have? Using the balanced equation, if you have that many mols, how many mols of product do you get? Now, you are told it is only a 76.1% yield, so what is 71.6% of that? Now, what is the mass of that?

[jonjon1324]

Yes. Err.

First, write a proper balanced equation.

Now, how many mols of Sodium Oxalate do you have? Using the balanced equation, if you have that many mols, how many mols of product do you get? Now, you are told it is only a 76.1% yield, so what is 71.6% of that? Now, what is the mass of that?

Okay so...I got 18.6 moles of Na2C2O4...and then set up an equation
(18.6 mol Na2C2O4)(1 mol CaC2O4/ 1 mol Na2C2O4) (128.097g CaC2O4/ 1 mol CaC2O4) = 2382.6042g Ca2O4...then multipled by .761 and got 1813.161798 g CaC2O4

[enahs]

Not quite. What are the units of molarity?

[jonjon1324]

Not quite. What are the units of molarity?

M = mol / L..oh..so the moles would be .0186...Okay so now I did that instead, and got 2.988028 x .761 = 1.8257 g CaC2O4???

[enahs]

Looks reasonable to me. I did not plug the numbers into a calculator my self, or check if the equation is balanced properly; but correct approach.

[Borek]

so the moles would be .0186

OK

Okay so now I did that instead, and got 2.988028 x .761 = 1.8257 g CaC2O4???

These are some weird numbers :-/ Check your math, as enahs wrote - approach is OK. When you have misiatken mL with L digits were OK, you just have displaced comma, now comma is OK (your result is in the correct range) but digits are wrong.