BaCl2+Na2SO4---->BaSO4+2NaCl Experiment Help.
BaCl2+Na2SO4---->BaSO4+2Na
We used a white powder which is a mixture of BaCl2 and NaSO4 and mixed it with water and HCL plus heat.
The liquid was then filtered, separating the precipitate.
Mass of white powder is 1.02 grams
Mass of Precipitate (BaSO4) is 0.17 grams
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.......... 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?
We used a white powder which is a mixture of BaCl2 and NaSO4 and mixed it with water and HCL plus heat.
The liquid was then filtered, separating the precipitate.
Mass of white powder is 1.02 grams
Mass of Precipitate (BaSO4) is 0.17 grams
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.......... 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?
ASKER
If there is 0.17 g of BaSO4 and 0.85 g NaCl then
Moles BaSO4 = 0.00073
Moles NaCl = 0.015
0.015/0.00073 = 20.5 moles NaCl per mole of BaSO4.
In the equation there is only 2 mole NaCl per mole BaSO4. I don't think you have to calculate moles for individual elements. All the questions I've done involve calculating moles of the molecules only, which works and I think in this case it's what's supposed to be done.
Moles BaSO4 = 0.00073
Moles NaCl = 0.015
0.015/0.00073 = 20.5 moles NaCl per mole of BaSO4.
In the equation there is only 2 mole NaCl per mole BaSO4. I don't think you have to calculate moles for individual elements. All the questions I've done involve calculating moles of the molecules only, which works and I think in this case it's what's supposed to be done.
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ASKER
Yes, I figured this out last night before bed here.
http://www.chemicalforums.com/index.php?topic=43733.0
Thanks for confirmation.
http://www.chemicalforums.com/index.php?topic=43733.0
Thanks for confirmation.
The way I would go about it is find the grams of SO4, Ba, Cl, and Na separately then add them up accordingly on the left hand side. Which ever has the least is the limiting reagent.
Hope this helps. I'll try to post an answer in a little while.