Thermal decomposition of calcium carbonate - problem

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Discussion Overview

The discussion revolves around a problem related to the thermal decomposition of calcium carbonate (CaCO3), specifically focusing on the mass loss during the reaction and the resulting molar fractions of the products formed. The scope includes mathematical reasoning and conceptual clarification regarding the decomposition process.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes that a 1/5 decrease in mass of calcium carbonate corresponds to 20 molecules decomposed out of 100.
  • Another participant challenges the assumption that the mass lost directly correlates to the amount of CaCO3 decomposed, suggesting that the reaction products must be considered.
  • A participant provides the balanced reaction equation (CaCO3 → CaO + CO2) and questions the interpretation of mass loss in terms of products formed.
  • There is a suggestion that if 20 g of mass is lost, it should be attributed to the formation of CO2, leading to a different calculation of the moles of CaCO3 decomposed.
  • One participant concludes that 0.45 moles of CaCO3 decomposed, resulting in a molar fraction of CaO in the final sample being 45%.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of mass loss and its implications for the decomposition of CaCO3. There is no clear consensus on the correct approach to the problem, as participants refine their calculations and reasoning throughout the discussion.

Contextual Notes

Participants highlight the importance of considering the products of the reaction and the need for clarity in the stoichiometric relationships involved. There are unresolved aspects regarding the assumptions made about mass loss and the resulting calculations.

Who May Find This Useful

This discussion may be useful for students studying chemical reactions, particularly those interested in stoichiometry and the thermal decomposition of compounds.

wuutoshi
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Hello, here goes the problem:
The mass of calcium carbonate upon thermal decomposition decreased by 1/5. (a) How many molecules of CaCO3 per 100 molecules were decomposed to CaO and CO2. (b) The content of CaO in the final sample express in molar fraction.

My solution:
I set the starting sample to 100g which corresponds to 1 mole of CaCO3. The decrease of mass by 1/5 corresponds to 20g or 0.2 mole. So the solution seems to me to be 20 molecules per 100 molecules.
Ad (b) If 0.2 mole of CaCO3 were decomposed, 0.2 mole of CaO and 0.2 mole CO2 should be formed. Thus the molar fraction of CaO in final samples should be 0.2/(0.2+0.2+0.8) = 1/6.

Yet the provided solutions from two scources differs. Where, if so, did I make mistake?
Thank you for your help.
 
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wuutoshi said:
I set the starting sample to 100g which corresponds to 1 mole of CaCO3.

OK

The decrease of mass by 1/5 corresponds to 20g

OK

or 0.2 mole.

No. Mass decreased by 20 g, but it doesn't mean 20 g decomposed and disappeared.

Write reaction equation - what is lost during decomposition? What is left?

And please, don't ignore the template.
 
Borek said:
No. Mass decreased by 20 g, but it doesn't mean 20 g decomposed and disappeared.
CaCO3 → CaO + CO2

Doesn't 20 g of the decomposed reactant corresponds to decomposition of 0.2 mole of that reactant? I don't assume that mass dissappeares. It appears as 20 g of products (0.4 moles of them). So now we are left with 0.8 moles of CaCO3, isn't that true?

And please, don't ignore the template.

Ok.
 
Last edited:
I already asked you - what is lost, what is left? Add state symbols to all three substances and it should become obvious.
 
Should I assume that 20 g went away as CO2? So from 1 to 1 mole ratio, 0.45 moles of CaCO3 decomposed, which corresponds to 45 molecules per 100 molecules.
We are left with 0.45 moles of CaO and 0.55 moles of CaCO3 which is 45 % of CaO in the final sample.
Ok, the solutions agree.
Thank you, Borek, for making me think and approaching the problem from more real and "chemical point of view".
 

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