Hello everyone, I would really appreciate your help on figuring out the issue of undesirable CO2 production in this situation. 1. The problem statement, all variables and given/known data In a mixture of CaCO3 + CaHPO4; I am observing undesirable CO2 production under conditions of 20-25°C (room temp) and an undesirable excess of CO2 production under conditions of 50+°C. The time of exposure to these temperatures is a repetitive 6 hour oscillation (to 50°C) and an 18 hour de-oscillation (to 25°C). High moisture content in the environment is a factor, but is outside the immediate physical contact with the material. Without moisture in the environment though, CO2 production still does occur. 1. I am trying to understand why the CO2 production is occuring and the equation responsible. 2. I am trying to understand possible solutions for stopping the CO2 producing reaction. 3. Does the high moisture content move the equation towards accelerating CO2 producing? 4. 2. Relevant equations 1. H3PO4 + CaCO3 → CaHPO4 + CO2 + H20 (not the exact equation, but the most informative) 2. CaCO3 + CaHPO4 + Heat → (CO2 producing equation)? 3. CaCO3 + CaHPO4 + Heat + H20 → (CO2 producing equation)? 4. CaCO3 releases CO2 in interactions with acids - and CaHPO4 is an acid. 5. 2(CaHPO4) + CaCO3 →300-900°C C2P2O7 + CaCO3 + H20 → further → 2(C2P2O7) + CaCO3 →>900°C Ca3(PO4)2 + CO2 (how does this compare to a repetitive exposure to 50°C) 3. The attempt at a solution Well...these are my thoughts 1. CaCO3 (thermal decomposition) → CaO + CO2 CaCO3 degrades towards CO2 production as a byproduct under high heat conditions (around 800°C), though in my problem I am experiencing CO2 production at a much lower temp. (≈50°C) 2. Would simply cooling down the mixture below room temp stop the CO2 producing reaction? Would a desiccant be effective in reducing the CO2 production by reducing the moisture content in an enclosed heated situation? 3. Is there a material that can be included within the solution to prevent this reaction from occurring? Any and all help will be appreciated. Thanks for your time.