shafaifer Posted May 30, 2014 Share Posted May 30, 2014 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). Link to comment Share on other sites More sharing options...
Enthalpy Posted June 9, 2014 Share Posted June 9, 2014 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+. Link to comment Share on other sites More sharing options...
rktpro Posted June 10, 2014 Share Posted June 10, 2014 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. Link to comment Share on other sites More sharing options...
John Cuthber Posted June 10, 2014 Share Posted June 10, 2014 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. Link to comment Share on other sites More sharing options...
shafaifer Posted June 12, 2014 Author Share Posted June 12, 2014 I thank you all very much. Link to comment Share on other sites More sharing options...
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