[Reviction]

Would you guys mine checking my work? Thanks.

Na2CO3(aq) + CaCl2.2H2O(aq) ----> CaCO3(s) + 2NaCl(aq) + 2H2O

CaCl2.2H2O is Calcium Chloride Dihydrate, that is not a typo above just to clarify

Based on the stoichiometry of your reaction preformed during this experiment, how many grams of sodium carbonate were you required to add in order to have completely reacted with 1.00 g of calcium chloride dihydrate?

Answer: 0.721 grams?

If, during your reaction, that you actually produced 0.482 g of the precipitate, what is the percent yield of your reaction?

Answer: 70.9%

On both of these questions above, I am only concerned because the 2H2O (the hydrate) is throwing me off and I have no examples to relate it to to verify I am doing the problem correctly.

Why was it necessary to wait until the filter paper containing the solid product from your reaction to be completely dried before weighing it? (note: this was during after filtering with the precipitate still inside the wet filter paper - why must you wait?)


1. Weighing it wet would cause the theoretical yield of product to be higher than it really is.
2. Weighing it wet would cause the actual yield of products to be higher than it really is.
3. Weighing it wet would cause the actual yield of product to be lower than it really is.
4. Weighing it wet would cause the theoretical yield of product to be lower than it really is.

I said 3, is that correct?

Also, this one I am unsure of I guess because I haven't ran into this problem.

Which of the following substances formed a precipitate when mixed with sodium sulfate?
Answers   :
copper (II) nitrate   
None of the above.   
cobalt (II) nitrate
barium (II) nitrate
no solid formed

Here is my problem with this. Barium Nitrate would make the precipitate but ...Barium II nitrate? Never heard of it and barium only has one ion and that is Barium 2+. Since Barium has no other ions, wouldn't barium (II) nitrate be considered incorrect due to the fact a roman numeral shouldn't be used? Second, just in my option, "None of the above" = "No solid formed," so that makes me think barium is the right answer than. I find this question confusing.

Thanks guys. I'm just preparing for my final and want to verify I am doing this still correctly.

[UG]

Answer: 0.721 grams?

If, during your reaction, that you actually produced 0.482 g of the precipitate, what is the percent yield of your reaction?

Answer: 70.9%

These two look good.

1. Weighing it wet would cause the theoretical yield of product to be higher than it really is.
2. Weighing it wet would cause the actual yield of products to be higher than it really is.
3. Weighing it wet would cause the actual yield of product to be lower than it really is.
4. Weighing it wet would cause the theoretical yield of product to be lower than it really is.

I said 3, is that correct?

No 3 can't be right, say your dry piece of filter paper weighs 1.0 g and your precipitate weighs 0.50 g, then say you had 0.20 g water on the filter paper you would weight the total as 1.70 g giving you an actual yield of 0.70 g which is not true since some of it is water.

Barium (II) nitrate means barium is in the +2 oxidation state, or Ba²⁺, it doesn't mean Ba₂NO₃ if that was what you were getting at.

[Reviction]

So then, Barium II nitrate does = BaNO3 which would cause a precipitate to form? I know none of the other compounds would cause a precipitate.

I just thought you don't use Roman numerously unless there are multiple ions states. Is that incorrect?

Thanks again I appreciate it!

[UG]

So then, Barium II nitrate does = BaNO3 which would cause a precipitate to form?

Well it would be Ba(NO₃)₂ and yes a precipitate would form.
I guess it is just an old habit writing compounds with Roman numerals even when they don't have any other common oxidation states.

[Reviction]

Thanks UG. Appreciate the time you've taken to help me verify it!