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Topic: BaCl2+Na2SO4---->BaSO4+2NaCl Experiment Help.  (Read 49741 times)

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Offline Boxxxed

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BaCl2+Na2SO4---->BaSO4+2NaCl Experiment Help.
« on: September 24, 2010, 05:10:43 PM »

BaCl2+Na2SO4---->BaSO4+2NaCl


We used a white powder which is a mixture of BaCl2 and NaSO4 and mixed it with water and HCL plus heat.

Mass of white powder is 1.02 grams
Mass of Precipitate (BaSO4) is 0.17 grams

I know that the limiting reagent is BaCl2 but I can't figure it out with calculations.

BaCl2 = 208.2 g/mol
Na2SO4 = 142.04 g/mol
BaSO4 = 233.4 g/mol

Moles of BaSO4 = 0.17/233.4 =  0.00073 moles

By looking at the ratios there are 0.00073 moles of BaCl2 and Na2SO4 as reactants.

Mass BaCl2 = Moles X Molar Mass = 0.00073 x 208.2 = 0.15 grams

Mass Na2SO4 = 0.00073 x 142.04                          = 0.10

The sum of both is 0.25 grams which does not equal the 1.02 gram mixture, it isn't even close. In addition there is more BaCl2 which implies that Na2SO4 is the limiting reagent which it is not. Am I doing something wrong? How do I find the limiting reagent?

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Re: BaCl2+Na2SO4---->BaSO4+2NaCl Experiment Help.
« Reply #1 on: September 24, 2010, 08:23:52 PM »
From the looks of it this was an experiment done in a lab rather than given by a textbook. If it is then the reaction did not have a 100% yield and, therefore, is the reason why you are not able to get 1.02 grams (your calculations looks right).

"In addition there is more BaCl2 which implies that Na2SO4 is the limiting reagent which it is not."

This statement is false, a sample of substance that is heavier than another does not always mean it has more molecules. You have shown that there are equal number of molecules of BaCl2 and Na2SO4, 0.00073 mol. When finding the limiting reagent, it is the number of molecules that matters not the weight of the substance. Looking at the reaction, BaCl2+Na2SO4---->BaSO4+2NaCl. For every mol of BaCl2, one mol of Na2SO4 is used. If you have equal numbers of molecules of BaCl2 and Na2SO4, then there is no limiting reagent.

If you want to find the limiting reagent you need to know how many moles of each reactant there is. Let's have a look at this example:

BaCl2+Na2SO4---->BaSO4+2NaCl

There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?

"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4

"Find the ratio of molecules of the reactants"
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2

"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.







Offline Boxxxed

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Re: BaCl2+Na2SO4---->BaSO4+2NaCl Experiment Help.
« Reply #2 on: September 24, 2010, 09:31:51 PM »

""This statement is false, a sample of substance that is heavier than another does not always mean it has more molecules.""


Ah, I realize my mistake there. The values in this question are indeed derived from a lab. I guess we didn't use enough heat to fuel the reaction.

To clarify on the experiment we filtered the liquid after heating the reaction, separating the precipitate (BaSO4). I know that the limiting reagent is BaCl2. I divided the filtrate into two portions and put a few drops of BaCl2 in one and a few drops of Na2SO4 in the other. The BaCl2 drops reacted and produced a precipitate while the Na2SO4 drops didn't react. Thus the limiting reagent is the BaCl2 because the filtrate was Na2SO4....

I thought there was a way to calculate it as well.

In the case that the reaction did not complete 100% how am I to calculate the mass percent of each molecule in the reaction?
I don't know the ratio of BaCl2 and Na2SO4 in the powder. It asks me to calculate the moles of the limiting reagent (BaCl2) from the mass of my precipitate (BaCl2 , 0.17g).






Offline Boxxxed

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Re: BaCl2+Na2SO4---->BaSO4+2NaCl Experiment Help.
« Reply #3 on: September 24, 2010, 10:03:35 PM »
I also now realize that if the reaction didn't complete to 100% there will be leftover BaCl2 in the filtrate as well as Na2SO4. If this is the case why didn't the Na2SO4 drops react with the BaCl2 when added to the filtrate? Isn't the fact that there was no reaction indicate that there was no BaCl2 left in the filtrate and thus the reaction completed to 100%?

Since the mass of the total reactants is 1.02 grams and only this much reacted....

Mass BaCl2 = Moles X Molar Mass = 0.00073 x 208.2 = 0.15 grams

Mass Na2SO4 = 0.00073 x 142.04                          = 0.10

Mass Total Reactants = 0.25 g

1.02-0.25= 0.77 grams.

Since we know that BaCl2 is the limiting reagent the 0.77 grams is the excess Na2SO4.

Am I right here?


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Re: BaCl2+Na2SO4---->BaSO4+2NaCl Experiment Help.
« Reply #4 on: September 25, 2010, 04:12:16 AM »
Reaction went to completion. If adding BaCl2 produced precipitate, how is it possible that PRESENCE of BaCl2 didn't produce precipitate?

Original sample contained: limiting reagent (you already know it was BaCl2), stoichiometric amount of the other reactant (Na2SO4) and excess of the other reactant (Na2SO4). After reaction you have solid that is a product of reaction, and solution containing just excess Na2SO4.

You have already calculated masses of both stoichiometric amounts of BaCl2 and Na2SO4, time to calculate mass of excess Na2SO4.
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