Hello,

I have these reactions (they should be correct):

CaCO₃(s) + HNO₃(aq) --> H₂O(l) + Ca²⁺(aq) + CO₂(g) + NO₃^-(aq) Eq. 1

 

Calcium cations will now form a complex with water according the following reaction:

 

Ca²⁺(aq) + 6H₂O(l) --> [Ca(H₂O)₆]²⁺(aq) + 6H₂O(l) Eq. 2

 

Now comes the combustion: This is a project of AAS (atomic absorption spectroscopy), and I want to measure the absorbance for calcium when irradiated with a hollow cathode lamp. I am only interested in calcium so I ignore water, nitrate ions and CO₂(g). As a combustion agent / oxidising agent, C₂H₂(g)/O₂(g) is used:

 

[Ca(H₂O)₆]²⁺(aq) + C₂H₂(g)/O₂(g) --> Eq. 3

 

What is the product of this equation? I am sure solid calcium is formed (thus reduced). But how can it be reduced when the compounds used for the combustion are oxidizing agents? I am considering C₂H₂(g)/O₂(g).

 

 

 

 

Same answer here as on the other forum...

 

Oxygen and acetylene serve to atomize calcium by heat, so you can observe its absorption. At 3000+°C there is no solid calcium, hydroxide, nitrate...

 

I wouldn't call acetylene an oxidiser, nor would I put a specific number of water molecules around Ca²⁺.

If you want calcium then use something that can reduce it like Lithium, Potassium, Barium etc,like they do in silver extraction from cynide complex. Although, side reactions would be there.

The reducing agent is something like hydrogen, or even a free electron, from thermal dissociation of water.

Flame chemistry does things that never happen in test tubes.

I thank you all very much.