Chemical Forums
Specialty Chemistry Forums => Citizen Chemist => Topic started by: Lonn on July 17, 2010, 10:46:16 AM
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Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)
suppose I was mixing this in a lab to get the desired products, with who of Pb(NO3)2(aq) or 2 NaCl(aq) would H2O be with and how many moles of H2O would it be.
I am gonna give an example so you better understand my question because I don't know how to word it exactly
In BaCl2(aq) + Na2SO4(aq) ------------> BaSO4(s) + 2 NaCl(aq)
if I where mixing this, BaCl2 would be [1 mole of BaCl2*2H2O] while Na2SO4 would be [1 mol of Na2SO4] at least stoichiometry wise)
so my question is: are any of the moles of the reactants in Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq) with *H2O like BaCl2 is?
also if you could tell me the why behind this so in the future I can figure it out by myself as I can't find any explanation in my books I would be grateful, but if you can answer only, please go ahead. If you can give me any useful comment at all please do. If you can explain how to know this so I'll do it myself please do as I understand those are the principles of this forum.
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From my understanding it just means the BaCl2 crystals are coordinated with water molecules. (Dry looking powder also can be a hydrated compound.)
My guess is that the answer to your question depends on the source of your reagent.
******Ehhhh old thread is old. sorry *******************
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I understand this to mean that you have aqueous solutions of the various reagents, which are then mixed to give the products, one of which precipitates out of the solution. Therefore the amount of water is unimportant.
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The water of hydration is important if it is part of your reactant, for example lead nitrate. It would have an effect on the quantity in grams that you will need.
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The water of hydration is important if it is part of your reactant, for example lead nitrate. It would have an effect on the quantity in grams that you will need.
Sure it will but with an aqueous solution this will be minimal
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Sure it will but with an aqueous solution this will be minimal
Well it would still affect your yield...
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Sure it will but with an aqueous solution this will be minimal
Well it would still affect your yield...
Yes it will, but minimal, H20 has a MWt of 18, compare that with lead nitrate, the effect will be minimal
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I understand this to mean that you have aqueous solutions of the various reagents, which are then mixed to give the products, one of which precipitates out of the solution. Therefore the amount of water is unimportant.
What if you are asked to prepare 1 mole of solid barium sulfate, starting from barium chloride and sulfuric acid? How much barium chloride would you weight?
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I understand this to mean that you have aqueous solutions of the various reagents, which are then mixed to give the products, one of which precipitates out of the solution. Therefore the amount of water is unimportant.
What if you are asked to prepare 1 mole of solid barium sulfate, starting from barium chloride and sulfuric acid? How much barium chloride would you weight?
Depends if you use the anhydrous form or the di-hydrate, 208.23 g/mol (anhydrous)
244.26 g/mol (dihydrate)