RogueRose
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Double displacement of Ca(NO3)2 + NaOH/NaCO3/KOH/NH4OH
I want to do a number of small experiments with double displacement with Ca(NO3)2 and the following:
NaOH, KOH, NH4OH, NaCO3 & (2Na2CO3 ยท 3H2O2 - sodium percarbonate)
I'm planning on trying something like 25g of Ca(NO3)2 per reaction. What I'm not sure of is how to mix the two. Do I dissolve both in as little water
as possible and then combine (basically dissolve to saturation of water with compound) or do I dissolve the CaNO3 in water then add the exchange
compound - or can I mix the two as powders and then add water.
I also remember reading in some double displacements that some things like HCl, H2SO4, NaHC03, H2O2 or NaCl are added for some reason to help with the
reaction. Is anything like this necessary? Does heat help with this or does it need cooling?
CalNit 164g/Mole (anhydrous)
NaCO3 100g/Mole
If I want to use 25g CalNit (25g/164g = 15.24%) then I would use 30.48g of NaCO3 ( 15.24*100 = 15.24 *2 = 30.48g) to get the equivalent for full
conversion - is that call correct stoichiometry?
Ca(NO3)2 + NaCO3
1M CalNit + 2M WashSoda
25g + 30.5g NaCO3
Ca(NO3)2 + NaOH
1M CalNit + 2M NaOH
25g + 12.2g NaOH
Ca(NO3)2 + KOH
1M CalNit + 2M KOH
25g + 17.1g KOH
Ca(NO3)2 + NH4OH
1M + 2M Amm Hydrox
25g + 10.7g NH4OH
Ca(NO3)2 + 2NA2CO3 * 3H2O2
1M CalNit + 1M Sod Percarb
25g + 23.9g 2NA2CO3 * 3H2O2
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Chemist_Cup_Noodles
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Hm, these are some rather mundane double replacements. Have you considered doing some of these same reactions but with cupric nitrate instead? It
should produce many more colors, and more easily observable results.
But one thing I need to say is that your formula for sodium carbonate is wrong. Sodium carbonate is Na2CO3 because the carbonate
anion has a -2 charge, not -1. So the correct molar mass for sodium carbonate is 106 g/mol (23+23+12+16+16+16=106). This would adjust the
stoichiometric amount you'd want to use to 32.31g. Your molar mass for the rest are correct I believe. Although your last reaction is a little odd,
I'm not sure that the products would be very different from the reaction with plain sodium carbonate. I know H2O2 is a very
strong oxidizer, but I just don't see how it would oxidize either product from the reaction. Someone correct me on this if I am wrong. In the reaction
you would actually be using two moles of Ca(NO3)2 to react with one mole of 2 Na2CO3* 3
H2O2. This would mean the stoichiometric mass you should use would be 47.85g.
To answer your question about adding things to the reaction, some people probably add certain acids in order to act as a catalyst in the reaction.
However, because many of these compounds are basic I would not add any acids to them. Actually for your first question, I don't really think it
matters when you add water/ dissolve the compounds too much for the compounds you are using. That may be something else for you to experiment with.
But for this maybe you could prepare the reagents as solutions first, and try them a second time by mixing the solid ingredients and adding water.
I'm not too sure you're going to get a reaction between ammonium hyrdoxide and calcium nitrate though. I know that in order to produce calcium
nitrate, ammonium nitrate and calcium hydroxide are often reacted. I don't know if this reaction is reversible or not.
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Bezaleel
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Quote: Originally posted by Chemist_Cup_Noodles  | | Actually for your first question, I don't really think it matters when you add water/ dissolve the compounds too much for the compounds you are using.
That may be something else for you to experiment with. But for this maybe you could prepare the reagents as solutions first, and try them a second
time by mixing the solid ingredients and adding water. |
I'd use strong solutions, or some of the soluble starting material may not react due to cluttering.
| Quote: Originally posted by Chemist_Cup_Noodles | | I'm not too sure you're going to get a reaction between ammonium hyrdoxide and calcium nitrate though. I know that in order to produce calcium
nitrate, ammonium nitrate and calcium hydroxide are often reacted. I don't know if this reaction is reversible or not. |
The solubility of Ca(OH)2 is highly temperature dependent. If you do it hot, HN3 will evaporate, and you'll be left with a solution of Ca(NO3)2, after
prolonged evaporation of NH3.
In the cold (0 C), Ca(OH)2 has a low solubility and will probably crystallise from solution.
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