How Many Moles of CO2 Are Produced in an Effervescent Antacid Lab?

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Discussion Overview

The discussion revolves around calculating the number of moles of CO2 produced in a lab experiment involving an effervescent antacid, specifically Alka-Seltzer. Participants explore the chemical reactions between aspirin, citric acid, and sodium bicarbonate in the presence of water, focusing on the stoichiometry and balancing of the reaction equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks help in determining the moles of CO2 produced from the reaction of aspirin, citric acid, and sodium bicarbonate.
  • Another participant emphasizes the importance of knowing the amount of sodium bicarbonate, as it is responsible for producing carbon dioxide.
  • There is a discussion about identifying which reactant (acid or base) is in excess, with suggestions to compare the active residues of the acids and bases.
  • Participants provide calculations for the moles of sodium bicarbonate, citric acid, and aspirin, with one participant expressing uncertainty about the significance of the other products of the reaction.
  • One participant proposes balanced equations for the reactions involving citric acid and sodium bicarbonate, as well as aspirin and sodium bicarbonate, while noting the need for clarity on the stoichiometry.
  • There is a mention of the concept of limiting reagents and the implications of having an excess of one reactant over the other.
  • Another participant points out the need to consider the properties of citric acid and aspirin in relation to their reactions with sodium bicarbonate.
  • Further calculations are presented regarding the volume of CO2 produced, with a correction made regarding the earlier calculation of moles.

Areas of Agreement / Disagreement

Participants express various viewpoints on the reactions and calculations, with no clear consensus on the balanced equations or the significance of the products. Multiple competing views remain regarding the stoichiometry and the interpretation of the results.

Contextual Notes

There are unresolved aspects regarding the balancing of the equations and the identification of limiting reagents. Some calculations are corrected, but the overall understanding of the reactions and their implications remains open to interpretation.

Who May Find This Useful

This discussion may be useful for students and educators involved in chemistry labs, particularly those focusing on stoichiometry, acid-base reactions, and experimental design in a high school setting.

Chemcrazy1
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Effervescent Antacid Lab! Need Help! Fast!

OK! I have a class where I need to design an Effervescent Antacid Lab. (Alka-seltzer). The point is to find how many Moles of CO2 are produced when 325 mg of aspirin (C9H8O4) 1000 mg of citric acid (C6H8O7) and 1916 mg Sodium Bicarbonate (NaHCO3) react with excess H20.

So, To my knowledge, my (unbalanced) equation should look something like this.

C9H8O4 + C6H8O7 + NaHCO3 + H2O --> CO2 +??

For the sake of calculations, I know that Aspirin = 180.187 g/Mol Citric Acid = 192.124 g/mol and Sodium Bicarbonate = 84.01 g/mol

I appreciate any help I could get As Soon As Possible! Thank you! :)
 
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Ok, first we need to know the amount of Sodium Bicarbonate because they produce carbon dioxide somehow.
Second, no matter aspirin or citric acid, it is the carboxylate that serves as acid for the production of carbon dioxide. we need to know which of them (acids or base) is overdosed.
 


1) Are you asking how many moles of the Sodium Bicarbonate there are?
2) how would I find out which one (acid or base) is overdosed?
 


1) yes
2) by counting and comparing acid's and base's active residues.
 


Ok, My (simple) calculations are saying that I have 22.806 mols of NaHCO3 (1916 mg)/(80.01 g/mol).
What exactly does does that second part mean? Sorry if I am a bit unknowing in this sort of thing, I am in high school, but I truly appreciate the help btw!
 
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Oh and I fixed my original post btw.. i forgot to type in the mass of Sodium Bicarbonate I have.. my b.
 


How exactly do I balance this equation?
 


I think my main problem in this lab is finding out how to balence the equation. I know that I have 22.806 mol of NaHCO3 5.204 mol of C6H8O7 and 1.803 mol of C9H8O4. But aside from the CO2 that I am trying to find, I don't know what the other products of the reaction are? are they even important (most likely yes)? If so, Can anyone explain how?
 


1) citric acid + 3 NaHCO3 == citric acid salt + 3 CO2 + 3 H2O
2) aspirin + NaHCO3 == aspirin salt + CO2 + H2O
btw how did you get 20.806 from 1916/84.01? It's 22.807mmol. (And the units)
n(citric acid)=5.205mmol; n(aspirin)=1.804mmol.
Since 3*n(citric acid)+n(aspirin)=17.419mmol<22.807mmol,
base is overdosed.
Then the amount of Carboxyl represents the amount of CO2, which is 17.419mmol, also 0.766g, also 0.778cm3.
 
  • #10


what is the balanced equation? By overdosed do you mean a limiting reagent or something to that nature? that is what my class refers to it as? and The molar mass of NaHCO3 was a typo.. lol
 
  • #11


That's all right!
I think equation 1)+2) will be the balanced equation you referred to.
Alka-seltzer is designed to relieve pain (the function of aspirin, which is also an acid) and neutralize excess stomach acid, you definitely don't want more acid from it, thus it's reasonable to be base overdosed.
The nature of neutralization reaction is its conservation. One base one acid, they will react until one of them is depleted. Thus to the reaction itself acid and base is balanced. Just remember the excess part does not involve in this reaction.
I hope this will be the answer you really needed, and good luck to your presentation!
 
  • #12


There is no one equation, as you have at least two separate reactions.

Check what are properties of citric acid and aspirin - how can they react with with a weak base (sodium carbonate is a weak base, isn't it?).
 
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  • #13


John_5696 said:
Then the amount of Carboxyl represents the amount of CO2, which is 17.419mmol, also 0.766g, also 0.778cm3.

Wow, didn't know carbon dioxide has density higher than water.

That is, assuming you calculated the volume correctly.
 
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  • #14


Thanks Borek that's an obvious mistake!

The right answer should be:

V(CO2)=n(CO2)(mol)*22.4(L/mol)=0.017419*22.4=0.39L=390cm3
 

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