Metathesis Reactions / Limiting Reactant

[jaded]

I have a worksheet with several rxns asking us to write the Ionic and Net Ionic equations, along with the theoretical yield and limiting reactant.

For all the worksheets, the volume of the samples are the same (2ml). When calculating the theoretical yield for two substances that have different volumes, does it have to be compensated for?

Ex) When 2.0mL of 1.50M Ammonium Phosphate is mixed with 1.000mL of 0.750 Ferrous Acetate, a precipitate forms.

I have written the balanced chemical eqn and the net ionic eqn. As for the theoretical yield, I am unsure what to do with the differing volumes.

Any help would be appreciated. =)

[jaded]

As usual, I asked a question and shortly after figured out a possible solution...but just in case, could someone reassure me that this is correct? I'll use names instead of formulas due to lack of subscripts and stack the normally horizontal calculations. Dashed lines signify fractions.

Of course, this is only the first equation, but as long as it's correct then I can fill in the rest.



.0020 L Amm. Phosphate
X

1.50M Amm. Phosphate
----------------------------
1L Amm. Phosphate
X

1M Iron (II) Phosphate
----------------------
2M Amm. Phosphate
X

357.48g Iron (II) Phosphate
-------------------------
1M Iron (II) Phospate


= 0.536 Iron (II) Phosphate

[RobJim]

2.0 mL is 0.00020 L, not 0.0020 L.

You correctly identified iron (II) phosphate as the precipitate. However phosphate as a -3 charge, and your equation suggests it has a -1 charge.

You also need to independently determine the moles of iron (II) and phosphate ion and figure out which one will run out first; in other words, which is the limiting reagent. To do this you need to take into account how many Fe2+ ions will react with how many phophate ions as well as the number of moles of each.

Then, once you've determined which one is the limiting reagent, you can figure out how many moles of precipitate there will be, and from there what the mass of the precipitate will be.